Quadratic Equations

(i) Given,

$x^2 - 5x + 6 = 0$

Substituting x = 2 in the LHS of the given equation, we get:

$$\begin{aligned}\text{LHS} = 2^2 - 5 \times 2 + 6 \\[0.5em] = 4 - 10 + 6 \\[0.5em] = 10 - 10 \\[0.5em] = 0 \\[0.5em] = \text{RHS}\end{aligned}$$

∴ 2 is a root of the given equation.

Substituting x = -3 in the LHS of the given equation, we get:

$$\begin{aligned}\text{LHS} = (-3)^2 - 5 \times (-3) + 6 \\[0.5em] = 9 + 15 + 6 \\[0.5em] = 30\end{aligned}$$

≠ $\text{RHS}$

∴ 3 is not a root of the given equation.

Hence, 2 is a root but -3 is not a root of the given equation.

(ii) Given,

$3x^2 - 13x - 10 = 0$

Substituting x = 5 in the LHS of the given equation, we get:

$$\begin{aligned}\text{LHS} = 3 \times (5)^2 - 13 \times 5 - 10 \\[0.5em] = 3 \times 25 - 65 -10 \\[0.5em] = 75 - 75 \\[0.5em] = 0 \\[0.5em] = \text{RHS}\end{aligned}$$

∴ 5 is a root of the equation

Substituting x = $-\frac{2}{3}$ in the LHS of the given equation, we get:

$$\begin{aligned}\text{LHS} = 3 \times \Big(-\frac{2}{3}\Big)^2 - 13 \times \Big( -\frac{2}{3} \Big) - 10 \\[1em] = 3 \times \frac{4}{9} + \frac{26}{3} - 10 \\[1em] = \frac{4}{3} + \frac{26}{3} - 10 \\[1em] = \frac{30}{3} - 10 \\[1em] = 10 - 10 \\[1em] = 0 \\[1em] = \text{RHS}\end{aligned}$$

∴ $-\frac{2}{3}$ is a root of the equation

Hence, both 5 and $-\frac{2}{3}$ are roots of the given equation.

(i) Given,

$x^2 - 3\sqrt{3}x + 6 = 0$

Substituting x = $\sqrt{3}$ in the LHS of the given equation, we get:

$$\begin{aligned}\text{LHS} = (\sqrt{3})^2 - 3\sqrt{3} \times \sqrt{3} + 6 \\[0.5em] = 3 - 9 + 6 \\[0.5em] = 9 - 9 \\[0.5em] = 0 \\[0.5em] = \text{RHS}\end{aligned}$$

∴ $\sqrt{3}$ is a solution of the given equation

Substituting x = $-2\sqrt{3}$ in the LHS of the given equation, we get:

$$\begin{aligned}\text{LHS} = (-2\sqrt{3})^2 - 3\sqrt{3} \times -2\sqrt{3} + 6 \\[0.5em] = 12 + 18 + 6 \\[0.5em] = 36\end{aligned}$$

≠ $\text{RHS}$

∴ $-2\sqrt{3}$ is not a solution of the given equation

Hence, $\sqrt{3}$ is a solution of the given equation but $-2\sqrt{3}$ is not.

(ii) Given,

$x^2 - \sqrt{2}x - 4 = 0$

Substituting x = $-\sqrt{2}$ in the LHS of the given equation, we get:

$$\begin{aligned}\text{LHS} = (-\sqrt{2})^2 - \sqrt{2} \times (-\sqrt{2}) - 4 \\[0.5em] = 2 + 2 - 4 \\[0.5em] = 0 \\[0.5em] = \text{RHS}\end{aligned}$$

∴ $-\sqrt{2}$ is a solution of the given equation

Substituting x = $2\sqrt{2}$ in the LHS of the given equation, we get:

$$\begin{aligned}\text{LHS} = (2\sqrt{2})^2 - \sqrt{2} \times 2\sqrt{2} - 4 \\[0.5em] = 8 - 4 - 4 \\[0.5em] = 8 - 8 \\[0.5em] = 0 \\[0.5em] = \text{RHS}\end{aligned}$$

∴ $2\sqrt{2}$ is a solution of the given equation

Hence, $-\sqrt{2}$ and $2\sqrt{2}$ both are solutions of the given equation.

Since, $-\frac{1}{2}$ is a solution of the equation 3x² + 2kx - 3 = 0, x = $-\frac{1}{2}$ satisfies the given equation.

Substituting x = $-\frac{1}{2}$ and solving for k we get:

$$\begin{aligned}3\Big(-\frac{1}{2}\Big)^2 + 2k \Big(-\frac{1}{2}\Big) -3 = 0 \\[1em] \Rightarrow 3 \times \frac{1}{4} - k - 3 = 0 \\[1em] \Rightarrow -k = 3 - \frac{3}{4} \\[1em] \Rightarrow k = \frac{3}{4} - 3 \\[1em] \Rightarrow k = \frac{3 - 12}{4} \\[1em] \Rightarrow k = -\frac{9}{4}\end{aligned}$$

Hence, the value of k is $-\frac{9}{4}$

Since, $\frac{2}{3}$ is the solution of the equation 7x² + kx - 3 = 0, x = $\frac{2}{3}$ satisfies the given equation.

Substituting x = $-\frac{2}{3}$ and solving for k we get:

$$\begin{aligned}\Rightarrow 7\Big(\frac{2}{3}\Big)^2 + k \times \frac{2}{3} -3 = 0 \\[1em] \Rightarrow 7 \times \frac{4}{9} + \frac{2k}{3} -3 = 0 \\[1em] \Rightarrow \frac{28}{9} - 3 + \frac{2k}{3} = 0 \\[1em] \Rightarrow \frac{28}{9} - \frac{27}{9} + \frac{6k}{9} = 0 \text{ (Taking L.C.M.) } \\[1em] \Rightarrow \frac{28 - 27 + 6k}{9} = 0 \\[1em] \Rightarrow 1 + 6k = 0 \\[1em] \Rightarrow 6k = -1 \\[1em] \Rightarrow k = -\frac{1}{6}\end{aligned}$$

Hence, the value of k is $-\frac{1}{6}.$

Since $\sqrt{2}$ is a root of the equation $kx^2 + \sqrt{2}x - 4 = 0$, $x = \sqrt{2}$ satisfies the given equation.

Substituting $x = \sqrt{2}$ in the given equation:

$$\begin{aligned}k ( \sqrt{2} )^2 + \sqrt{2} \times \sqrt{2} - 4 = 0 \\[0.5em] \Rightarrow 2k + 2 - 4 = 0 \\[0.5em] \Rightarrow 2k - 2 = 0 \\[0.5em] \Rightarrow 2k = 2 \\[0.5em] \Rightarrow k = 1\end{aligned}$$

Hence, the value of k is 1

Since, a is the root of the equation x² - (a + b)x + k = 0 , x = a satisfies the given equation.

Substituting x = a in the given equation:

$$\begin{aligned}\Rightarrow a^2 - (a+b)a + k = 0 \\[0.5em] \Rightarrow a^2 - (a^2 + ab) + k = 0 \\[0.5em] \Rightarrow a^2 - a^2 - ab + k = 0 \\[0.5em] \Rightarrow k - ab = 0 \\[0.5em] \Rightarrow k = ab\end{aligned}$$

Hence, the value of k is ab.

The given equation is px² + 7x + q = 0.

As $\frac{2}{3}$ is a root of the equation, so x = $\frac{2}{3}$ satisfies the given equation.

Substituting x = $\frac{2}{3}$ in the given equation:

$$\begin{aligned}p\big(\frac{2}{3} \big)^2 + 7 \times \frac{2}{3} + q = 0 \\[1em] \Rightarrow \frac{4p}{9} + \frac{14}{3} + q = 0 \\[1em] \Rightarrow q = -\frac{4p}{9} - \frac{14}{3} \\[1em] \Rightarrow q = -\big(\frac{4p + 42}{9}\big) \space \text{...(i)} \\[1em]\end{aligned}$$

Also -3 is a root of the given equation, so x = -3 satisfies the given equation.

Substituting x = -3 in the given equation:

$$\begin{aligned}p(-3)^2 + 7 \times (-3) + q = 0 \\[0.5em] \Rightarrow 9p - 21 + q = 0 \space \text{...(ii)}\end{aligned}$$

Putting value of q from eqn (i) into eqn (ii)

$9p - 21 - \Big(\frac{4p + 42}{9}\Big) = 0$

Taking 9 as the LCM

$$\begin{aligned}\Rightarrow \frac{81p - 189 - 4p - 42}{9} = 0 \\[1em] \Rightarrow 77p - 231 = 0 \\[1em] \Rightarrow 77p = 231 \\[1em] \Rightarrow p = \frac{231}{77} \\[1em] \Rightarrow p = 3\end{aligned}$$

Substituting p = 3 in (i), we get

$$\begin{aligned}q = -\Big( \frac{4\times 3 + 42}{9} \Big) \\[1em] \Rightarrow q= -\Big(\frac{12 + 42}{9}\Big) \\[1em] \Rightarrow q= -\Big(\frac{54}{9}\Big) \\[1em] \Rightarrow q= -6\end{aligned}$$

Hence, p = 3 and q = -6

Given,

$$\begin{aligned}x^2 - 3x - 10 = 0 \\[0.5em] \Rightarrow x^2 - 5x + 2x - 10 = 0 \\[0.5em] \Rightarrow x(x - 5) + 2(x - 5) = 0 \\[0.5em] \Rightarrow (x + 2)(x - 5) = 0 \text{ (Factorising left side) } \\[0.5em] \Rightarrow x + 2 = 0 \text{ or } x - 5 = 0 \text{ ( Zero - product rule) }\\[0.5em] \Rightarrow x = -2 \text{ or } x = 5\end{aligned}$$

Hence, the roots of given equation are -2, 5.

Given,

$$\begin{aligned}x(2x + 5) = 3 \\[0.5em] \Rightarrow 2x^2 + 5x = 3 \\[0.5em] \Rightarrow 2x^2 + 5x - 3 = 0 \text{ (Writing as } ax^2 + bx + c = 0) \\[0.5em] \Rightarrow 2x^2 + 6x - x - 3 = 0 \\[0.5em] \Rightarrow 2x(x + 3) - 1(x + 3) = 0 \\[0.5em] \Rightarrow (2x - 1)(x + 3) = 0 \text{ (Factorising left side) } \\[0.5em] \Rightarrow 2x - 1 = 0 \text{ or } x + 3 = 0 \text{ (Zero product rule) } \\[0.5em] \Rightarrow 2x = 1 \text{ or } x = -3 \\[0.5em] \Rightarrow x = \frac{1}{2} \text{ or } x = -3 \\[0.5em]\end{aligned}$$

Hence, the roots of given equation are $\frac{1}{2}$, -3.

Given,

$$\begin{aligned}3x^2 - 5x - 12 = 0 \\[0.5em] \Rightarrow 3x^2 - 9x + 4x - 12 = 0 \\[0.5em] \Rightarrow 3x(x - 3) + 4(x - 3) = 0 \\[0.5em] \Rightarrow (x - 3)(3x + 4) = 0 \text{ (Factorising left side) } \\[0.5em] \Rightarrow x - 3 = 0 \text{ or } 3x + 4 = 0 \text{ (Zero-product rule) }\\[0.5em] \Rightarrow x = 3 \text{ or } x = -\frac{4}{3}\end{aligned}$$

Hence, the roots of given equation are 3, $-\frac{4}{3}$.

Given,

$$\begin{aligned}21x^2 - 8x - 4 = 0 \\[0.5em] \Rightarrow 21x^2 - 14x + 6x - 4 = 0 \\[0.5em] \Rightarrow 7x(3x - 2) + 2(3x - 2) = 0 \\[0.5em] \Rightarrow (7x + 2)(3x - 2) = 0 \text{ (Factorising left side) } \\[0.5em] \Rightarrow 7x + 2 = 0 \text{ or } 3x - 2 = 0 \text{ (Zero-product rule) }\\[0.5em] \Rightarrow x = -\frac{2}{7} \text{ or } x = \frac{2}{3}\end{aligned}$$

Hence, the roots of given equation are $-\frac{2}{7}$, $\frac{2}{3}$.

Given,

$$\begin{aligned}3x^2 = x + 4 \\[0.5em] \Rightarrow 3x^2 - x - 4 = 0 \text{ (Writing as } ax^2 + bx + c = 0)\\[0.5em] \Rightarrow 3x^2 - 4x + 3x - 4 = 0 \\[0.5em] \Rightarrow x(3x - 4) + 1(3x - 4) = 0 \\[0.5em] \Rightarrow (x + 1)(3x - 4) = 0 \text{ (Factorising left side) } \\[0.5em] \Rightarrow x + 1 = 0 \text{ or } 3x - 4 = 0 \text{ (Zero-product rule) }\\[0.5em] \Rightarrow x = -1 \text{ or } x = \frac{4}{3}\end{aligned}$$

Hence, the roots of given equation are -1, $\frac{4}{3}$.

Given,

$$\begin{aligned}x(6x - 1) = 35 \\[0.5em] \Rightarrow 6x^2 - x - 35 = 0 \text{ (Writing as } ax^2 + bx + c = 0) \\[0.5em] \Rightarrow 6x^2 - 15x + 14x - 35 = 0 \\[0.5em] \Rightarrow 3x(2x - 5) + 7(2x - 5) = 0 \\[0.5em] \Rightarrow (3x + 7)(2x - 5) = 0 \text{ (Factorising left side) } \\[0.5em] \Rightarrow 3x + 7 = 0 \text{ or } 2x - 5 = 0 \text{ (Zero-product rule) }\\[0.5em] \Rightarrow x = -\frac{7}{3} \text{ or } x = \frac{5}{2}\end{aligned}$$

Hence, the roots of given equation are $-\frac{7}{3}$, $\frac{5}{2}$.

Given,

$$\begin{aligned}6p^2 + 11p - 10 = 0 \\[0.5em] \Rightarrow 6p^2 + 15p - 4p - 10 = 0 \\[0.5em] \Rightarrow 3p(2p + 5) - 2(2p + 5) = 0 \\[0.5em] \Rightarrow (2p + 5)(3p - 2) = 0 \text{ (Factorising left side) } \\[0.5em] \Rightarrow 2p + 5 = 0 \text{ or } 3p - 2 = 0 \text{ (Zero-product rule) }\\[0.5em] \Rightarrow 2p = -5 \text{ or } 3p = 2 \\[0.5em] \Rightarrow p = -\frac{5}{2} \text{ or } p = \frac{2}{3}\end{aligned}$$

Hence, the roots of given equation are $-\frac{5}{2}$, $\frac{2}{3}$.

Given,

$$\begin{aligned}\frac{2}{3}x^2 - \frac{1}{3}x = 1 \\[1em] \Rightarrow \frac{2}{3}x^2 - \frac{1}{3}x - 1 = 0 \text{ (Writing as } ax^2 + bx + c = 0) \\[1em] \Rightarrow \frac{2}{3}x^2 \times 3 - \frac{1}{3}x \times 3 - 1 \times 3 = 0 \times 3 \text{ (Multiplying the equation by 3) } \\[1em] \Rightarrow 2x^2 - x - 3 = 0 \\[1em] \Rightarrow 2x^2 - 3x + 2x - 3 = 0 \\[1em] \Rightarrow x(2x - 3) + 1(2x - 3) = 0 \\[1em] \Rightarrow (x + 1)(2x - 3) = 0 \text{ (Factorising left side) } \\[1em] \Rightarrow x + 1 = 0 \text{ or } 2x - 3 = 0 \text{ (Zero-product rule) } \\[1em] \Rightarrow x = -1 \text{ or } x = \frac{3}{2}\end{aligned}$$

Hence, the roots of given equation are -1, $\frac{3}{2}$.

Given,

$$\begin{aligned}3(x-2)^2 = 147 \\[1em] \Rightarrow 3( x^2 - 4x + 4) = 147 \\[1em] \Rightarrow 3x^2 - 12x + 12 = 147 \\[1em] \Rightarrow 3x^2 - 12x + 12 - 147 = 0 \text{ (Writing as } ax^2 + bx + c = 0) \\[1em] \Rightarrow 3x^2 - 12x - 135 = 0 \\[1em] \Rightarrow \frac{3x^2 - 12x - 135 }{3} = \frac{0}{3} \\[1em] \text{ (Dividing the complete equation by 3) } \\[1em] \Rightarrow x^2 - 4x - 45 = 0 \\[1em] \Rightarrow x^2 - 9x + 5x - 45 = 0 \\[1em] \Rightarrow x(x - 9) + 5(x - 9) = 0 \\[1em] \Rightarrow (x - 9)(x + 5) = 0 \text{ (Factorising left side) }\\[1em] \Rightarrow x - 9 = 0 \text{ or } x + 5 = 0 \text{ (Zero-product rule) } \\[1em] \Rightarrow x = 9 \text { or } x = -5.\end{aligned}$$

Hence, the roots of given equation are 9, -5.

Given,

$$\begin{aligned}\frac{1}{7}(3x-5)^2 = 28 \\[1em] \Rightarrow \frac{1}{7}( 9x^2 - 30x + 25) = 28 \\[1em] \Rightarrow \frac{9}{7}x^2 - \frac{30}{7}x + \frac{25}{7} = 28 \\[1em] \Rightarrow \frac{9}{7}x^2 \times 7 - \frac{30}{7}x \times 7 + \frac{25}{7} \times 7 = 28 \times 7 \text{ (Multiplying the complete equation by 7) } \\[1em] \Rightarrow 9x^2 - 30x + 25 = 196 \\[1em] \Rightarrow 9x^2 - 30x + 25 - 196 = 0 \text{ (Writing as } ax^2 + bx + c = 0) \\[1em] \Rightarrow 9x^2 - 30x - 171 = 0 \\[1em] \Rightarrow 9x^2 - 57x + 27x - 171 = 0 \\[1em] \Rightarrow 3x(3x - 19) + 9(3x - 19) = 0 \\[1em] \Rightarrow (3x + 9)(3x - 19) = 0 \text{ (Factorising left side) } \\[1em] \Rightarrow 3x + 9 = 0 \text{ or } 3x - 19 = 0 \text{ (Zero - product rule) }\\[1em] \Rightarrow 3x = -9 \text { or } 3x = 19. \\[1em] \Rightarrow x = -\frac{9}{3} \text{ or } x = \frac{19}{3} \\[1em] \Rightarrow x = -3 \text{ or } x = \frac{19}{3} \\[1em]\end{aligned}$$

Hence, the roots of given equation are -3, $\frac{19}{3}$.

Given,

$$\begin{aligned}x^2 - 4x - 12 = 0 \\[0.5em] \Rightarrow x^2 - 6x + 2x - 12 = 0 \\[0.5em] \Rightarrow x(x - 6) + 2(x - 6) = 0 \\[0.5em] \Rightarrow (x + 2)(x - 6) = 0 \text{ (Factorising left side) }\\[0.5em] \Rightarrow x + 2 = 0 \text{ or } x - 6 = 0 \text{ (Zero-product rule) } \\[0.5em] \Rightarrow x = -2 \text{ or } x = 6\end{aligned}$$

Since x ∈ N hence x = -2 is not the root.
Hence, the root of given equation is 6.

Given,

$$\begin{aligned}2x^2 - 9x + 10 = 0 \\[0.5em] \Rightarrow 2x^2 - 5x - 4x + 10 = 0 \\[0.5em] \Rightarrow x(2x - 5) - 2(2x - 5) = 0 \\[0.5em] \Rightarrow (x - 2)(2x - 5) = 0 \text{ (Factorising left side) } \\[0.5em] \Rightarrow x - 2 = 0 \text{ or } 2x - 5 = 0 \text{ (Zero-product rule) } \\[0.5em] \Rightarrow x = 2 \text{ or } x =\frac{5}{2}\end{aligned}$$

(i) Hence, the root of given equation is 2 , when x ∈ N

(ii) Hence, the root of given equation is 2, $\frac{5}{2}$ , when x ∈ Q

Given,

$$\begin{aligned}a^2x^2 + 2ax + 1 = 0 \\[0.5em] \Rightarrow a^2x^2 + ax + ax + 1 = 0 \\[0.5em] \Rightarrow ax(ax + 1) + 1(ax + 1) = 0 \\[0.5em] \Rightarrow (ax + 1)(ax + 1) = 0 \text{ (Factorising left side) } \\[0.5em] \Rightarrow ax + 1 = 0 \text{ (Zero-product rule) }\\[0.5em] \Rightarrow ax = -1 \\[0.5em] \Rightarrow x = -\frac{1}{a}\end{aligned}$$

Hence, the roots of given equation are $-\frac{1}{a} , -\frac{1}{a}$

Given,

$$\begin{aligned}x^2 - (p + q)x + pq = 0\\[0.5em] \Rightarrow x^2 - px - qx + pq = 0 \\[0.5em] \Rightarrow x(x - p) - q(x - p) = 0 \\[0.5em] \Rightarrow (x - q)(x - p) = 0 \text{ (Factorising left side) }\\[0.5em] \Rightarrow x -q = 0 \text{ or } x - p = 0 \text{ (Zero-product rule)}\\[0.5em] \Rightarrow x = q \text{ or } x = p. \\[0.5em]\end{aligned}$$

Hence, the roots of given equation are p, q.

Given,

$$\begin{aligned}a^2x^2 + (a^2 + b^2)x + b^2 = 0 \\[0.5em] \Rightarrow a^2x^2 + a^2x + b^2x + b^2 = 0 \\[0.5em] \Rightarrow a^2x(x + 1) + b^2(x + 1) = 0 \\[0.5em] \Rightarrow (a^2x + b^2)(x + 1) = 0 \text{ (Factorising left side) } \\[0.5em] \Rightarrow a^2x + b^2 = 0 \text{ or } x + 1 = 0 \text{ (Zero-product rule) } \\[0.5em] \Rightarrow a^2x = -b^2 \text{ or } x = -1 \\[0.5em] \Rightarrow x = -\frac{b^2}{a^2} \text{ or } x =-1 \\[0.5em]\end{aligned}$$

Hence, the roots of given equation are $-\frac{b^2}{a^2}, -1.$

Given,

$$\begin{aligned}\sqrt{3}x^2 + 10x + 7\sqrt{3} = 0 \\[1em] \Rightarrow \sqrt{3}x^2 + 7x + 3x + 7\sqrt{3} = 0 \\[1em] \Rightarrow x(\sqrt{3}x + 7) + \sqrt{3}(\sqrt{3}x + 7) = 0 \\[1em] \Rightarrow (x + \sqrt{3})(\sqrt{3}x + 7) = 0 \text{ (Factorising left side) } \\[1em] \Rightarrow x + \sqrt{3} = 0 \text{ or } \sqrt{3}x + 7 = 0 \text{ (Zero- product rule) } \\[1em] \Rightarrow x = -\sqrt{3} \text{ or } \sqrt{3}x + 7 = 0 \\[1em] \Rightarrow x = -\sqrt{3} \text{ or } \sqrt{3}x = -7 \\[1em] \Rightarrow x = -\sqrt{3} \text{ or } x = -\frac{7}{\sqrt{3}} \\[1em] x = -\sqrt{3} \text{ or } x = -\frac{7\sqrt{3}}{3}\end{aligned}$$

Hence, the roots of given equation are $-\sqrt{3}, -\frac{7\sqrt{3}}{3}$

Given,

$$\begin{aligned}4\sqrt{3}x^2 + 5x - 2\sqrt{3} = 0 \\[1em] \Rightarrow 4\sqrt{3}x^2 + 8x - 3x - 2\sqrt{3} = 0 \\[1em] \Rightarrow 4x(\sqrt{3}x + 2) - \sqrt{3}(\sqrt{3}x + 2) = 0 \\[1em] \Rightarrow (\sqrt{3}x + 2)(4x - \sqrt{3}) = 0 \text{ (Factorising left side) } \\[1em] \Rightarrow \sqrt{3}x + 2 = 0 \text{ or } 4x - \sqrt{3} = 0 \text{ (Zero-product rule) } \\[1em] \Rightarrow \sqrt{3}x = -2 \text{ or } 4x = \sqrt{3} \\[1em] \Rightarrow x = -\frac{2}{\sqrt{3}} \text{ or } x = \frac{\sqrt{3}}{4} \\[1em] \Rightarrow x = -\frac{2}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} \text{ or } x = \frac{\sqrt{3}}{4} \\[1em] x = -\frac{2\sqrt{3}}{3} \text{ or } x = \frac{\sqrt{3}}{4}\end{aligned}$$

Hence, the roots of given equation are $-\frac{2\sqrt{3}}{3}$, $\frac{\sqrt{3}}{4}.$

Given,

$$\begin{aligned}x^2 - (1 + \sqrt{2})x + \sqrt{2} = 0 \\[1em] \Rightarrow x^2 - x - \sqrt{2}x + \sqrt{2} = 0 \\[1em] \Rightarrow x(x - 1) - \sqrt{2}(x - 1) = 0 \\[1em] \Rightarrow (x - \sqrt{2})(x - 1) = 0 \text{ (Factorising left side) } \\[1em] x - \sqrt{2} = 0 \text{ or } x - 1 = 0 \text{ (Zero-product rule) } \\[1em] x = \sqrt{2} \text{ or } x = 1\end{aligned}$$

Hence, the roots of given equation are $\sqrt{2}$ , 1.

Given,

$$\begin{aligned}x + \frac{1}{x} = 2\frac{1}{20} \\[1em] \Rightarrow x \times x + \frac{1}{x} \times x = \frac{41}{20} \times x \\[1em] \Rightarrow x^2 + 1 = \frac{41}{20}x \\[1em] \Rightarrow 20(x^2 + 1) = 41x \\[1em] \Rightarrow 20x^2 + 20 = 41x \\[1em] \Rightarrow 20x^2 - 41x + 20 = 0 \text{ (Writing as } ax^2 + bx + c = 0) \\[1em] \Rightarrow 20x^2 - 25x - 16x + 20 = 0 \\[1em] \Rightarrow 5x(4x - 5) - 4(4x - 5) = 0 \\[1em] \Rightarrow (5x - 4)(4x - 5) = 0 \text{ (Factorising left side) } \\[1em] \Rightarrow 5x - 4 = 0 \text{ or } 4x - 5 = 0 \text{ (Zero-product rule) } \\[1em] \Rightarrow 5x = 4 \text{ or } 4x = 5 \\[1em] x = \frac{4}{5} \text{ or } x = \frac{5}{4} \\[1em]\end{aligned}$$

Hence, the roots of given equation are $\frac{4}{5}$ , $\frac{5}{4}$.

Given,

$$\begin{aligned}\frac{2}{x^2} - \frac{5}{x} + 2 = 0 \\[1em] \Rightarrow \frac{2 - 5x + 2x^2}{x^2} = 0 \\[1em] \Rightarrow 2 - 5x + 2x^2 = 0 \times x^2 \\[1em] \Rightarrow 2x^2 - 5x + 2 = 0 \\[1em] \Rightarrow 2x^2 - 4x - x + 2 = 0 \\[1em] \Rightarrow2x(x - 2) - 1(x - 2) = 0 \\[1em] \Rightarrow (2x - 1)(x - 2) = 0 \text{ (Factorising left side) } \\[1em] \Rightarrow 2x - 1 = 0 \text{ or } x - 2 = 0 \text{ (Zero-product rule) } \\[1em] \Rightarrow 2x = 1 \text{ or } x = 2 \\[1em] x = \frac{1}{2} \text{ or } x = 2 \\[1em]\end{aligned}$$

Hence, the roots of given equation are $\frac{1}{2}$ , 2.

Given,

$$\begin{aligned}\frac{x^2}{15} - \frac{x}{3} -10 = 0 \\[1em] \Rightarrow \frac{x^2 - x \times 5 - 10 \times 15}{15} = 0 \\[1em] \Rightarrow x^2 - 5x - 150 = 0 \\[1em] \Rightarrow x^2 - 15x + 10x - 150 = 0 \\[1em] \Rightarrow x(x - 15) + 10(x - 15) = 0 \\[1em] \Rightarrow (x + 10)(x - 15) = 0 \text{ (Factorising left side) } \\[1em] \Rightarrow x + 10 = 0 \text{ or } x - 15 = 0 \text{ (Zero-product rule) } \\[1em] x = -10 \text{ or } x = 15 \\[1em]\end{aligned}$$

Hence, the roots of given equation are -10 , 15.

Given,

$$\begin{aligned}3x - \frac{8}{x} = 2 \\[1em] \Rightarrow \frac{3x^2 - 8}{x} = 2 \\[1em] \Rightarrow 3x^2 - 8 = 2x \\[1em] \Rightarrow 3x^2 - 2x - 8 = 0 \text{ (Writing as } ax^2 + bx + c = 0) \\[1em] \Rightarrow 3x^2 - 6x + 4x - 8 = 0 \\[1em] \Rightarrow 3x(x - 2) + 4(x - 2) = 0 \\[1em] \Rightarrow (3x + 4)(x - 2) = 0 \text{ (Factorising left side) } \\[1em] \Rightarrow 3x + 4 = 0 \text{ or } x - 2 = 0 \text{ (Zero-product rule) } \\[1em] \Rightarrow 3x = -4 \text { or } x = 2 \\[1em] x = -\frac{4}{3} \text{ or } x = 2 \\[1em]\end{aligned}$$

Hence, the roots of given equation are $-\frac{4}{3}$ , 2.

Given,

$$\begin{aligned}\frac{x + 2}{x + 3} = \frac{2x - 3}{3x - 7} \\[1em] \Rightarrow (x + 2) \times (3x - 7) = (2x - 3) \times (x + 3) \\[1em] \Rightarrow 3x^2 - 7x + 6x - 14 = 2x^2 + 6x - 3x - 9 \\[1em] \Rightarrow 3x^2 - x - 14 = 2x^2 + 3x - 9 \\[1em] \Rightarrow 3x^2 - 2x^2 - x - 3x - 14 + 9 = 0 \\[1em] \Rightarrow x^2 - 4x - 5 = 0 \text{ (Writing as } ax^2 + bx + c = 0) \\[1em] \Rightarrow x^2 - 5x + x - 5 = 0 \\[1em] \Rightarrow x(x - 5) + 1(x - 5) = 0 \\[1em] \Rightarrow (x - 5)(x + 1) = 0 \text{ (Factorising left side) } \\[1em] \Rightarrow x - 5 = 0 \text{ or } x + 1 = 0 \text{ (Zero-product rule) } \\[1em] x = 5 \text{ or } x = -1 \\[1em]\end{aligned}$$

Hence, the roots of given equation are -1, 5.

Given,

$$\begin{aligned}\frac{8}{x + 3} - \frac{3}{2 - x} = 2 \\[1em] \Rightarrow \frac{8(2 - x) - 3(x + 3)}{(x + 3)(2 - x)} = 2 \\[1em] \Rightarrow 8(2 - x) - 3(x + 3) = 2(x + 3)(2 - x) \\[1em] \Rightarrow 16 - 8x - 3x - 9 = 2(2x - x^2 + 6 - 3x) \\[1em] \Rightarrow 7 - 11x = 2(- x^2 - x + 6) \\[1em] \Rightarrow 7 - 11x = -2x^2 - 2x + 12 \\[1em] \Rightarrow 7 - 11x + 2x^2 + 2x - 12 = 0 \\[1em] \Rightarrow 2x^2 - 9x - 5 = 0 \text{ (Writing as } ax^2 + bx + c = 0) \\[1em] \Rightarrow 2x^2 - 10x + x - 5 = 0 \\[1em] \Rightarrow 2x(x - 5) + 1(x - 5) = 0 \\[1em] \Rightarrow (2x + 1)(x - 5) = 0 \text{ (Factorising left side) } \\[1em] \Rightarrow 2x + 1 = 0 \text{ or } x - 5 = 0 \text{ (Zero-product rule) } \\[1em] x = -\frac{1}{2} \text{ or } x = 5\end{aligned}$$

Hence, the roots of given equation are $-\frac{1}{2}$, 5.

Given,

$$\begin{aligned}\frac{x}{x - 1} + \frac{x - 1}{x} = 2\frac{1}{2} \\[1em] \Rightarrow \frac{x \times x + (x - 1)(x - 1)}{x(x - 1)} = 2\frac{1}{2} \\[1em] \Rightarrow x^2 + x^2 - x - x + 1 = \frac{5x(x - 1)}{2} \\[1em] \Rightarrow 2x^2 - 2x + 1 = \frac{5x^2 - 5x}{2} \\[1em] \Rightarrow 2(2x^2 - 2x + 1) = 5x^2 - 5x \\[1em] \Rightarrow 4x^2 - 4x + 2 = 5x^2 - 5x \\[1em] \Rightarrow 4x^2 - 5x^2 - 4x + 5x + 2 = 0 \\[1em] \Rightarrow -x^2 + x + 2 = 0 \text{ (Writing as } ax^2 + bx + c = 0) \\[1em] \Rightarrow x^2 - x - 2 = 0 \text{ (Multiplying the equation by -1) } \\[1em] \Rightarrow x^2 - 2x + x - 2 = 0 \\[1em] \Rightarrow x(x - 2) + 1(x - 2) = 0 \\[1em] \Rightarrow (x + 1)(x - 2) \text{ (Factorising left side) } \\[1em] \Rightarrow x + 1 = 0 \text{ or } x - 2 = 0 \text{ (Zero-product rule) } \\[1em] x = -1 \text{ or } x = 2 \\[1em]\end{aligned}$$

Hence, the roots of given equation are -1, 2.

Given,

$$\begin{aligned}\frac{x + 1}{x - 1} + \frac{x - 2}{x + 2} = 3 \\[1em] \Rightarrow \frac{(x + 1)(x + 2) + (x - 2)(x - 1)}{(x - 1)(x + 2)} = 3 \\[1em] \Rightarrow \frac{x^2 + 2x + x + 2 + x^2 - x - 2x + 2}{(x - 1)(x + 2)} = 3 \\[1em] \Rightarrow x^2 + 2x + x + 2 + x^2 - x - 2x + 2 = 3(x - 1)(x + 2) \\[1em] \Rightarrow x^2 + x^2 + 3x - 3x + 2 + 2 = 3(x^2 + 2x - x - 2) \\[1em] \Rightarrow 2x^2 + 4 = 3(x^2 + x - 2) \\[1em] \Rightarrow 2x^2 + 4 = 3x^2 + 3x - 6 \\[1em] \Rightarrow 2x^2 - 3x^2 - 3x + 4 + 6 = 0 \\[1em] \Rightarrow -x^2 - 3x + 10 = 0 \text{ (Writing as } ax^2 + bx + c = 0) \\[1em] \Rightarrow x^2 + 3x - 10 = 0 \text{ (Multiplying the equation by -1) } \\[1em] \Rightarrow x^2 + 5x - 2x - 10 = 0 \\[1em] \Rightarrow x(x + 5) - 2(x + 5) = 0 \\[1em] \Rightarrow (x - 2)(x + 5) = 0 \text{ (Factorising left side) } \\[1em] \Rightarrow x - 2 = 0 \text{ or } x + 5 = 0 \text{ (Zero-product rule) } \\[1em] x = 2 \text{ or } x = -5 \\[1em]\end{aligned}$$

Hence, the roots of given equation are 2 , -5.

Given,

$$\begin{aligned}\frac{1}{x - 3} - \frac{1}{x + 5} = \frac{1}{6} \\[1em] \Rightarrow \frac{(x + 5) - (x - 3)}{(x - 3)(x + 5)} = \frac{1}{6} \\[1em] \Rightarrow x + 5 - x + 3 = \frac{(x - 3)(x + 5)}{6} \\[1em] \Rightarrow 8 = \frac{x^2 + 5x - 3x - 15 }{6} \\[1em] \Rightarrow 8 \times 6 = x^2 + 2x - 15 \\[1em] \Rightarrow 48 = x^2 + 2x - 15 \\[1em] \Rightarrow x^2 + 2x - 15 = 48 \\[1em] \Rightarrow x^2 + 2x - 15 - 48 = 0 \\[1em] \Rightarrow x^2 + 2x - 63 = 0 \text{ (Writing as } ax^2 + bx + c = 0) \\[1em] \Rightarrow x^2 + 9x - 7x - 63 = 0 \\[1em] \Rightarrow x(x + 9) - 7(x + 9) = 0 \\[1em] \Rightarrow (x - 7)(x + 9) = 0 \text{ (Factorising left side) } \\[1em] x - 7 = 0 \text{ or } x + 9 = 0 \text{ (Zero-product rule) } \\[1em] x = 7 \text{ or } x = -9 \\[1em]\end{aligned}$$

Hence, the roots of given equation are -9 , 7.

Given,

$$\begin{aligned}\frac{a}{ax - 1} + \frac{b}{bx - 1} = a + b \\[1em] \Rightarrow \frac{a}{ax - 1} + \frac{b}{bx - 1} - a - b = 0 \\[1em] \Rightarrow \big( \frac{a}{ax - 1} - b \big) + \big(\frac{b}{bx -1} - a \big) = 0 \\[1em] \Rightarrow \frac{a - b(ax - 1)}{ax - 1} + \frac{b - a(bx - 1)}{bx -1} = 0 \\[1em] \Rightarrow \frac{a - abx + b}{ax -1} + \frac{b - abx + a}{bx - 1} = 0 \\[1em] \Rightarrow (a - abx + b) \big( \frac{1}{ax - 1} + \frac{1}{bx - 1}\big) = 0 \text{ (Factorising left side) } \\[1em] \Rightarrow a - abx + b = 0 \text{ or } \frac{1}{ax - 1} + \frac{1}{bx - 1} = 0 \text{ (Zero-product rule) }\\[1em] \Rightarrow a + b = abx \text{ or } \frac{bx - 1 + ax - 1}{(ax - 1)(bx - 1)} = 0 \\[1em] \Rightarrow x = \frac{a + b}{ab} \text{ or } ax + bx - 2 = 0 \times (ax - 1)(bx - 1) \\[1em] \Rightarrow x = \frac{a + b}{ab} \text{ or } ax + bx - 2 = 0 \\[1em] \Rightarrow x = \frac{a + b}{ab} \text{ or } x(a + b) = 2 \\[1em] x = \frac{a + b}{ab} \text{ or } x = \frac{2}{(a + b)} \\[1em]\end{aligned}$$

Hence, the roots of given equation are $\frac{a + b}{ab} , \frac{2}{(a + b)}$.

Given,

$$\begin{aligned}\frac{1}{2a + b + 2x} = \frac{1}{2a} + \frac{1}{b} + \frac{1}{2x} \\[1em] \Rightarrow \frac{1}{2a + b + 2x} - \frac{1}{2x} = \frac{1}{2a} + \frac{1}{b} \\[1em] \Rightarrow \frac{2x - (2a + b + 2x)}{(2a + b + 2x)(2x)} = \frac{b + 2a}{2ab} \\[1em] \Rightarrow \frac{ -(2a + b )}{(2a + b + 2x)(2x)} = \frac{b + 2a}{2ab}\\[1em] \Rightarrow \frac{ -1}{(2a + b + 2x)(2x)} = \frac{1}{2ab}\\[1em] \Rightarrow -2ab = (2a + b + 2x)(2x) \\[1em] \Rightarrow -2ab = 4ax + 2bx + 4x^2 \\[1em] \Rightarrow -ab = 2ax + bx + 2x^2 \text{ (Dividing the complete equation by 2) } \\[1em] \Rightarrow 2ax + bx + 2x^2 + ab = 0 \\[1em] \Rightarrow 2x^2 + 2ax + bx + ab = 0 \\[1em] \Rightarrow 2x(x + a) + b(x + a) = 0 \\[1em] \Rightarrow (2x + b)(x + a) = 0 \text{ (Factorising left side) } \\[1em] \Rightarrow 2x + b = 0 \text{ or } x + a = 0 \text{ (Zero-product rule) } \\[1em] x = -\frac{b}{2} \text{ or } x = -a \\[1em]\end{aligned}$$

Hence, the roots of given equation are $-\frac{b}{2}$, -a.

Given,

$$\begin{aligned}\frac{1}{x + 6} + \frac{1}{x - 10} = \frac{3}{x - 4} \\[1em] \Rightarrow \frac{x - 10 + x + 6}{(x + 6)(x - 10)} = \frac{3}{x - 4} \\[1em] \Rightarrow \frac{2x - 4}{(x + 6)(x - 10)} = \frac{3}{x - 4} \\[1em] \Rightarrow (2x - 4)(x - 4) = 3(x + 6)(x - 10) \\[1em] \Rightarrow 2x^2 - 8x - 4x + 16 = 3(x^2 - 10x + 6x - 60) \\[1em] \Rightarrow 2x^2 - 12x + 16 = 3(x^2 - 4x - 60) \\[1em] \Rightarrow 2x^2 - 12x + 16 = 3x^2 - 12x - 180 \\[1em] \Rightarrow 2x^2 - 3x^2 -12x + 12x + 16 + 180 = 0 \\[1em] \Rightarrow -x^2 + 196 = 0 \\[1em] \Rightarrow x^2 = 196 \\[1em] \Rightarrow x = \sqrt{196} \\[1em] x = 14 , -14 \\[1em]\end{aligned}$$

Hence, the roots of given equation are 14 , -14.

Given,

$\sqrt{3x + 4} = x$

On squaring both sides, we get

$$\begin{aligned}3x + 4 = x^2 \\[0.5em] \Rightarrow x^2 - 3x - 4 = 0 \\[0.5em] \Rightarrow x^2 - 4x + x - 4 = 0 \\[0.5em] \Rightarrow x(x - 4) + 1(x - 4) = 0 \\[0.5em] \Rightarrow (x + 1)(x - 4) = 0 \text{ (Factorising left side) }\\[0.5em] \Rightarrow x + 1 = 0 \text{ or } x - 4 = 0 \text{ (Zero-product rule) }\\[0.5em] x = -1 \text{ or } x = 4\end{aligned}$$

As equation is squared so roots need to be checked so putting x = -1 and x = 4 in the equation $\sqrt{3x + 4} = x.$
Checking for x = -1

$$\begin{aligned}\Rightarrow \sqrt{3 \times -1 + 4} = -1 \\[0.5em] \Rightarrow \sqrt{-3 + 4} = -1 \\[0.5em] \Rightarrow \sqrt{1} = -1 \text{ (This equation is false) }\end{aligned}$$

Checking for x = 4

$$\begin{aligned}\Rightarrow \sqrt{3 \times 4 + 4} = 4 \\[0.5em] \Rightarrow \sqrt{12 + 4} = 4 \\[0.5em] \Rightarrow \sqrt{16} = 4 \text{ (This equation is true ) }\\[0.5em]\end{aligned}$$

Since for x = -1 equation is false hence x = -1 is not the root of the given equation.
Hence, the root of given equation is 4.

Given,

$\sqrt{x(x - 7)} = 3\sqrt{2}$

On squaring both sides, we get

$$\begin{aligned}x(x - 7) = 18 \\[0.5em] \Rightarrow x^2 - 7x = 18 \\[0.5em] \Rightarrow x^2 - 7x - 18 = 0 \\[0.5em] \Rightarrow x^2 - 9x + 2x - 18 = 0 \\[0.5em] \Rightarrow x(x - 9) + 2(x - 9) = 0 \\[0.5em] \Rightarrow (x + 2)(x - 9) = 0 \text{ (Factorising left side) } \\[0.5em] \Rightarrow x + 2 = 0 \text{ or } x - 9 = 0 \text{ (Zero-product rule) } \\[0.5em] x = -2 \text{ or } x = 9.\end{aligned}$$

As equation is squared so roots need to be checked so putting x = -2 and x = 9 in the equation $\sqrt{x(x - 7)} = 3\sqrt{2}$
Checking for x = -2

$$\begin{aligned}\Rightarrow \sqrt{-2(-2 - 7)} = 3\sqrt{2} \\[0.5em] \Rightarrow \sqrt{-2(-9)} = 3\sqrt{(2)}\\[0.5em] \Rightarrow \sqrt{18} = 3\sqrt{2} \text{ (This equation is true) }\\[0.5em]\end{aligned}$$

Checking for x = 9

$$\begin{aligned}\Rightarrow \sqrt{9(9 - 7)} = 3\sqrt{2} \\[0.5em] \Rightarrow \sqrt{9 \times 2} = 3\sqrt{2} \\[0.5em] \Rightarrow \sqrt{18} = 3\sqrt{2} \text{ (This equation is true) }\end{aligned}$$

As the above two equations are true,

∴ The roots of given equation are -2, 9.

Given,

$$\begin{aligned}5(3x + 1)^2 + 6(3x + 1) - 8 = 0 \\[1em] \Rightarrow 5y^2 + 6y - 8 = 0 \text{ (Putting 3x + 1 = y) } \\[1em] \Rightarrow 5y^2 + 10y - 4y - 8 = 0 \\[1em] \Rightarrow 5y(y + 2) - 4(y + 2) = 0 \\[1em] \Rightarrow (5y - 4)(y + 2) = 0 \text{ (Factorising left side) }\\[1em] \Rightarrow 5y - 4 = 0 \text{ or } y + 2 = 0 \text{ (Zero-product rule) } \\[1em] \Rightarrow y = \frac{4}{5} \text{ or } y = -2 \\[1em] \text{When } y = \frac{4}{5}, 3x + 1 = \frac{4}{5} \\[1em] \Rightarrow 3x = \frac{4}{5} - 1 \\[1em] \Rightarrow 3x = -\frac{1}{5} \\[1em] \Rightarrow x = -\frac{1}{15} \\[1em] \text{When } y = -2 , 3x + 1 = -2 \\[1em] \Rightarrow 3x = -3 \\[1em] \Rightarrow x = -1\end{aligned}$$

Hence, the roots of given equation are $-\frac{1}{15}$ , -1.

Since, p + 1 = 0 it means p = -1.

Given,

$$\begin{aligned}x^2 + px - 6 = 0 \\[0.5em] \Rightarrow x^2 + (-1)x - 6 = 0 \\[0.5em] \Rightarrow x^2 - x - 6 = 0 \\[0.5em] \Rightarrow x^2 - 3x + 2x - 6 = 0 \\[0.5em] \Rightarrow x(x - 3) + 2(x - 3) = 0 \\[0.5em] \Rightarrow (x + 2)(x - 3) = 0 \text{ (Factorising left side) } \\[0.5em] \Rightarrow x + 2 = 0 \text{ or } x - 3 = 0 \text{ (Zero-product rule) } \\[0.5em] x = -2 \text{ or } x = 3\end{aligned}$$

Hence, the values of x are -2 , 3.

Since , p + 7 = 0, q - 12 = 0 it means p = -7 and q = 12.

Given,

$$\begin{aligned}x^2 + px + q = 0 \\[0.5em] \Rightarrow x^2 + (-7)x + 12 = 0 \\[0.5em] \Rightarrow x^2 - 7x + 12 = 0 \\[0.5em] \Rightarrow x^2 - 4x - 3x + 12 = 0 \\[0.5em] \Rightarrow x(x - 4) - 3(x - 4) = 0 \\[0.5em] \Rightarrow (x - 3)(x - 4) = 0 \text{ (Factorising left side) } \\[0.5em] \Rightarrow x - 3 = 0 \text{ or } x - 4 = 0 \text{ (Zero-product rule) } \\[0.5em] x = 3 \text{ or } x = 4\end{aligned}$$

Hence, the values of x are 3 , 4.

If x = p is a solution of the equation x(2x + 5) = 3 , then x = p satisfies the equation.
Putting x = p in equation,

$$\begin{aligned}p(2p + 5) = 3 \\[0.5em] \Rightarrow 2p^2 + 5p = 3 \\[0.5em] \Rightarrow 2p^2 + 5p - 3 = 0 \\[0.5em] \Rightarrow 2p^2 + 6p - p - 3 = 0 \\[0.5em] \Rightarrow 2p(p + 3) - 1(p + 3) = 0 \\[0.5em] \Rightarrow (2p - 1)(p + 3) = 0 \text{ (Factorising left side) }\\[0.5em] \Rightarrow 2p - 1 = 0 \text{ or } p + 3 = 0 \text{ (Zero-product rule) } \\[0.5em] \Rightarrow 2p = 1 \text{ or } p = -3 \\[0.5em] p = \frac{1}{2} \text{ or } p = -3\end{aligned}$$

Hence, the values of p are -3 , $\frac{1}{2}$.

If x = 3 is a solution of the equation (k + 2)x² - kx + 6 = 0 , then x = 3 satisfies the equation.
Putting x = 3 in equation,

$$\begin{aligned}(k + 2)3^2 - 3k + 6 = 0 \\[0.5em] \Rightarrow 9(k + 2) - 3k + 6 = 0 \\[0.5em] \Rightarrow 9k + 18 - 3k + 6 = 0 \\[0.5em] \Rightarrow 6k + 24 = 0 \\[0.5em] \Rightarrow 6k = -24 \\[0.5em] \Rightarrow k = -\frac{24}{6} \\[0.5em] k = -4 \\[0.5em]\end{aligned}$$

Putting value of k in equation in order to find other root

$$\begin{aligned}\Rightarrow (-4 + 2)x^2 - (-4)x + 6 = 0 \\[0.5em] \Rightarrow -2x^2 + 4x + 6 = 0 \\[0.5em] \Rightarrow 2x^2 - 4x - 6 = 0 \text{ (Multiplying equation by -1) }\\[0.5em] \Rightarrow 2x^2 - 6x + 2x - 6 = 0 \\[0.5em] \Rightarrow 2x(x - 3) + 2(x - 3) = 0 \\[0.5em] \Rightarrow (2x + 2)(x - 3) = 0 \text{ (Factorising left side) } \\[0.5em] \Rightarrow 2x + 2 = 0 \text{ or } x - 3 = 0 \text{ (Zero-product rule) } \\[0.5em] \Rightarrow 2x = -2 \text{ or } x = 3 \\[0.5em] x = -1 \text{ or } x = 3 .\end{aligned}$$

Hence, the values of k is -4 ,and the other root is -1.

The given equation is 2x² - 7x + 6 = 0.

Comparing it with ax² + bx + c = 0, we get
a = 2 , b = -7 , c = 6

By using formula,

$x = \frac{-b ± \sqrt{b^2 - 4ac}}{2a}$

we obtain:

$$\begin{aligned}\Rightarrow x = \frac{-(-7) ± \sqrt{(-7^2) - 4\times 2 \times 6}}{2 \times 2} \\[1em] \Rightarrow x = \frac{7 ± \sqrt{49 - 48}}{4} \\[1em] \Rightarrow x = \frac{7 ± \sqrt{1}}{4} \\[1em] \Rightarrow x = \frac{7 + 1}{4} \text{ or } \frac{7 - 1}{4} \\[1em] \Rightarrow x = \frac{8}{4} \text { or } \frac{6}{4} \\[1em] x = 2 \text{ or } \frac{3}{2}\end{aligned}$$

Hence roots of the given equation are 2 , $\frac{3}{2}$.

The given equation is 2x² - 6x + 3 = 0.

Comparing it with ax² + bx + c = 0, we get
a = 2 , b = -6 , c = 3

By using formula,

$x = \frac{-b ± \sqrt{b^2 - 4ac}}{2a}$

we obtain:

$$\begin{aligned}\Rightarrow x = \frac{-(-6) ± \sqrt{(-6^2) - 4\times 2 \times 3}}{2 \times 2} \\[1em] \Rightarrow x = \frac{6 ± \sqrt{36 - 24}}{4} \\[1em] \Rightarrow x = \frac{6 ± \sqrt{12}}{4} \\[1em] \Rightarrow x = \frac{6 + \sqrt{12}}{4} \text{ or } \frac{6 - \sqrt{12}}{4} \\[1em] \Rightarrow x = \frac{6 + 2\sqrt{3}}{4} \text { or } \frac{6 - 2\sqrt{3}}{4} \\[1em] x = \frac{3 + \sqrt{3}}{2} \text{ or } \frac{3 - \sqrt{3}}{2}\end{aligned}$$

Hence roots of the given equation are $\frac{3 + \sqrt{3}}{2} , \frac{3 - \sqrt{3}}{2}$.

The given equation is 256x² - 32x + 1 = 0.

Comparing it with ax² + bx + c = 0, we get
a = 256 , b = -32 , c = 1

By using formula,

$x = \frac{-b ± \sqrt{b^2 - 4ac}}{2a}$

we obtain:

$$\begin{aligned}\Rightarrow x = \frac{-(-32) ± \sqrt{(-32)^2 - 4\times 256 \times 1}}{2 \times 256} \\[1em] \Rightarrow x = \frac{32 ± \sqrt{1024 - 1024}}{512} \\[1em] \Rightarrow x = \frac{32 ± \sqrt{0}}{512} \\[1em] \Rightarrow x = \frac{32 + 0}{512} \text{ or } \frac{32 - 0}{512} \\[1em] \Rightarrow x = \frac{32}{512} \text{ or } \frac{32 - 0}{512} \\[1em] x = \frac{1}{16} \text{ or } \frac{1}{16}\end{aligned}$$

Hence roots of the given equation are $\frac{1}{16} , \frac{1}{16}$.

The given equation is 25x² + 30x + 7 = 0.

Comparing it with ax² + bx + c = 0, we get
a = 25 , b = 30 , c = 7

By using formula,

$x = \frac{-b ± \sqrt{b^2 - 4ac}}{2a}$

we obtain:

$$\begin{aligned}\Rightarrow x = \frac{-(30) ± \sqrt{(30^2) - 4\times 25 \times 7}}{2 \times 25} \\[1em] \Rightarrow x = \frac{-30 ± \sqrt{900 - 700}}{50} \\[1em] \Rightarrow x = \frac{-30 ± \sqrt{200}}{50} \\[1em] \Rightarrow x = \frac{-30 + \sqrt{200}}{50} \text{ or } \frac{-30 - \sqrt{200}}{50} \\[1em] \Rightarrow x = \frac{-30 + 10\sqrt{2}}{50} \text { or } \frac{-30 - 10\sqrt{2}}{50} \\[1em] x = \frac{-3 + \sqrt{2}}{5} \text{ or } \frac{-3 - \sqrt{2}}{5}\end{aligned}$$

Hence roots of the given equation are $\frac{-3 + \sqrt{2}}{5} , \frac{-3 - \sqrt{2}}{5}$.

The given equation is $2x^2 + \sqrt{5}x - 5 = 0$

Comparing it with ax² + bx + c = 0, we get
a = 2 , b = $\sqrt{5}$ , c = -5

By using formula,

$x = \frac{-b ± \sqrt{b^2 - 4ac}}{2a}$

we obtain:

$$\begin{aligned}\Rightarrow x = \frac{-(\sqrt{5}) ± \sqrt{(\sqrt{5})^2 - 4\times 2 \times -5}}{2 \times 2} \\[1em] \Rightarrow x = \frac{-\sqrt{5} ± \sqrt{5 + 40}}{4} \\[1em] \Rightarrow x = \frac{-\sqrt{5} ± \sqrt{45}}{4} \\[1em] \Rightarrow x = \frac{-\sqrt{5} + \sqrt{45}}{4} \text{ or } \frac{-\sqrt{5} - \sqrt{45}}{4} \\[1em] \Rightarrow x = \frac{-\sqrt{5} + 3\sqrt{5}}{4} \text { or } \frac{-\sqrt{5} - 3\sqrt{5}}{4} \\[1em] \Rightarrow x = \frac{\sqrt{5}(-1 + 3)}{4} \text{ or } \frac{\sqrt{5}(-1 - 3)}{4} \\[1em] x = \frac{\sqrt{5}}{2} \text{ or } -\sqrt{5}\end{aligned}$$

Hence roots of the given equations are $\frac{\sqrt{5}}{2} , -\sqrt{5}$.

The given equation is $\sqrt{3}x^2 + 10x - 8\sqrt{3} = 0$

Comparing it with ax² + bx + c = 0, we get
$a = \sqrt{3}, \space b =10, \space c = -8\sqrt{3}$

By using formula, $x = \frac{-b ± \sqrt{b^2 - 4ac}}{2a}$

we obtain:

$$\begin{aligned}\Rightarrow x = \frac{-(10) ± \sqrt{(10)^2 - 4\times \sqrt{3} \times -8\sqrt{3}}}{2 \times \sqrt{3}} \\[1em] \Rightarrow x = \frac{-10 ± \sqrt{100 + 96}}{2\sqrt{3}} \\[1em] \Rightarrow x = \frac{-10 ± \sqrt{196}}{2\sqrt{3}} \\[1em] \Rightarrow x = \frac{-10 + 14}{2\sqrt{3}} \text{ or } \frac{-10 - 14}{2\sqrt{3}}\\[1em] \Rightarrow x = \frac{4}{2\sqrt{3}} \text { or } \frac{-24}{2\sqrt{3}} \\[1em] \Rightarrow x = \frac{2}{\sqrt{3}} \text{ or } \frac{-12}{\sqrt{3}} \\[1em] \Rightarrow x = \frac{2}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} \text{ or } \frac{-12}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} \text{ (Multiplying both roots by } \frac{\sqrt{3}}{\sqrt{3}}) \\[1em] \Rightarrow x = \frac{2\sqrt{3}}{3} \text{ or } -4\sqrt{3}.\end{aligned}$$

Hence roots of the given equations are $\frac{2\sqrt{3}}{3} , -4\sqrt{3}$.

Given,

$$\begin{aligned}\frac{x - 2}{x + 2} + \frac{x + 2}{x - 2} = 4 \\[1em] \Rightarrow \frac{(x - 2)^2 + (x + 2)^2}{(x - 2)(x + 2)} = 4 \\[1em] \Rightarrow \frac{x^2 + 4 - 4x + x^2 + 4 + 4x}{x^2 - 2x + 2x - 4} = 4 \\[1em] \Rightarrow \frac{2x^2 + 8}{x^2 - 4} = 4 \\[1em] \Rightarrow 2x^2 + 8 = 4(x^2 - 4) \\[1em] \Rightarrow 2x^2 + 8 = 4x^2 - 16 \\[1em] \Rightarrow 2x^2 - 4x^2 + 8 + 16 = 0 \\[1em] \Rightarrow -2x^2 + 24 = 0 \\[1em] 2x^2 - 24 = 0 \text{ (Multiplying equation by -1) }\end{aligned}$$

The equation is $2x^2 - 24 = 0$

Comparing it with ax² + bx + c = 0, we get
a = 2 , b = 0 , c = -24

By using formula, $x = \frac{-b ± \sqrt{b^2 - 4ac}}{2a}$

we obtain:

$$\begin{aligned}\Rightarrow \frac{-0 ± \sqrt{0^2 - 4 \times 2 \times (-24)}}{2 \times 2} \\[1em] \Rightarrow \frac{ ±\sqrt{192}}{4} \\[1em] \Rightarrow \frac{±8\sqrt{3}}{4} \\[1em] +2\sqrt{3} \text { or } -2\sqrt{3} \\[1em]\end{aligned}$$

Hence roots of the given equations are $2\sqrt{3} , -2\sqrt{3}$.

Given,

$$\begin{aligned}\frac{x + 1}{x + 3} = \frac{3x + 2}{2x + 3} \\[0.5em] \Rightarrow (x + 1)(2x + 3) = (3x + 2)(x + 3) \text{ On cross multiplication}\\[0.5em] \Rightarrow 2x^2 + 3x + 2x + 3 = 3x^2 + 9x + 2x + 6 \\[0.5em] \Rightarrow 2x^2 + 5x + 3 = 3x^2 + 11x + 6 \\[0.5em] \Rightarrow 2x^2 - 3x^2 + 5x - 11x + 3 - 6 = 0 \\[0.5em] \Rightarrow -x^2 - 6x - 3 = 0 \\[0.5em] \Rightarrow x^2 + 6x + 3 = 0 \text{ (Multiplying equation by -1) } \\[0.5em]\end{aligned}$$

The equation is x² + 6x + 3 = 0

Comparing it with ax² + bx + c = 0, we get
a = 1 , b = 6 , c = 3

By using formula, $x = \frac{-b ± \sqrt{b^2 - 4ac}}{2a}$

we obtain:

$$\begin{aligned}\Rightarrow \frac{-6 ± \sqrt{6^2 - 4 \times 1 \times 3}}{2 \times 1} \\[1em] \Rightarrow \frac{-6 ±\sqrt{24}}{2} \\[1em] \Rightarrow \frac{-6 + \sqrt{24}}{2} \text { or } \frac{-6 - \sqrt{24}}{2} \\[1em] \Rightarrow \frac{-6 + 2\sqrt{6}}{2} \text{ or } \frac{-6 - 2\sqrt{6}}{2} \\[1em] -3 + \sqrt{6} \text{ or } -3 - \sqrt{6}\end{aligned}$$

Hence roots of the given equations are $-3 + \sqrt{6} , -3 - \sqrt{6}$.

Given,

a(x² + 1) = (a² + 1)x

First converting the equation in the form ax² + bx + c = 0.

$$\begin{aligned}\Rightarrow ax^2 + a = a^2x + x \\[0.5em] \Rightarrow ax^2 - a^2x - x + a = 0 \\[0.5em] \Rightarrow ax^2 -(a^2 + 1)x + a = 0 \\[0.5em]\end{aligned}$$

Comparing it with ax² + bx + c = 0, we get
a = a , b = -(a² + 1) , c = a

By using formula, $x = \frac{-b ± \sqrt{b^2 - 4ac}}{2a}$

we obtain:

$$\begin{aligned}\Rightarrow \frac{-(-(a^2 + 1)) ± \sqrt{(-(a^2 + 1))^2 - 4 \times a \times a}}{2 \times a} \\[1em] \Rightarrow \frac{a^2 + 1 ± \sqrt{(a^2 + 1)^2 - 4a^2}}{2a} \\[1em] \Rightarrow \frac{a^2 + 1 ± \sqrt{a^4 + 1 + 2a^2 - 4a^2}}{2a} \\[1em] \Rightarrow \frac{a^2 + 1 ± \sqrt{a^4 + 1 - 2a^2}}{2a} \\[1em] \Rightarrow \frac{a^2 + 1 ± \sqrt{(a^2 - 1)^2}}{2a} \\[1em] \Rightarrow \frac{a^2 + 1 ± (a^2 - 1) }{2a} \\[1em] \Rightarrow \frac{a^2 + 1 + a^2 - 1}{2a} \text{ or } \frac{a^2 + 1 - a^2 + 1}{2a} \\[1em] \Rightarrow \frac{2a^2}{2a} \text{ or } \frac{2}{2a} \\[1em] \Rightarrow a \text{ or } \frac{1}{a}\end{aligned}$$

Hence roots of the given equations are a , $\frac{1}{a}$.

The given equation is 4x² - 4ax + (a² - b²) = 0.

Comparing it with ax² + bx + c = 0, we get
a = 4 , b = -4a , c = a² - b²

By using formula,

$x = \frac{-b ± \sqrt{b^2 - 4ac}}{2a}$

we obtain:

$$\begin{aligned}\Rightarrow \frac{-(-4a) ± \sqrt{(-4a)^2 - 4 \times 4 \times (a^2 - b^2)}}{2 \times 4} \\[1em] \Rightarrow \frac{4a ± \sqrt{16a^2 - 16(a^2 - b^2)}}{8} \\[1em] \Rightarrow \frac{4a ± \sqrt{16a^2 - 16a^2 + 16b^2}}{8} \\[1em] \Rightarrow \frac{4a ± \sqrt{16b^2}}{8} \\[1em] \Rightarrow \frac{4a ± 4b}{8} \\[1em] \Rightarrow \frac{4a + 4b}{8} \text{ or } \frac{4a - 4b}{8} \\[1em] \Rightarrow \frac{a + b}{2} \text{ or } \frac{a - b}{2} \\[1em]\end{aligned}$$

Hence roots of the given equations are $\frac{a + b}{2} , \frac{a - b}{2}$.

Given,

$x - \frac{1}{x} = 3 , x$ ≠ 0

$$\begin{aligned}\Rightarrow \frac{x^2 - 1}{x} = 3 \text{ (By taking L.C.M) } \\[0.5em] \Rightarrow x^2 - 1 = 3x \\[0.5em] \Rightarrow x^2 - 3x - 1 = 0 \\[0.5em]\end{aligned}$$

Comparing it with ax² + bx + c = 0, we get
a = 1 , b = -3 , c = -1

By using formula, $x = \frac{-b ± \sqrt{b^2 - 4ac}}{2a}$

we obtain:

$$\begin{aligned}\Rightarrow \frac{-(-3) ± \sqrt{(-3)^2 - 4 \times 1 \times -1}}{2 \times 1} \\[1em] \Rightarrow \frac{3 ± \sqrt{9 - (-4)}}{2} \\[1em] \Rightarrow \frac{3 ± \sqrt{13}}{2} \\[1em] \Rightarrow \frac{3 ± \sqrt{13}}{2} \\[1em] \Rightarrow \frac{3 + \sqrt{13}}{2} \text{ or } \frac{3 - \sqrt{13}}{2} \\[1em]\end{aligned}$$

Hence roots of the given equations are $\frac{3 + \sqrt{13}}{2} , \frac{3 - \sqrt{13}}{2}$.

Given,

$$\begin{aligned}\frac{1}{x} + \frac{1}{x - 2} = 3 \\[1em] \Rightarrow \frac{x - 2 + x }{x(x - 2)} = 3 \\[1em] \Rightarrow \frac{2x - 2}{x^2 - 2x} = 3 \\[1em] \Rightarrow 2x - 2 = 3(x^2 - 2x) \\[1em] \Rightarrow 2x - 2 = 3x^2 - 6x \\[1em] \Rightarrow 3x^2 - 6x - 2x + 2 = 0 \\[1em] \Rightarrow 3x^2 - 8x + 2 = 0 \\[1em]\end{aligned}$$

Comparing it with ax² + bx + c = 0, we get
a = 3 , b = -8 , c = 2

By using formula, $x = \frac{-b ± \sqrt{b^2 - 4ac}}{2a}$

we obtain:

$$\begin{aligned}\Rightarrow \frac{-(-8) ± \sqrt{(-8)^2 - 4 \times 3 \times 2}}{2 \times 3} \\[1em] \Rightarrow \frac{-(-8) ± \sqrt{(-8)^2 - 24}}{6} \\[1em] \Rightarrow \frac{8 ± \sqrt{64 - 24}}{6} \\[1em] \Rightarrow \frac{8 ± \sqrt{40}}{6} \\[1em] \Rightarrow \frac{8 ± 2\sqrt{10}}{6} \\[1em] \Rightarrow \frac{4 ± \sqrt{10}}{3} \\[1em] \Rightarrow \frac{4 + \sqrt{10}}{3} \text{ or } \frac{4 - \sqrt{10}}{3}\end{aligned}$$

Hence roots of the given equations are $\frac{4 + \sqrt{10}}{3} , \frac{4 - \sqrt{10}}{3}$.

Given,

$$\begin{aligned}2\Big(\frac{2x - 1}{x + 3}\Big) - 3\Big(\frac{x + 3}{2x - 1}\Big) = 5 \\[1em] \Rightarrow \text{ Taking } y = \frac{2x - 1}{x + 3} \text{ the equation becomes} \\[1em] \Rightarrow 2y - \frac{3}{y} = 5 \\[1em] \Rightarrow 2y^2 - 3 = 5y \\[1em] \Rightarrow 2y^2 - 5y - 3 = 0 \\[1em]\end{aligned}$$

Comparing it with ax² + bx + c = 0, we get
a = 2 , b = -5 , c = -3

By using formula, $x = \frac{-b ± \sqrt{b^2 - 4ac}}{2a}$

we obtain:

$$\begin{aligned}\Rightarrow \frac{-(-5) ± \sqrt{(-5)^2 - 4 \times 2 \times -3}}{2 \times 2} \\[1em] \Rightarrow \frac{5 ± \sqrt{25 + 24}}{4} \\[1em] \Rightarrow \frac{5 ± \sqrt{49}}{4} \\[1em] \Rightarrow \frac{5 ± 7}{4} \\[1em] \Rightarrow \frac{5 + 7}{4} \text{ or } \frac{5 - 7}{4}\\[1em] \Rightarrow \frac{12}{4} \text{ or } \frac{-2}{4} \\[1em] \Rightarrow 3 \text{ or } -\frac{1}{2} \\[1em]\end{aligned}$$

But,

$$\begin{aligned}y = \frac{2x - 1}{x + 3} \\[1em] \therefore 3 = \frac{2x - 1}{x + 3} \text{ or } -\frac{1}{2} = \frac{2x - 1}{x + 3} \\[1em] \Rightarrow 3(x + 3) = 2x - 1 \text{ or } -(x + 3) = 2(2x - 1) \\[1em] \Rightarrow 3x + 9 = 2x - 1 \text{ or } -x - 3 = 4x - 2 \\[1em] \Rightarrow 3x - 2x = -1 - 9 \text{ or } -x - 4x = -2 + 3 \\[1em] \Rightarrow x = -10 \text{ or } -5x = 1 \\[1em] \Rightarrow x = -10 \text{ or } x = -\frac{1}{5} \\[1em]\end{aligned}$$

Hence roots of the given equations are -10 , $-\frac{1}{5}$.

(i) The given equation is x² - 5x - 10 = 0

Comparing it with ax² + bx + c = 0, we get
a = 1 , b = -5 , c = -10

By using formula, $x = \frac{-b ± \sqrt{b^2 - 4ac}}{2a}$

we obtain:

$$\begin{aligned}\Rightarrow x = \frac{-(-5) ± \sqrt{(-5)^2 - 4\times 1 \times -10}}{2 \times 1} \\[1em] \Rightarrow x = \frac{5 ± \sqrt{25 + 40}}{2} \\[1em] \Rightarrow x = \frac{5 ± \sqrt{65}}{2} \\[1em] \Rightarrow x = \frac{5 + \sqrt{65}}{2} \text{ or } \frac{5 - \sqrt{65}}{2} \\[1em] \text{ Also } \sqrt{65} = 8.062 (\text{From tables}) \\[1em] \Rightarrow x = \frac{5 + 8.062}{2} \text { or } \frac{5 - 8.062}{2} \\[1em] \Rightarrow x = \frac{13.062}{2} \text{ or } \frac{-3.062}{2} \\[1em] \Rightarrow x = 6.531 \text{ or } -1.531 \\[1em] x = 6.53 \text{ or } -1.53 \text{ (correct to two decimal places) }\end{aligned}$$

Hence roots of the given equations are 6.53 , -1.53.

(ii) Given, x² + 7x = 7

or , x² + 7x - 7 = 0

The given equation is x² + 7x - 7 = 0

Comparing it with ax² + bx + c = 0, we get
a = 1 , b = 7 , c = -7

By using formula, $x = \frac{-b ± \sqrt{b^2 - 4ac}}{2a}$

we obtain:

$$\begin{aligned}\Rightarrow x = \frac{-(7) ± \sqrt{(7)^2 - 4\times 1 \times -7}}{2 \times 1} \\[1em] \Rightarrow x = \frac{-7 ± \sqrt{49 + 28}}{2} \\[1em] \Rightarrow x = \frac{-7 ± \sqrt{77}}{2} \\[1em] \Rightarrow x = \frac{-7 + \sqrt{77}}{2} \text{ or } \frac{-7 - \sqrt{77}}{2} \\[1em] \text{ Also } \sqrt{77} = 8.775 (\text{From tables}) \\[1em] \Rightarrow x = \frac{-7 + 8.775}{2} \text { or } \frac{-7 - 8.775}{2} \\[1em] \Rightarrow x = \frac{1.775}{2} \text{ or } \frac{-15.775}{2} \\[1em] \Rightarrow x = 0.885 \text{ or } -7.885 \\[1em] \Rightarrow x = 0.89\text{ or } -7.89 \text{ (correct to two decimal places) }\end{aligned}$$

Hence roots of the given equations are 0.89 , -7.89.

(i) The given equation is 4x² - 5x - 3 = 0

Comparing it with ax² + bx + c = 0, we get
a = 4 , b = -5 , c = -3

By using formula, $x = \frac{-b ± \sqrt{b^2 - 4ac}}{2a}$

we obtain:

$$\begin{aligned}\Rightarrow x = \frac{-(-5) ± \sqrt{(-5)^2 - 4\times 4 \times -3}}{2 \times 4} \\[1em] \Rightarrow x = \frac{5 ± \sqrt{25 + 48}}{8} \\[1em] \Rightarrow x = \frac{5 ± \sqrt{73}}{8} \\[1em] \Rightarrow x = \frac{5 + \sqrt{73}}{8} \text{ or } \frac{5 - \sqrt{73}}{8} \\[1em] \text{ Also } \sqrt{73} = 8.54 \text{ (From tables)} \\[1em] \Rightarrow x = \frac{5 + 8.54}{8} \text { or } \frac{5 - 8.54}{8} \\[1em] \Rightarrow x = \frac{13.54}{8} \text{ or } \frac{-3.54}{8} \\[1em] \Rightarrow x = 1.69 \text{ or } -0.44\end{aligned}$$

Hence roots of the given equations are 1.69, -0.44.

(ii) For a quadratic equation in the form :

ax² + bx + c = 0

The solutions are :

x = $\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Comparing equation x² - 7x + 3 = 0, with ax² + bx + c = 0, we get :

a = 1, b = -7 and c = 3

$$\begin{aligned}\Rightarrow x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4 \times 1 \times 3}}{2 \times 1} \\[1em] = \frac{7 \pm \sqrt{49 - 12}}{2} \\[1em] = \frac{7 \pm \sqrt{37}}{2} \\[1em] = \frac{7 \pm 6.08}{2} \\[1em] = \frac{13.08}{2} \text{ or } \frac{0.92}{2}\\[1em] = 6.54 \text{ or } 0.46\end{aligned}$$

Hence, the value of x = 6.54 or 0.46

(i) The given equation is x² - 4x - 8 = 0

Comparing it with ax² + bx + c = 0, we get
a = 1 , b = -4 , c = -8

By using formula, $x = \frac{-b ± \sqrt{b^2 - 4ac}}{2a}$

we obtain:

$$\begin{aligned}\Rightarrow x = \frac{-(-4) ± \sqrt{(-4)^2 - 4\times 1 \times -8}}{2 \times 1} \\[1em] \Rightarrow x = \frac{4 ± \sqrt{16 + 32}}{2} \\[1em] \Rightarrow x = \frac{4 ± \sqrt{48}}{2} \\[1em] \Rightarrow x = \frac{4 + \sqrt{48}}{2} \text{ or } \frac{4 - \sqrt{48}}{2} \\[1em] \text{ Also } \sqrt{48} = 6.928 (\text{From tables}) \\[1em] \Rightarrow x = \frac{4 + 6.928}{2} \text { or } \frac{4 - 6.928}{2} \\[1em] \Rightarrow x = \frac{10.928}{2} \text{ or } \frac{-2.928}{2} \\[1em] \Rightarrow x = 5.464 \text{ or } -1.464 \\[1em] x = 5.5 \text{ or } -1.5 \text{ (correct to two significant figures) }\end{aligned}$$

Hence roots of the given equations are 5.5 , -1.5.

(ii) Given,

$$\begin{aligned}\Rightarrow x - \frac{18}{x} = 6 \\[1em] \Rightarrow \frac{x^2 - 18}{x} = 6 \\[1em] \Rightarrow x^2 - 18 = 6x \\[1em] \Rightarrow x^2 - 6x - 18 = 0 \\[1em]\end{aligned}$$

Comparing it with ax² + bx + c = 0, we get
a = 1 , b = -6 , c = -18

By using formula, $x = \frac{-b ± \sqrt{b^2 - 4ac}}{2a}$

we obtain:

$$\begin{aligned}\Rightarrow x = \frac{-(-6) ± \sqrt{(-6)^2 - 4\times 1 \times -18}}{2 \times 1} \\[1em] \Rightarrow x = \frac{6 ± \sqrt{36 + 72}}{2} \\[1em] \Rightarrow x = \frac{6 ± \sqrt{108}}{2} \\[1em] \Rightarrow x = \frac{6 + \sqrt{108}}{2} \text{ or } \frac{6 - \sqrt{108}}{2} \\[1em] \text{ Also } \sqrt{108} = 10.392 (\text{From tables}) \\[1em] \Rightarrow x = \frac{6 + 10.392}{2} \text { or } \frac{6 - 10.392}{2} \\[1em] \Rightarrow x = \frac{16.392}{2} \text{ or } \frac{-4.392}{2} \\[1em] \Rightarrow x = 8.196 \text{ or } -2.195 \\[1em] \Rightarrow x = 8.2 \text{ or } -2.2 \text{ (correct to two significant figures) }\end{aligned}$$

Hence roots of the given equations are 8.2 , -2.2.

Comparing equation 2x² - 10x + 5 = 0 with ax² + bx + c = 0, we get :

a = 2, b = -10 and c = 5

By formula,

⇒ x = $\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substituting values we get :

$$\begin{aligned}\Rightarrow x = \frac{-(-10) \pm \sqrt{(-10)^2 - 4 \times 2 \times 5}}{2 \times 2} \\[1em] = \frac{10 \pm \sqrt{100 - 40}}{4} \\[1em] = \frac{10 \pm \sqrt{60}}{4} \\[1em] = \frac{10 \pm 2\sqrt{15}}{4} \\[1em] = \frac{2(5 \pm \sqrt{15})}{4} \\[1em] = \frac{5 \pm \sqrt{15}}{2} \\[1em] = \frac{5 \pm 3.87}{2} \\[1em] = \frac{5 + 3.87}{2}, \frac{5 - 3.87}{2} \\[1em] = \frac{8.87}{2}, \frac{1.13}{2} \\[1em] = 4.44, 0.565\end{aligned}$$

Hence, x = 4.44, 0.565

(i) The given equation is 3x² - 5x - 2 = 0.

Comparing it with ax² + bx + c = 0, we get
a = 3 , b = -5 , c = -2

∴ Discriminant = b² - 4ac

Putting values of a, b, c in formula

$$\begin{aligned}(-5)^2 - 4 \times 3 \times -2 \\[0.5em] = 25 + 24 \\[0.5em] = 49 \gt 0\end{aligned}$$

Discriminant = 49; Hence, the given equation has two distinct real roots.

(ii) The given equation is 2x² - 3x + 5 = 0.

Comparing it with ax² + bx + c = 0, we get
a = 2 , b = -3 , c = 5
∴ Discriminant = b² - 4ac
Putting values of a, b, c in formula

$$\begin{aligned}(-3)^2 - 4 \times 2 \times 5 \\[0.5em] = 9 - 40 \\[0.5em] = -31 \lt 0\end{aligned}$$

Discriminant = -31; Hence, the given equation has no real roots.

(iii) The given equation is 16x² - 40x + 25 = 0.
Comparing it with ax² + bx + c = 0, we get
a = 16 , b = -40 , c = 25
∴ Discriminant = b² - 4ac
Putting values of a, b, c in formula

$$\begin{aligned}(-40)^2 - 4 \times 16 \times 25 \\[0.5em] = 1600 - 1600 \\[0.5em] = 0\end{aligned}$$

Discriminant = 0; Hence, the given equation has two equal real roots.

(iv) The given equation is 2x² + 15x + 30 = 0.
Comparing it with ax² + bx + c = 0, we get
a = 2 , b = 15 , c = 30
∴ Discriminant = b² - 4ac
Putting values of a, b, c in formula

$$\begin{aligned}(15)^2 - 4 \times 2 \times 30 \\[0.5em] = 225 - 240 \\[0.5em] = -15 \lt 0\end{aligned}$$

Discriminant = -15; Hence, the given equation has no real roots.

(i) The given equation is $3x^2 - 4\sqrt{3}x + 4 = 0$.

Comparing it with ax² + bx + c = 0, we get
a = 3 , b = $-4\sqrt{3}$ , c = 4
∴ Discriminant = b² - 4ac
Putting values of a, b, c in formula

$$\begin{aligned}(-4\sqrt{3})^2 - 4 \times 3 \times 4 \\[0.5em] = 48 - 48 \\[0.5em] = 0\end{aligned}$$

Hence the given equation has two equal real roots.

(ii) The given equation is $x^2 - \frac{1}{2}x + 4 = 0$.
Comparing it with ax² + bx + c = 0, we get
a = 1 , b = $-\frac{1}{2}$ , c = 4
∴ Discriminant = b² - 4ac
Putting values of a, b, c in formula

$$\begin{aligned}\Big(-\frac{1}{2}\Big)^2 - 4 \times 1 \times 4 \\[1em] = \frac{1}{4} - 16 \\[1em] = \frac{1 - 64}{4} \text{ Taking L.C.M. } \\[1em] = -\frac{63}{4} \lt 0\end{aligned}$$

Hence the given equation has no real roots.

(iii)The given equation is -2x² + x + 1 = 0.
Comparing it with ax² + bx + c = 0, we get
a = -2 , b = 1 , c = 1
∴ Discriminant = b² - 4ac
Putting values of a, b, c in formula

$$\begin{aligned}(1)^2 - 4 \times -2 \times 1 \\[0.5em] = 1 + 8 \\[0.5em] = 9 \gt 0 \\[0.5em]\end{aligned}$$

Hence, the given equation has two distinct real roots.

(iv) The given equation is $2\sqrt{3}x^2 - 5x + \sqrt{3} = 0$.
Comparing it with ax² + bx + c = 0, we get
a = $2\sqrt{3}$ , b = -5 , c = $\sqrt{3}$
∴ Discriminant = b² - 4ac
Putting values of a, b, c in formula

$$\begin{aligned}(-5)^2 - 4 \times 2\sqrt{3} \times \sqrt{3} \\[0.5em] = 25 - 24 \\[0.5em] = 1 \gt 0\end{aligned}$$

Hence, the given equation has two distinct real roots.

(i) The given equation is $x^2 - \frac{1}{2}x - \frac{1}{2} = 0$

Comparing it with ax² + bx + c = 0, we get
a = 1 , b = $-\frac{1}{2}$ , c = $-\frac{1}{2}$

∴ Discriminant = b² - 4ac

Putting values of a, b, c in formula

$$\begin{aligned}\Big(-\frac{1}{2}\Big)^2 - 4 \times 1 \times -\frac{1}{2} \\[1em] = \frac{1}{4} + 2 \\[1em] = \frac{1 + 8}{4} \\[1em] = \frac{9}{4} \\[1em]\end{aligned}$$

Discriminant = $\frac{9}{4}$
Since Discriminant > 0, hence the given equation has two distinct real roots.

The roots of the equation are given by:

$$\begin{aligned}x = \frac{-b ± \sqrt{b^2 - 4ac}}{2a} \\[1em] \Rightarrow x = \frac{-\Big(-\frac{1}{2}\Big) ± \sqrt{\Big(-\frac{1}{2}\Big)^2 - 4\times 1 \times -\frac{1}{2}}}{2 \times 1} \\[1em] \Rightarrow x = \frac{\frac{1}{2} ± \sqrt{\frac{1}{4} + 2}}{2} \\[1em] \Rightarrow x = \frac{\frac{1}{2} ± \sqrt{\frac{9}{4}}}{2} \\[1em] \Rightarrow x = \frac{\frac{1}{2} + \sqrt{\frac{9}{4}}}{2} \text{ or } \frac{\frac{1}{2} - \sqrt{\frac{9}{4}}}{2} \\[1em] \Rightarrow x = \frac{\frac{1}{2} + \frac{3}{2}}{2} \text{ or } \frac{\frac{1}{2} - \frac{3}{2}}{2} \\[1em] \Rightarrow x = \frac{\frac{4}{2}}{2} \text{ or } \frac{-\frac{2}{2}}{2} \\[1em] \Rightarrow x = \frac{4}{4} \text{ or } -\frac{2}{4} \\[1em] \Rightarrow x = 1 \text{ or } -\frac{1}{2}\end{aligned}$$

Hence roots of the given equations are 1 , $-\frac{1}{2}$.

(ii) The given equation is $x^2 - 2\sqrt{3}x - 1 = 0$
Comparing it with ax² + bx + c = 0, we get
a = 1 , b = $-2\sqrt{3}$ , c = -1
∴ Discriminant = b² - 4ac
Putting values of a, b, c in formula

$$\begin{aligned}(-2\sqrt{3})^2 - 4 \times 1 \times -1 \\[1em] = 12 + 4 \\[1em] = 16 \\[1em]\end{aligned}$$

Discriminant = 16 ,
Since Discriminant > 0, hence given equation have real and distinct roots.

The roots of the equation are given by

$$\begin{aligned}x = \frac{-b ± \sqrt{b^2 - 4ac}}{2a} \\[1em] \Rightarrow x = \frac{-(-2\sqrt{3}) ± \sqrt{(-2\sqrt{3})^2 - 4\times 1 \times -1}}{2 \times 1} \\[1em] \Rightarrow x = \frac{2\sqrt{3} ± \sqrt{12 + 4}}{2} \\[1em] \Rightarrow x = \frac{2\sqrt{3} ± \sqrt{16}}{2} \\[1em] \Rightarrow x = \frac{2\sqrt{3} + 4}{2} \text{ or } \frac{2\sqrt{3} - 4}{2} \\[1em] \Rightarrow x = \sqrt{3} + 2 \text{ or } \sqrt{3} - 2 \\[1em]\end{aligned}$$

Hence roots of the given equations are , $\sqrt{3} + 2 , \sqrt{3} - 2$ .

(i) The given equation is px² - 4x + 3 = 0

Comparing it with ax² + bx + c = 0, we get
a = p , b = -4 , c = 3

$$\begin{aligned}\therefore \text{Discriminant} = b^2 - 4ac \\[1em] = (-4)^2 - 4 \times p \times 3 \\[1em] = 16 - 12p\end{aligned}$$

For equal roots, discriminant = 0

$$\begin{aligned}\Rightarrow 16 - 12p = 0 \\[1em] \Rightarrow 16 = 12p \\[1em] \Rightarrow p = \frac{16}{12} \\[1em] p = \frac{4}{3}\end{aligned}$$

Hence the value of p is $\frac{4}{3}$ .

(ii) The given equation is x² + (p - 3)x + p = 0.

Comparing with ax² + bx + c = we obtain,
a = 1 , b = (p - 3) , c = p

$$\begin{aligned}\therefore \text{Discriminant} = b^2 - 4ac \\[1em] = (p - 3)^2 - 4 \times 1 \times p \\[1em] = p^2 + 9 - 6p - 4p \\[1em] = p^2 + 9 - 10p\end{aligned}$$

For equal roots, discriminant = 0

$$\begin{aligned}\Rightarrow p^2 + 9 - 10p = 0 \\[1em] \Rightarrow p^2 - 10p + 9 = 0 \\[1em] \Rightarrow p^2 - 9p - p + 9 = 0 \\[1em] \Rightarrow p(p - 9) - 1(p - 9) = 0 \\[1em] \Rightarrow (p - 1)(p - 9) = 0 \\[1em] \Rightarrow (p - 1) = 0 \text{ or } p - 9 = 0 \\[1em] \Rightarrow p = 1 \text{ or } p = 9 .\end{aligned}$$

Hence the value of p is 1, 9.

(i) The given equation is x² + 4kx + (k² - k + 2) = 0.

Comparing with ax² + bx + c = we obtain,
a = 1 , b = 4k , c = (k² - k + 2)

$$\begin{aligned}\therefore \text{Discriminant} = b^2 - 4ac \\[1em] = (4k)^2 - 4 \times 1 \times (k^2 - k + 2) \\[1em] = 16k^2 - 4(k^2 - k + 2) \\[1em] = 16k^2 - 4k^2 + 4k - 8 \\[1em] = 12k^2 + 4k - 8\end{aligned}$$

For equal roots, discriminant = 0

$$\begin{aligned}\Rightarrow 12k^2 + 4k - 8 = 0 \\[1em] \Rightarrow 12k^2 + 12k - 8k - 8 = 0 \\[1em] \Rightarrow 12k(k + 1) - 8(k + 1) = 0 \\[1em] \Rightarrow (k + 1) - (12k - 8) = 0 \\[1em] \Rightarrow (k + 1)(12k - 8) = 0 \\[1em] \Rightarrow k + 1 = 0 \text{ or } 12k - 8 = 0 \\[1em] \Rightarrow k = -1 \text{ or } k = \frac{8}{12} \\[1em] \Rightarrow k = -1 \text{ or } k = \frac{2}{3}\end{aligned}$$

Hence, the value of k is -1, $\frac{2}{3}$.

(ii) The given equation is (k - 4)x² + 2(k - 4)x + 4 = 0.

Comparing with ax² + bx + c = we obtain,
a = k - 4 , b = 2(k - 4) , c = 4

$$\begin{aligned}\therefore \text{Discriminant} = b^2 - 4ac \\[1em] = (2k - 8)^2 - 4 \times k - 4 \times 4 \\[1em] = 4k^2 + 64 - 32k - 16(k - 4) \\[1em] = 4k^2 - 32k - 16k + 64 + 64 \\[1em] = 4k^2 - 48k + 128\end{aligned}$$

For equal roots, discriminant = 0

$$\begin{aligned}\Rightarrow 4k^2 - 48k + 128 = 0 \\[1em] \Rightarrow 4(k^2 - 12k + 32) = 0 \\[1em] \Rightarrow k^2 - 12k + 32 = 0 \\[1em] \Rightarrow k^2 - 8k - 4k + 32 = 0 \\[1em] \Rightarrow k(k - 8) - 4(k - 8) = 0 \\[1em] \Rightarrow k - 4 = 0 \text{ or } k - 8 = 0 \\[1em] \Rightarrow k = 4 \text{ or } k = 8 \\[1em]\end{aligned}$$

k ≠ 4 , as that will make a = (k - 4) = 0 and thus roots will become = ∞ .

Hence, the value of k is 8.

(i) The given equation is (3m + 1)x² + 2(m + 1)x + m = 0.

Comparing with ax² + bx + c = we obtain,
a = 3m + 1 , b = 2(m + 1) , c = m

$$\begin{aligned}\therefore \text{Discriminant} = b^2 - 4ac \\[0.5em] = (2m + 2)^2 - 4 \times 3m + 1 \times m \\[0.5em] = 4m^2 + 4 + 8m - 4m(3m + 1) \\[0.5em] = 4m^2 + 4 + 8m - 12m^2 - 4m \\[0.5em] = 4m^2 - 12m^2 + 8m - 4m + 4 \\[0.5em] = -8m^2 + 4m + 4 \\[0.5em]\end{aligned}$$

For equal roots, discriminant = 0

$$\begin{aligned}\Rightarrow -8m^2 + 4m + 4 = 0 \\[0.5em] \Rightarrow -4(2m^2 - m - 1) = 0 \\[0.5em] \Rightarrow 2m^2 - m - 1 = 0 \\[0.5em] \Rightarrow 2m^2 - 2m + m - 1 = 0 \\[0.5em] \Rightarrow 2m(m - 1) + 1(m - 1) = 0 \\[0.5em] \Rightarrow (2m + 1)(m - 1) = 0 \\[0.5em] \Rightarrow 2m + 1 = 0 \text{ or } m - 1 = 0 \\[0.5em] \Rightarrow m = -\frac{1}{2} \text{ or } m = 1\end{aligned}$$

Hence, the value of m is $-\frac{1}{2}$ and 1.

(ii) The given equation is x² + 2(m - 1)x + (m + 5) = 0.

Comparing with ax² + bx + c = we obtain,
a = 1 , b = 2(m - 1) , c = m + 5

$$\begin{aligned}\therefore \text{Discriminant} = b^2 - 4ac \\[0.5em] = (2m - 2)^2 - 4 \times 1 \times m + 5 \\[0.5em] = 4m^2 + 4 - 8m - 4(m + 5) \\[0.5em] = 4m^2 + 4 - 8m - 4m - 20 \\[0.5em] = 4m^2 - 12m - 16 \\[0.5em] = 4m^2 - 12m - 16 \\[0.5em]\end{aligned}$$

For equal roots, discriminant = 0

$$\begin{aligned}\Rightarrow 4m^2 - 12m - 16 = 0 \\[0.5em] \Rightarrow 4(m^2 - 3m - 4) = 0 \\[0.5em] \Rightarrow m^2 - 3m - 4 = 0 \\[0.5em] \Rightarrow m^2 - 4m + m - 4 = 0 \\[0.5em] \Rightarrow m(m - 4) + 1(m - 4) = 0 \\[0.5em] \Rightarrow (m - 4)(m + 1) = 0 \\[0.5em] \Rightarrow m - 4 = 0 \text{ or } m + 1 = 0 \\[0.5em] \Rightarrow m = 4 \text{ or } m = -1\end{aligned}$$

Hence, the value of m is 4, -1.

(i) The given equation is 9x² + kx + 1 = 0.

Comparing with ax² + bx + c = we obtain,
a = 9 , b = k , c = 1

$$\begin{aligned}\therefore \text{Discriminant} = b^2 - 4ac \\[0.5em] = (k)^2 - 4 \times 9 \times 1 \\[0.5em] = k^2 - 36 \\[0.5em]\end{aligned}$$

For equal roots, discriminant = 0

$$\begin{aligned}\Rightarrow k^2 - 36 = 0 \\[0.5em] \Rightarrow k^2 - (6)^2 = 0 \\[0.5em] \Rightarrow (k + 6)(k - 6) = 0 \\[0.5em] \Rightarrow k + 6 = 0 \text{ or } k - 6 = 0 \\[0.5em] \Rightarrow k = -6 \text{ or } k = 6 \\[0.5em]\end{aligned}$$

When k = 6 the equation becomes 9x² + 6x + 1 = 0 so the roots are :

The roots of the equation are given by

$$\begin{aligned}x = \frac{-b ± \sqrt{b^2 - 4ac}}{2a} \\[1em] \Rightarrow x = \frac{-(6) ± \sqrt{(6)^2 - 4\times 9 \times 1}}{2 \times 9} \\[1em] \Rightarrow x = \frac{-6 ± \sqrt{36 - 36}}{18} \\[1em] \Rightarrow x = \frac{-6 ± \sqrt{0}}{18} \\[1em] \Rightarrow x = \frac{-6 +0}{18} \text{ or } \frac{-6 - 0}{18} \\[1em] \Rightarrow x = -\frac{1}{3} \text{ or } -\frac{1}{3} \\[1em]\end{aligned}$$

When k = -6 the equation becomes 9x² - 6x + 1 = 0 so the roots are :

The roots of the equation are given by

$$\begin{aligned}x = \frac{-b ± \sqrt{b^2 - 4ac}}{2a} \\[1em] \Rightarrow x = \frac{-(-6) ± \sqrt{(-6)^2 - 4\times 9 \times 1}}{2 \times 9} \\[1em] \Rightarrow x = \frac{6 ± \sqrt{36 - 36}}{18} \\[1em] \Rightarrow x = \frac{6 ± \sqrt{0}}{18} \\[1em] \Rightarrow x = \frac{6 +0}{18} \text{ or } \frac{6 - 0}{18} \\[1em] \Rightarrow x = \frac{1}{3} \text{ or } \frac{1}{3} \\[1em]\end{aligned}$$

Hence, the values of k are 6, -6 ; when k = 6, roots are $-\frac{1}{3}, -\frac{1}{3}$ and when k = -6, roots are $\frac{1}{3}, \frac{1}{3}$.

(ii) The given equation is x² - 2kx + 7k - 12 = 0.

Comparing with ax² + bx + c = we obtain,
a = 1 , b = -2k , c = 7k - 12

$$\begin{aligned}\therefore \text{Discriminant} = b^2 - 4ac \\[0.5em] = (-2k)^2 - 4 \times 1 \times (7k - 12) \\[0.5em] = 4k^2 - 4(7k - 12) \\[0.5em] = 4k^2 - 28k + 48\end{aligned}$$

For equal roots, discriminant = 0

$$\begin{aligned}\Rightarrow 4k^2 - 28k + 48 = 0 \\[0.5em] \Rightarrow 4(k^2 - 7k + 12) = 0 \\[0.5em] \Rightarrow k^2 - 7k + 12 = 0 \\[0.5em] \Rightarrow k^2 - 4k - 3k + 12 = 0 \\[0.5em] \Rightarrow k(k - 4) - 3(k - 4) = 0 \\[0.5em] \Rightarrow (k - 3)(k - 4) = 0 \\[0.5em] \Rightarrow k - 3 = 0 \text{ or } k - 4 = 0 \\[0.5em] \Rightarrow k = 3 \text{ or } k = 4 \\[0.5em]\end{aligned}$$

When k = 3 the equation becomes x² - 6x + 9 = 0 so the roots are :

The roots of the equation are given by

$$\begin{aligned}x = \frac{-b ± \sqrt{b^2 - 4ac}}{2a} \\[1em] \Rightarrow x = \frac{-(-6) ± \sqrt{(-6)^2 - 4\times 1 \times 9}}{2 \times 1} \\[1em] \Rightarrow x = \frac{6 ± \sqrt{36 - 36}}{2} \\[1em] \Rightarrow x = \frac{6 ± \sqrt{0}}{12} \\[1em] \Rightarrow x = \frac{6 +0}{2} \text{ or } \frac{6 - 0}{2} \\[1em] \Rightarrow x = 3 \text{ or } 3 \\[1em]\end{aligned}$$

When k = 4 the equation becomes x² - 8x + 16 = 0 so the roots are :

The roots of the equation are given by

$$\begin{aligned}x = \frac{-b ± \sqrt{b^2 - 4ac}}{2a} \\[1em] \Rightarrow x = \frac{-(-8) ± \sqrt{(-8)^2 - 4\times 1 \times 16}}{2 \times 1} \\[1em] \Rightarrow x = \frac{8 ± \sqrt{64 - 64}}{2} \\[1em] \Rightarrow x = \frac{8 ± \sqrt{0}}{2} \\[1em] \Rightarrow x = \frac{8 + 0}{2} \text{ or } \frac{8 - 0}{2} \\[1em] \Rightarrow x = 4 \text{ or } 4 \\[1em]\end{aligned}$$

Hence the values of k are 3, 4 ; when k = 3, roots are 3, 3 and when k = 4, roots are 4, 4.

The given equation is (2p + 1)x² - (7P + 2)x + (7p - 3) = 0.

Comparing with ax² + bx + c = we obtain,
a = 2p + 1 , b = -(7p + 2) , c = 7p - 3

$$\begin{aligned}\therefore \text{Discriminant} = b^2 - 4ac \\[1em] = (-(7p + 2))^2 - 4 \times (2p + 1) \times (7p - 3) \\[1em] = 49p^2 + 4 + 28p - 4(2p + 1)(7p - 3) \\[1em] = 49p^2 + 4 + 28p - 4(14p^2 - 6p + 7p - 3) \\[1em] = 49p^2 - 56p^2 + 28p - 4p + 4 + 12 \\[1em] = -7p^2 + 24p + 16\end{aligned}$$

For equal roots, discriminant = 0

$$\begin{aligned}\Rightarrow -7p^2 + 24p + 16 \\[1em] \Rightarrow 7p^2 - 24p - 16 = 0 \text{ (Multiplying the equation by -1) } \\[1em] \Rightarrow 7p^2 - 28p + 4p - 16 = 0 \\[1em] \Rightarrow 7p(p - 4) + 4(p - 4) = 0 \\[1em] \Rightarrow (7p + 4)(p - 4) = 0 \\[1em] \Rightarrow 7p + 4 = 0 \text{ or } p - 4 = 0 \\[1em] \Rightarrow 7p = -4 \text{ or } p = 4 \\[1em] \Rightarrow p = -\frac{4}{7} \text{ or } p = 4 \\[1em]\end{aligned}$$

When p = $-\frac{4}{7}$ the equation becomes $(2 \times -\frac{4}{7} + 1)x^2 - (7 \times -\frac{4}{7} + 2)x + (7 \times -\frac{4}{7} - 3) = 0$ so the roots are :

$$\begin{aligned}\Rightarrow -\frac{1}{7}x^2 + 2x - 7 = 0 \\[1em] \Rightarrow x^2 - 14x + 49 = 0 \text{ (On multiplying complete equation by -7) } \\[1em] \Rightarrow x^2 - 14x + 49 = 0 \\[1em] \Rightarrow x^2 - 7x - 7x + 49 = 0 \\[1em] \Rightarrow x(x - 7) - 7(x - 7) = 0 \\[1em] \Rightarrow (x - 7)(x - 7) = 0 \\[1em] \Rightarrow x - 7 = 0 \text{ or } x - 7 = 0 \\[1em] \Rightarrow x = 7 \text{ or } x = 7 \\[1em]\end{aligned}$$

When p = 4 the equation becomes $(2 \times 4 + 1)x^2 - (7 \times 4 + 2)x + (7 \times 4 - 3) = 0$ so the roots are :

$$\begin{aligned}\Rightarrow 9x^2 - 30x + 25 = 0 \\[1em] \Rightarrow 9x^2 - 15x - 15x + 25 = 0 \\[1em] \Rightarrow 3x(3x - 5) - 5(3x - 5) = 0 \\[1em] \Rightarrow (3x - 5)(3x - 5) = 0 \\[1em] \Rightarrow 3x - 5 = 0 \text{ or } 3x - 5 = 0 \\[1em] \Rightarrow x = \frac{5}{3} \text{ or } x = \frac{5}{3} \\[1em]\end{aligned}$$

(Ans.) 4, $-\frac{4}{7}$ ; when p = $-\frac{4}{7}$, roots are 7, 7 and when p = 4, roots are $\frac{5}{3} , \frac{5}{3}.$

For real roots, Discriminant ≥ 0

or , b² - 4ac ≥ 0

The above equation is 2x² + 3x + p = 0
Comparing with ax² + bx + c = we obtain,
a = 2 , b = 3 , c = p

Putting values in b² - 4ac ≥ 0 we get,

$$\begin{aligned}= 3^2 - 4 \times 2 \times p \ge 0 \\[0.5em] = 9 - 8p \ge 0 \\[0.5em] \Rightarrow 9 \ge 8p \\[0.5em] \Rightarrow 8p \le 9 \\[0.5em] \Rightarrow p \le \frac{9}{8} \\[0.5em]\end{aligned}$$

Hence the value of p is ≤ $\frac{9}{8}$.

For real roots, Discriminant ≥ 0

or , b² - 4ac ≥ 0

The above equation is x² + kx + 4 = 0
Comparing with ax² + bx + c = we obtain,
a = 1 , b = k , c = 4

Putting values in b² - 4ac ≥ 0 we get,

$$\begin{aligned}= k^2 - 4 \times 1 \times 4 \ge 0 \\[0.5em] = k^2 - 16 \ge 0 \\[0.5em] \Rightarrow k^2 - 4^2 \ge 0 \\[0.5em] \Rightarrow (k - 4)(k + 4) \ge 0 \\[0.5em] \Rightarrow k \ge 4 \\[0.5em]\end{aligned}$$

Hence least positive value for which equation has real roots is 4 .

For real roots,

Discriminant ≥ 0

or, b² - 4ac ≥ 0

The above equation is 3x² - px + 5 = 0
Comparing with ax² + bx + c = we obtain,
a = 3 , b = -p , c = 5

Putting values in b² - 4ac ≥ 0 we get,

$$\begin{aligned}= (-p)^2 - 4 \times 3 \times 5 \ge 0 \\[1em] = p^2 - 60 \ge 0 \\[1em] \Rightarrow p^2 - (\sqrt{60})^2 \ge 0 \\[1em] \Rightarrow (p + \sqrt{60})(p -\sqrt{60}) \ge 0 \\[1em] \Rightarrow (p + 2\sqrt{15})(p - 2\sqrt{15}) \ge 0 \\[1em] \text{(Using) } (a + b)(a - b) \ge 0 = a \ge b \text{ or } a \le -b \text{ we get, } \\[1em] \Rightarrow p \ge 2\sqrt{15} \text{ or } p \le -2\sqrt{15} \\[1em]\end{aligned}$$

Hence the values of p are $p \le -2\sqrt{15} \text{ or } p\ge 2\sqrt{15}$ .

Let the required numbers be = x , x + 1

Given, sum of squares of the numbers = 61

$$\begin{aligned}\Rightarrow x^2 + (x + 1)^2 = 61 \\[1em] \Rightarrow x^2 + (x^2 + 1 + 2x) = 61 \\[1em] \Rightarrow x^2 + x^2 + 1 + 2x = 61 \\[1em] \Rightarrow 2x^2 + 2x + 1 = 61 \\[1em] \Rightarrow 2x^2 + 2x + 1 - 61 = 0 \\[1em] \Rightarrow 2x^2 + 2x - 60 = 0 \\[1em] \Rightarrow 2(x^2 + x - 30) = 0 \\[1em] \Rightarrow x^2 + x - 30 = 0 \\[1em] \Rightarrow x^2 + 6x - 5x - 30 = 0 \\[1em] \Rightarrow x(x + 6) - 5(x + 6) = 0 \\[1em] \Rightarrow (x + 6)(x - 5) = 0 \text{ (Factorising left side) } \\[1em] \Rightarrow x + 6 = 0 \text{ or } x - 5 = 0 \text{ (Zero-product rule) } \\[1em] \Rightarrow x = -6 \text { or } x = 5\end{aligned}$$

Since the numbers are natural number so x ≠ -6.

∴ x = 5 , x + 1 = 6.

Hence, the required natural numbers are 5 , 6.

Let the required numbers be = x , x + 1

Given, sum of squares of the numbers = 61

$$\begin{aligned}\Rightarrow x^2 + (x + 1)^2 = 61 \\[1em] \Rightarrow x^2 + (x^2 + 1 + 2x) = 61 \\[1em] \Rightarrow x^2 + x^2 + 1 + 2x = 61 \\[1em] \Rightarrow 2x^2 + 2x + 1 = 61 \\[1em] \Rightarrow 2x^2 + 2x + 1 - 61 = 0 \\[1em] \Rightarrow 2x^2 + 2x - 60 = 0 \\[1em] \Rightarrow 2(x^2 + x - 30) = 0 \\[1em] \Rightarrow x^2 + x - 30 = 0 \\[1em] \Rightarrow x^2 + 6x - 5x - 30 = 0 \\[1em] \Rightarrow x(x + 6) - 5(x + 6) = 0 \\[1em] \Rightarrow (x + 6)(x - 5) = 0 \text{ (Factorising left side) } \\[1em] \Rightarrow x + 6 = 0 \text{ or } x - 5 = 0 \text{ (Zero-product rule) } \\[1em] \Rightarrow x = -6 \text { or } x = 5\end{aligned}$$

∴ When x = -6 , x + 1 = -5 and when x = 5 , x + 1 = 6.

Hence, required integers are 5,6 or -6,-5

Let the required two positive consecutive even integers be x , x + 2

Given, product of two consecutive even integers = 288

$$\begin{aligned}\Rightarrow x(x + 2) = 288 \\[1em] \Rightarrow x^2 + 2x = 288 \\[1em] \Rightarrow x^2 + 2x - 288 = 0 \\[1em] \Rightarrow x^2 + 18x - 16x - 288 = 0 \\[1em] \Rightarrow x(x + 18) - 16(x + 18) = 0 \\[1em] \Rightarrow (x - 16)(x + 18) = 0 \text{ (Factorising left side) } \\[1em] \Rightarrow x - 16 = 0 \text{ or } x + 18 = 0 \text{ (Zero-product rule) } \\[1em] \Rightarrow x = 16 \text{ or } x = -18\end{aligned}$$

Since the numbers are natural number so x ≠ -18.

∴ x = 16 , x + 2 = 18.

Hence, required integers are 16, 18.

Let the required two consecutive even integers be x , x + 2

Given, product of two consecutive even integers = 224

$$\begin{aligned}\Rightarrow x(x + 2) = 224 \\[1em] \Rightarrow x^2 + 2x = 224 \\[1em] \Rightarrow x^2 + 2x - 224 = 0 \\[1em] \Rightarrow x^2 + 16x - 14x - 224 = 0 \\[1em] \Rightarrow x(x + 16) - 14(x + 16) = 0 \\[1em] \Rightarrow (x + 16)(x - 14) = 0 \text{ (Factorising left side) } \\[1em] \Rightarrow x + 16 = 0 \text{ or } x - 14 = 0. \\[1em] \Rightarrow x = -16 \text{ or } x = 14 \\[1em]\end{aligned}$$

∴ When x = 14 , x + 2 = 16 and when x = -16 , x + 2 = -14.

Hence, required integers are 14 , 16 or -16, -14 .

Let the required two consecutive even integers be x , x + 2

Given, sum of squares of two consecutive even natural numbers = 340

$$\begin{aligned}\Rightarrow x^2 + (x + 2)^2 = 340 \\[1em] \Rightarrow x^2 + x^2 + 4 + 4x = 340 \\[1em] \Rightarrow 2x^2 + 4x + 4 -340 = 0 \\[1em] \Rightarrow 2x^2 + 4x - 336 = 0 \\[1em] \Rightarrow 2(x^2 + 2x - 168) = 0 \\[1em] \Rightarrow x^2 + 2x - 168 = 0 \\[1em] \Rightarrow x^2 + 14x - 12x - 168 = 0 \\[1em] \Rightarrow x(x + 14) - 12(x + 14) = 0 \\[1em] \Rightarrow (x - 12)(x + 14) = 0 \text{ (Factorising left side) } \\[1em] \Rightarrow x - 12 = 0 \text{ or } x + 14 = 0 \text{ (Zero-product rule) } \\[1em] \Rightarrow x = 12 \text{ or } x = -14\end{aligned}$$

Since the numbers are natural number so x ≠ -14

∴ x = 12 , x + 2 = 14

Hence, required natural numbers are 12 , 14 .

Let the required two consecutive odd integers be x , x + 2

Given, sum of squares of two consecutive odd integers = 394

$$\begin{aligned}\Rightarrow x^2 + (x + 2)^2 = 394 \\[1em] \Rightarrow x^2 + x^2 + 4 + 4x = 394 \\[1em] \Rightarrow 2x^2 + 4x + 4 -394 = 0 \\[1em] \Rightarrow 2x^2 + 4x - 390 = 0 \\[1em] \Rightarrow 2(x^2 + 2x - 195) = 0 \\[1em] \Rightarrow x^2 + 2x - 195 = 0 \\[1em] \Rightarrow x^2 + 15x - 13x - 195 = 0 \\[1em] \Rightarrow x(x + 15) - 13(x + 15) = 0 \\[1em] \Rightarrow (x + 15)(x - 13) = 0 \text{ (Factorising left side) } \\[1em] \Rightarrow x + 15 = 0 \text{ or } x - 13 = 0 \text{ (Zero-product rule) } \\[1em] \Rightarrow x = -15 \text{ or } x = 13\end{aligned}$$

∴ When x = -15 , x + 2 = -13 and when x = 13 , x + 2 = 15.

Hence required integers are -15 , -13 or 13, 15 .

Let the first number be x.

Since the sum of two numbers is 9, so other number is 9 - x.

Given, the sum of squares of numbers = 41

⇒ x² + (9 - x)² = 41

⇒ x² + x² + 81 - 18x = 41

⇒ 2x² - 18x + 81 - 41 = 0

⇒ 2x² - 18x + 40 = 0

⇒ 2(x² - 9x + 20) = 0

⇒ x² - 9x + 20 = 0

⇒ x² - 4x - 5x + 20 = 0

⇒ x(x - 4) - 5(x - 4) = 0

⇒ (x - 4)(x - 5) = 0

⇒ x - 4 = 0 or x - 5 = 0

⇒ x = 4 or x = 5

∴ x = 5, 9 - x = 4

∴ x = 4, 9 - x = 5

Hence, the numbers are 4 and 5.

Let first number be x.

Since the difference of two numbers is 7, so other number is x + 7.

Given, the products of numbers = 450

⇒ x(x + 7) = 450

⇒ x² + 7x = 450

⇒ x² + 7x - 450 = 0

⇒ x² + 25x - 18x - 450 = 0

⇒ x(x + 25) - 18(x + 25) = 0

⇒ (x + 25)(x - 18) = 0

⇒ x + 25 = 0 or x - 18 = 0

⇒ x = -25 or x = 18.

Since, the numbers are natural numbers,

∴ x ≠ -25

∴ x = 18, x + 7 = 25

Hence, the numbers are 18 and 25.

Let the number be x

Given, 5 times the number = 3 less than twice the square of the number

$$\begin{aligned}\Rightarrow 5x = 2x^2 - 3 \\[0.5em] \Rightarrow 5x - 2x^2 + 3 = 0 \\[0.5em] \Rightarrow 2x^2 - 5x - 3 = 0 \text{ (on multiplying the equation by -1) } \\[0.5em] \Rightarrow 2x^2 - 6x + x - 3 = 0 \\[0.5em] \Rightarrow 2x(x - 3) + 1(x - 3) = 0 \\[0.5em] \Rightarrow (2x + 1)(x - 3) = 0 \text{ (Factorising left side) } \\[0.5em] \Rightarrow 2x + 1 = 0 \text{ or } x - 3 = 0 \text{ (Zero-product rule) } \\[0.5em] \Rightarrow x = -\frac{1}{2} \text{ or } x = 3 \\[0.5em]\end{aligned}$$

Since the number is a whole number x ≠ $-\frac{1}{2}$

∴ x = 3

Hence, the required whole number is 3.

Let the first number be x

Since, the sum of two numbers is 8, so other number is 8 - x.

Given, the difference of reciprocal of numbers = $\frac{2}{15}$

$$\begin{aligned}\Rightarrow \frac{1}{x} - \frac{1}{8 - x} = \frac{2}{15} \\[0.5em] \Rightarrow \frac{8 - x - x}{x(8 - x)} = \frac{2}{15} \text{ (On taking L.C.M.) } \\[0.5em] \Rightarrow 15(8 - 2x) = 2x(8 - x) \text{ (On cross multiplication ) } \\[0.5em] \Rightarrow 120 - 30x = 16x - 2x^2 \\[0.5em] \Rightarrow 120 - 30x - 16x + 2x^2 = 0 \\[0.5em] \Rightarrow 2x^2 - 46x + 120 = 0 \\[0.5em] \Rightarrow 2(x^2 - 23x + 60) = 0 \\[0.5em] \Rightarrow x^2 - 20x - 3x + 60 = 0 \\[0.5em] \Rightarrow x(x - 20) - 3(x - 20) = 0 \\[0.5em] \Rightarrow (x - 3)(x - 20) = 0 \text{ (Factorising left side) } \\[0.5em] \Rightarrow x - 3 = 0 \text{ (or) } x - 20 = 0 \text{ (Zero-product rule) } \\[0.5em] \Rightarrow x = 3 \text{ or } x = 20.\end{aligned}$$

If x = 20 , 8 - x = -12 , Since both are natural numbers hence x ≠ 20.

∴ x = 3 , 8 - x = 5

Hence, the required natural numbers are 3 , 5.

Let the larger number be x .

Given, the square of the smaller number is 4 times the larger number.

Hence, square of smaller number = 4x

Given, the difference of squares of two numbers is = 45

$$\begin{aligned}\Rightarrow x^2 - 4x = 45 \\[0.5em] \Rightarrow x^2 - 4x - 45 = 0 \\[0.5em] \Rightarrow x^2 - 9x + 5x - 45 = 0 \\[0.5em] \Rightarrow x(x - 9) + 5(x - 9) = 0 \\[0.5em] \Rightarrow (x + 5)(x - 9) = 0 \\[0.5em] \Rightarrow x + 5 = 0 \text{ or } x - 9 = 0 \\[0.5em] \Rightarrow x = -5 \text{ or } x = 9\end{aligned}$$

If x = -5 , $\sqrt{4x} = \sqrt{-20}$ , which is not valid as there is no real value of square root of a negative value.

∴ x = 9 , $\sqrt{4x} = \sqrt{36} = 6 \text{ or } -6$.

Hence, the two numbers are 9, 6 or 9, -6.

Let the numbers be x , x + 1 , x + 2.

Given, sum of the square of the first and the product of the other two is = 154

$$\begin{aligned}\Rightarrow x^2 + (x + 1)(x + 2) = 154 \\[1em] \Rightarrow x^2 + (x^2 + 2x + x + 2) = 154 \\[1em] \Rightarrow 2x^2 + 3x + 2 - 154 = 0 \\[1em] \Rightarrow 2x^2 + 3x - 152 = 0 \\[1em] \Rightarrow 2x^2 + 19x - 16x - 152 = 0 \\[1em] \Rightarrow x(2x + 19) - 8(2x + 19) = 0 \\[1em] \Rightarrow (x - 8)(2x + 19) = 0 \\[1em] \Rightarrow x - 8 = 0 \text{ or } 2x + 19 = 0 \\[1em] x = 8 \text{ or } x = -\frac{19}{2}\end{aligned}$$

Since the integers are positive hence , x ≠ $-\frac{19}{2}$

∴ x = 8 , x + 1 = 9 , x + 2 = 10.

Hence , the required numbers are 8, 9, 10.

Let the required numbers be x, x + 2, x + 4.

Given, the sum of squares of these numbers = 308

$$\begin{aligned}\Rightarrow x^2 + (x + 2)^2 + (x + 4)^2 = 308 \\[1em] \Rightarrow x^2 + x^2 + 4 + 4x + x^2 + 16 + 8x = 308 \\[1em] \Rightarrow 3x^2 + 12x + 20 = 308 \\[1em] \Rightarrow 3x^2 + 12x - 288 = 0 \\[1em] \Rightarrow 3(x^2 + 4x - 96) = 0 \\[1em] \Rightarrow x^2 + 4x - 96 = 0 \\[1em] \Rightarrow x^2 + 12x - 8x - 96 = 0 \\[1em] \Rightarrow x(x + 12) - 8(x + 12) = 0 \\[1em] \Rightarrow (x - 8)(x + 12) = 0 \\[1em] \Rightarrow x - 8 = 0 \text{ or } x + 12 = 0 \\[1em] x = 8 \text{ or } x = -12\end{aligned}$$

Since the numbers are natural hence , x ≠ -12

∴ x = 8 , x + 2 = 10 , x + 4 = 12.

Hence , the required numbers are 8, 10, 12.

Let the required numbers be x , x + 2 , x + 4.

Given, the sum of squares of these numbers = 308

$$\begin{aligned}\Rightarrow x^2 + (x + 2)^2 + (x + 4)^2 = 83 \\[1em] \Rightarrow x^2 + x^2 + 4 + 4x + x^2 + 16 + 8x = 83 \\[1em] \Rightarrow 3x^2 + 12x + 20 = 83 \\[1em] \Rightarrow 3x^2 + 12x - 63 = 0 \\[1em] \Rightarrow 3(x^2 + 4x - 21) = 0 \\[1em] \Rightarrow x^2 + 4x - 21 = 0 \\[1em] \Rightarrow x^2 + 7x - 3x - 21 = 0 \\[1em] \Rightarrow x(x + 7) - 3(x + 7) = 0 \\[1em] \Rightarrow (x + 7)(x - 3) = 0 \\[1em] \Rightarrow x + 7 = 0 \text{ or } x - 3 = 0 \\[1em] x = -7 \text{ or } x = 3\end{aligned}$$

∴ When x = -7 , x + 2 = -5 , x + 4 = -3 and when x = 3 , x + 2 = 5 , x + 4 = 7.

Hence , the required numbers are -7, -5, -3 and 3, 5, 7.

Let the numerator of the fraction be x

Given, the denominator is greater than numerator by 3 hence, denominator = x + 3

Fraction = $\frac{x}{x + 3}$

Given, if 1 is subtracted from both the numerator and denominator , the fraction is decreased by $\frac{1}{14}$

$$\begin{aligned}\Rightarrow \frac{x}{x + 3} - \frac{x - 1}{x + 3 - 1} = \frac{1}{14} \\[1em] \Rightarrow \frac{x}{x + 3} - \frac{x - 1}{x + 2} = \frac{1}{14} \\[1em] \Rightarrow \frac{x(x + 2) - (x - 1)(x + 3)}{(x + 3)(x + 2)} = \frac{1}{14} \text{ (On taking L.C.M) } \\[1em] \Rightarrow \frac{x^2 + 2x - (x^2 + 3x - x - 3)}{(x^2 + 2x + 3x + 6)} = \frac{1}{14} \\[1em] \Rightarrow \frac{x^2 - x^2 + 2x - 2x + 3}{x^2 + 5x + 6} = \frac{1}{14} \\[1em] \Rightarrow 3 \times 14 = x^2 + 5x + 6 \text{ (Cross multiplying) } \\[1em] \Rightarrow x^2 + 5x + 6 = 42 \\[1em] \Rightarrow x^2 + 5x + 6 - 42 = 0 \\[1em] \Rightarrow x^2 + 5x - 36 = 0 \\[1em] \Rightarrow x^2 + 9x - 4x - 36 = 0 \\[1em] \Rightarrow x(x + 9) - 4(x + 9) = 0 \\[1em] \Rightarrow (x - 4)(x + 9) = 0 \\[1em] x = 4 \text{ or } x = -9 \\[1em]\end{aligned}$$

If x = -9 , Fraction = $\frac{x}{x + 3} = \frac{9}{6}$ In this case numerator > denominator which is incorrect according to the question hence, x ≠ -9.

∴ x = 4 , Fraction = $\frac{x}{x + 3} = \frac{4}{7}$

Hence, the fraction is $\frac{4}{7}$.

Let the denominator be = x so, numerator = 8 - x.

Fraction = $\frac{8 - x}{x}$

Given , if 2 is added to both the numerator and denominator, the fraction is increased by $\frac{4}{35}.$

$$\begin{aligned}\Rightarrow \frac{8 - x + 2}{x + 2} - \frac{8 - x}{x} = \frac{4}{35} \\[1em] \Rightarrow \frac{x(10 - x) - (x + 2)(8 - x)}{x(x + 2)} = \frac{4}{35} \text{ (On taking L.C.M) }\\[1em] \Rightarrow \frac{10x - x^2 - (8x - x^2 + 16 - 2x)}{x^2 + 2x} = \frac{4}{35} \\[1em] \Rightarrow 35(-x^2 + x^2 - 8x + 10x + 2x - 16) = 4(x^2 + 2x) \\[1em] \Rightarrow 35(4x - 16) = 4x^2 + 8x \\[1em] \Rightarrow 140x - 560 = 4x^2 + 8x \\[1em] \Rightarrow 4x^2 + 8x - 140x + 560 = 0 \\[1em] \Rightarrow 4x^2 - 132x + 560 = 0 \\[1em] \Rightarrow 4(x^2 - 33x + 140) = 0 \\[1em] \Rightarrow x^2 - 33x + 140 = 0 \\[1em] \Rightarrow x^2 - 28x - 5x + 140 = 0 \\[1em] \Rightarrow x(x - 28) - 5(x - 28) = 0 \\[1em] \Rightarrow (x - 28)(x - 5) = 0 \\[1em] \Rightarrow x - 28 = 0 \text{ or } x - 5 = 0 \\[1em] x = 28 \text{ or } x = 5\end{aligned}$$

If x = 28 , 8 - x = -20 which will make fraction = $-\frac{20}{28}$ negative hence, x ≠ 28

∴ x = 5 , 8 - x = 3 ,fraction = $\frac{3}{5}$

Hence, the fraction is $\frac{3}{5}$.

Let the unit's digit be x,
so, ten's digit be = x + 6.

Number = 10(x + 6) + x = 10x + 60 + x = 11x + 60

Given, product of digits is 27

$$\begin{aligned}\Rightarrow x(x + 6) = 27 \\[1em] \Rightarrow x^2 + 6x = 27 \\[1em] \Rightarrow x^2 + 6x - 27 = 0 \\[1em] \Rightarrow x^2 + 9x - 3x - 27 = 0 \\[1em] \Rightarrow x(x + 9) - 3(x + 9) = 0 \\[1em] \Rightarrow (x - 3)(x + 9) = 0 \\[1em] \Rightarrow x - 3 = 0 \text{ or } x + 9 = 0 \\[1em] \Rightarrow x = 3 \text{ or } x = -9\end{aligned}$$

When x = -9,
Number = 11x + 60
= 11(-9) + 60
= -99 + 60
= -39
In this case the ten's digit is smaller than unit's digit hence x ≠ -9

When x = 3,
Number = 11(x) + 60
= 11(3) + 60
= 33 + 60
= 93

Hence, the required number is 93 .

Let the digit at unit's place be x .

Since, the product of digits is 6 , it's ten's digit = $\frac{6}{x}$

$$\begin{aligned}\therefore \text{ Number } = 10 \times \frac{6}{x} + x \\[0.5em] = \frac{60}{x} + x \\[0.5em] = \frac{60 + x^2}{x} \\[0.5em] = \frac{x^2 + 60}{x} \\[0.5em]\end{aligned}$$

Given, if 9 is added to the number , the digits interchange their place

On interchanging the digits, number becomes = $10 \times x + \frac{6}{x}$

Acccording to given,

$$\begin{aligned}\Rightarrow \frac{x^2 + 60}{x} + 9 = 10 \times x + \frac{6}{x} \\[1em] \Rightarrow \frac{x^2 + 9x + 60}{x} = 10x + \frac{6}{x} \\[1em] \Rightarrow \frac{x^2 + 9x + 60}{x} = \frac{10x^2 + 6}{x} \\[1em] \Rightarrow x^2 + 9x + 60 =10x^2 + 6 \\[1em] \Rightarrow 10x^2 - x^2 - 9x + 6 - 60 = 0 \\[1em] \Rightarrow 9x^2 - 9x - 54 = 0 \\[1em] \Rightarrow 9(x^2 - x - 6) = 0 \\[1em] \Rightarrow x^2 - 3x + 2x - 6 = 0 \\[1em] \Rightarrow x(x - 3) + 2(x - 3) = 0 \\[1em] \Rightarrow (x + 2)(x - 3) = 0 \\[1em] \Rightarrow x + 2 = 0 \text{ or } x - 3 = 0 \\[1em] \Rightarrow x = -2 \text{ or } x = 3\end{aligned}$$

Since the number is positive hence x ≠ -2.

If x = 3 ,

$$\begin{aligned}\text{ Number } = \frac{x^2 + 60}{x} \\[1em] = \frac{3^2 + 60}{3} \\[1em] = \frac{69}{3} \\[1em] = 23\end{aligned}$$

Hence, the required number is 23.

Length of rectangle = x

Since the area of rectangle = 105 cm², breadth = $\frac{105}{x}$ cm.

∴ Perimeter = 2(length + breadth)

$$\begin{aligned}= 2(x + \frac{105}{x}) \\[0.5em] = 2(\frac{x^2 + 105}{x}) \\[0.5em]\end{aligned}$$

Given, Perimeter = 44 cm

$$\begin{aligned}\Rightarrow 2(\frac{x^2 + 105}{x}) = 44 \\[0.5em] \Rightarrow \frac{x^2 + 105}{x} = 22 \\[0.5em] \Rightarrow x^2 + 105 = 22x \\[0.5em] \Rightarrow x^2 - 22x + 105 = 0 \\[0.5em] \Rightarrow x^2 - 15x - 7x + 105 = 0 \\[0.5em] \Rightarrow x(x - 15) - 7(x - 15) = 0 \\[0.5em] \Rightarrow (x - 7)(x - 15) = 0 \\[0.5em] \Rightarrow x - 7 = 0 \text{ or } x - 15 = 0 \\[0.5em] \Rightarrow x = 7 \text{ or } x = 15\end{aligned}$$

Breadth = $\frac{105}{x}$ cm
Equation in x for perimeter, 2(x + $\frac{105}{x}$) = 44
Length = 7 cm , Breadth = 15 cm

Given,

Length of rectangular garden = 10 m

Breadth of rectangular garden = 16 m

Width of walk = x

So, length of garden and walk combined = (10 + x + x) m

Breadth of garden and walk combined = (16 + x + x) m

∴ Area of garden and walk combined = Length $\times$ Breadth = (10 + 2x)(16 + 2x) m²

Given, area of walk = 120m²

Area of walk = Area of combined - Area of garden

$$\begin{aligned}\Rightarrow (10 + 2x)(16 + 2x) - 10 \times 16 = 120 \\[0.5em] \Rightarrow 160 + 20x + 32x + 4x^2 - 160 = 120 \\[0.5em] \Rightarrow 4x^2 + 52x = 120 \\[0.5em] \Rightarrow 4x^2 + 52x - 120 = 0 \\[0.5em] \Rightarrow 4(x^2 + 13x - 30) = 0 \\[0.5em] \Rightarrow x^2 + 13x - 30 = 0 \\[0.5em] \Rightarrow x^2 + 15x - 2x - 30 = 0 \\[0.5em] \Rightarrow x(x + 15) - 2(x + 15) = 0 \\[0.5em] \Rightarrow (x - 2)(x + 15) = 0 \\[0.5em] \Rightarrow x - 2 \text{ or } x + 15 = 0 \\[0.5em] x = 2 \text{ or } x = -15\end{aligned}$$

Since, width cannot be negative hence x ≠ -15.

The equation in x = (10 + 2x)(16 + 2x) - 10 x 16 = 120.
Value of x = 2m.

Let breadth of rectangle be x meters

Since , length of rectangle exceeds breadth by 5 meters so, length = (x + 5) meters

Given, if breadth were doubled and the length reduced by 9m, the area of the rectangle would have increased by 140 m

∴ Area of new rectangle = Length $\times$ Breadth = (x + 5 - 9)(2x)

$$\begin{aligned}\Rightarrow (x + 5 - 9)(2x) - x(x + 5) = 140 \\[0.5em] \Rightarrow (x - 4)(2x) - (x^2 + 5x) = 140 \\[0.5em] \Rightarrow 2x^2 - 8x - x^2 - 5x = 140 \\[0.5em] \Rightarrow x^2 - 13x - 140 = 0 \\[0.5em] \Rightarrow x^2 - 20x + 7x - 140 = 0 \\[0.5em] \Rightarrow x(x - 20) + 7(x - 20) = 0 \\[0.5em] \Rightarrow (x + 7)(x - 20) = 0 \\[0.5em] \Rightarrow x + 7 = 0 \text{ or } x - 20 = 0 \\[0.5em] x = -7 \text{ or } x = 20\end{aligned}$$

Since, breadth cannot be negative hence x ≠ -7

∴ x = 20 and x + 5 = 25

Length of rectangle = 25 m , Breadth of rectangle = 20 m.

Length of rectangular plot = x meters

Perimeter = 2(length + breadth)

$$\begin{aligned}\Rightarrow 2(x + \text{Breadth}) = 180 \\[0.5em] \Rightarrow x + \text{Breadth} = 90 \\[0.5em] \Rightarrow \text{Breadth} = 90 - x\end{aligned}$$

Area of the rectangle = Length $\times$ Breadth

Given, Area of rectangle = 1800 m²

$$\begin{aligned}\Rightarrow x(90 - x) = 1800 \\[0.5em] \Rightarrow 90x - x^2 = 1800 \\[0.5em] \Rightarrow x^2 - 90x + 1800 = 0 \\[0.5em] \Rightarrow x^2 - 30x - 60x + 1800 = 0 \\[0.5em] \Rightarrow x(x - 30) - 60(x - 30) = 0 \\[0.5em] \Rightarrow (x - 30)(x - 60) = 0 \\[0.5em] \Rightarrow x - 30 = 0 \text{ or } x - 60 = 0 \\[0.5em] \Rightarrow x = 30 \text{ or } x = 60\end{aligned}$$

Breadth = (90 - x) meters
Equation in x : x(90 - x) = 1800
Length of rectangle = 60 m , Breadth of rectangle = 30 m

Given ,

Length of first parallel side = (x + 9) cm

Length of second parallel side = (2x - 3) cm

Distance between parallel side = (x + 4) cm

Area of trapezium = 540 cm²

Area of trapezium is given by,

$$\begin{aligned}=\frac{1}{2} \times \text{ (sum of parallel sides) } \times \text{ (distance between them) } \\[1em] \Rightarrow \frac{1}{2} \times (x + 9 + 2x - 3) \times (x + 4) = 540 \\[1em] \Rightarrow \frac{1}{2} \times (3x + 6)(x + 4) = 540 \\[1em] \Rightarrow \frac{1}{2} \times (3x^2 + 12x + 6x + 24) = 540 \\[1em] \Rightarrow \frac{1}{2} \times (3x^2 + 18x + 24) = 540 \\[1em] \Rightarrow 3x^2 + 18x + 24 = 540 \times 2 \text{ (On cross multiplication) } \\[1em] \Rightarrow 3x^2 + 18x + 24 = 1080 \\[1em] \Rightarrow 3x^2 + 18x + 24 - 1080 = 0 \\[1em] \Rightarrow 3x^2 + 18x - 1056 = 0 \\[1em] \Rightarrow 3(x^2 + 6x - 352) = 0 \\[1em] \Rightarrow x^2 + 6x - 352 = 0 \\[1em] \Rightarrow x^2 + 22x - 16x - 352 = 0 \\[1em] \Rightarrow x(x + 22) - 16(x + 22) = 0 \\[1em] \Rightarrow (x - 16)(x + 22) = 0 \\[1em] \Rightarrow x - 16 = 0 \text{ or } x + 22 = 0 \\[1em] x = 16 \text{ or } x = -22\end{aligned}$$

If x = -22 , Length = x + 9 = -22 + 9 = -13 , Breadth = (2x - 3) = -44 - 3 = -47
Since length and breadth cannot be negative hence , x ≠ -22

∴ x = 16

The value of x is 16.

Taking length = l and breadth = b

Perimeter of rectangle = 2(l + b)

Length of diagonal of a rectangle = $\sqrt{l^2 + b^2}$

Given,

Perimeter = 68 m

$$\begin{aligned}\Rightarrow 2(l + b) = 68 \\[0.5em] \Rightarrow l + b = 34 \\[0.5em] \Rightarrow l = 34 - b \qquad \text{ Equation (a)}\end{aligned}$$

Given,

Diagonal of a rectangle = 26 m

$\Rightarrow \sqrt{l^2 + b^2} = 26$

On squaring both sides,

$\Rightarrow l^2 + b^2 = 26^2$

Putting values of l from equation a,

$$\begin{aligned}\Rightarrow (34 - b)^2 + b^2 = 676 \\[1em] \Rightarrow 1156 + b^2 - 68b + b^2 = 676 \\[1em] \Rightarrow 2b^2 - 68b + 1156 - 676 = 0 \\[1em] \Rightarrow 2b^2 - 68b + 480 = 0 \\[1em] \Rightarrow 2(b^2 - 34b + 240) = 0 \\[1em] \Rightarrow b^2 - 34b + 240 = 0 \\[1em] \Rightarrow b^2 - 24b - 10b + 240 = 0 \\[1em] \Rightarrow b(b - 24) - 10(b - 24) = 0 \\[1em] \Rightarrow (b - 24)(b - 10) = 0 \\[1em] \Rightarrow b - 24 = 0 \text{ or } b - 10 = 0 \\[1em] b = 24 \text{ or } b = 10\end{aligned}$$

∴ If b = 24 ,l = 34 - b = 34 - 24 = 10

If b = 10 , l = 34 - b = 34 - 10 = 24

Area of rectangle = Length $\times$ Breadth = 24 $\times$ 10 = 240 m²

Hence, the area of rectangle is 240 m².

Let one of the two smaller sides be x cm, then the other side is (17 - x) cm.

Length of hypotenuse = perimeter - sum of other two sides = 30cm - 17cm = 13cm

In right angled triangle

Perpendicular² + Base² = Hypotenuse²

∴ x² + (17 - x)² = 13²

$$\begin{aligned}\Rightarrow x^2 + 289 + x^2 - 34x = 169 \\[1em] \Rightarrow 2x^2 - 34x + 289 - 169 = 0 \\[1em] \Rightarrow 2x^2 - 34x + 120 = 0 \\[1em] \Rightarrow 2(x^2 - 17x + 60) = 0 \\[1em] \Rightarrow x^2 - 17x + 60 = 0 \\[1em] \Rightarrow x^2 - 12x - 5x + 60 = 0 \\[1em] \Rightarrow x(x - 12) - 5(x - 12) = 0 \\[1em] \Rightarrow (x - 5)(x - 12) = 0 \\[1em] \Rightarrow x - 5 = 0 \text{ or } x - 12 = 0 \\[1em] \Rightarrow x = 5 \text{ or } x = 12\end{aligned}$$

If x = 5 , 17 - x = 12 and if x = 12 , 17 - x = 5.

Hence, two small sides are 5 , 12.

Area of right angled triangle = $\frac{1}{2} \times \text{base} \times \text{height}$

∴ $\frac{1}{2} \times 5\text{cm} \times 12\text{cm} = 30 cm^2$

Hence the area of triangle is 30cm².

Let the shortest side be x metres

So, hypotenuse = (2x + 1) metres and third side = (x + 7) metres

Hypotenuse² = Perpendicular² + Base²

∴ (2x + 1)² = x² + (x + 7)²

$$\begin{aligned}\Rightarrow 4x^2 + 1 + 4x = x^2 + x^2 + 49 + 14x \\[1em] \Rightarrow 4x^2 + 1 + 4x = 2x^2 + 49 + 14x \\[1em] \Rightarrow 4x^2 - 2x^2 + 1 - 49 + 4x - 14x = 0 \\[1em] \Rightarrow 2x^2 - 48 - 10x = 0 \\[1em] \Rightarrow 2(x^2 - 24 - 5x) = 0 \\[1em] \Rightarrow x^2 - 5x - 24 = 0 \\[1em] \Rightarrow x^2 - 8x + 3x - 24 = 0 \\[1em] \Rightarrow x(x - 8) + 3(x - 8) = 0 \\[1em] \Rightarrow (x + 3)(x - 8) = 0 \\[1em] \Rightarrow x + 3 = 0 \text{ or } x - 8 = 0 \\[1em] x = -3 \text{ or } x = 8\end{aligned}$$

Since no side of a triangle can be negative hence, x ≠ -3

If x = 8 , (2x + 1) = 17 , (x + 7) = 15

Hence, the hypotenuse of the triangle is 17 metres while the shorter sides are 8 metres and 15 metres.

Let the length of hypotenuse be x cm,

So, longer side = (x - 2) cm and shortest side = (x - 4) cm

Hypotenuse² = Sum of the squares of other two sides

∴ x² = (x - 2)² + (x - 4)²

$$\begin{aligned}\Rightarrow x^2 = x^2 + 4 - 4x + x^2 + 16 - 8x \\[1em] \Rightarrow x^2 = 2x^2 - 12x + 20 \\[1em] \Rightarrow 2x^2 - x^2 - 12x + 20 = 0 \\[1em] \Rightarrow x^2 - 12x + 20 = 0 \\[1em] \Rightarrow x^2 - 10x - 2x + 20 = 0 \\[1em] \Rightarrow x(x - 10) - 2(x - 10) = 0 \\[1em] \Rightarrow (x - 2)(x - 10) = 0 \\[1em] \Rightarrow x - 2 = 0 \text{ or } x - 10 = 0 \\[1em] x = 2 \text{ or } x = 10\end{aligned}$$

x ≠ 2 as that will make shortest side negative (x - 4) = -2.

If x = 10 , x - 2 = 8 , x - 4 = 6.

Hence, the dimensions of triangle are Hypotenuse = 10 cm , Longer side = 8 cm , Shortest side = 6 cm.

Let the number of rows be x

So, the number of students in each row = x + 4

Given, total students = 480

$$\begin{aligned}\Rightarrow x(x + 4) = 480 \\[1em] \Rightarrow x^2 + 4x = 480 \\[1em] \Rightarrow x^2 + 4x - 480 = 0 \\[1em] \Rightarrow x^2 + 24x - 20x - 480 = 0 \\[1em] \Rightarrow x(x + 24) - 20(x + 24) = 0 \\[1em] \Rightarrow (x - 20)(x + 24) = 0 \\[1em] \Rightarrow x - 20 = 0 \text{ or } x + 24 = 0 \\[1em] x = 20 \text{ or } x = -24.\end{aligned}$$

Since number of rows cannot be negative hence, x ≠ -24.

If x = 20 , x + 4 = 24.

Hence the number of students in each row are 24.

Let number of rows be = $x$ = number of seats in each row

Hence, total number of seats = $x \times x = x^2$

Given, if the number of rows is doubled and the number of seats in each row is reduced by 5, then the total number of seats is increased by 375

$\therefore 2x \times (x - 5) - x^2 = 375$

$$\begin{aligned}\Rightarrow 2x(x - 5) - x^2 = 375 \\[1em] \Rightarrow 2x^2 - 10x - x^2 = 375 \\[1em] \Rightarrow x^2 - 10x - 375 = 0 \\[1em] \Rightarrow x^2 - 25x + 15x - 375 = 0 \\[1em] \Rightarrow x(x - 25) + 15(x - 25) = 0 \\[1em] \Rightarrow (x - 25)(x + 15) = 0 \\[1em] \Rightarrow x - 25 = 0 \text{ or } x + 15 = 0 \\[1em] \Rightarrow x = 25 \text{ or } x = -15\end{aligned}$$

Since number of rows cannot be negative hence, x ≠ -15.

Hence the number rows were 25.

Let the number of students be x

If each student gives gift to every other student so each student gives gift to (x - 1) students

So, x students gives gifts to total = x(x - 1) students.

According to given,

$$\begin{aligned}\Rightarrow x(x - 1) = 1980 \\[1em] \Rightarrow x^2 - x = 1980 \\[1em] \Rightarrow x^2 - x - 1980 = 0 \\[1em] \Rightarrow x^2 - 45x + 44x - 1980 \\[1em] \Rightarrow x(x - 45) + 44(x - 45) \\[1em] \Rightarrow (x - 45)(x + 44) \\[1em] \Rightarrow x - 45 = 0 \text{ or } x + 44 = 0 \\[1em] x = 45 \text{ or } x = -44\end{aligned}$$

Since number of students cannot be negative hence, x ≠ -44.

Hence, the number of students are 45.

Uniform speed of bus = x km/h

Due to heavy rain the speed reduces to = (x - 10) km/h

Given, due to decrease in speed it takes two hours longer to cover the distance

Since, Time = $\frac{\text{Distance}}{\text{Speed}}$

$$\begin{aligned}\therefore \frac{240}{x - 10} - \frac{240}{x} = 2 \\[1em] \Rightarrow \frac{240x - 240(x - 10)}{x(x - 10)} = 2 \text{ (On taking L.C.M.) } \\[1em] \Rightarrow 240x - 240x + 2400 = 2x(x - 10) \\[1em] \Rightarrow 2400 = 2x^2 - 20x \\[1em] \Rightarrow 2x^2 - 20x - 2400 = 0 \\[1em] \Rightarrow 2(x^2 - 10x - 1200) = 0 \\[1em] \Rightarrow x^2 - 10x - 1200 = 0 \\[1em] \Rightarrow x^2 - 40x + 30x - 1200 = 0 \\[1em] \Rightarrow x(x - 40) + 30(x - 40) = 0 \\[1em] \Rightarrow (x + 30)(x - 40) = 0 \\[1em] \Rightarrow x + 30 = 0 \text{ or } x - 40 = 0 \\[1em] x = -30 \text{ or } x = 40\end{aligned}$$

Since, speed cannot be negative hence, x ≠ -30.

The equation in x is $\rightarrow \frac{240}{x - 10} - \frac{240}{x} = 2$
Hence, the value of uniform speed is 40 km/h.

Speed of an express train is x km/h

So, the speed of ordinary train is (x - 12) km/h

Given, ordinary train takes one hour longer than express train to cover 240 km

Since, Time = $\frac{\text{Distance}}{\text{Speed}}$

$$\begin{aligned}\therefore \frac{240}{x - 12} - \frac{240}{x} = 1 \\[1em] \Rightarrow \frac{240x - 240(x - 12)}{x(x - 12)} = 1 \\[1em] \Rightarrow 240x - 240x + 2880 = x(x - 12) \\[1em] \Rightarrow 2880 = x^2 - 12x \\[1em] \Rightarrow x^2 - 12x - 2880 = 0 \\[1em] \Rightarrow x^2 - 60x + 48x - 2880 = 0 \\[1em] \Rightarrow x(x - 60) + 48(x - 60) = 0 \\[1em] \Rightarrow (x - 60)(x + 48) = 0 \\[1em] \Rightarrow x - 60 = 0 \text{ or } x + 48 = 0 \\[1em] \Rightarrow x = 60 \text{ or } x = -48\end{aligned}$$

Since speed of train cannot be negative hence, x ≠ -48

Hence, the speed of express train is 60 km/h.

Let the speed of car be x km/h

Given, if speed been 12 km/h more, the time taken for the journey would have been 1 hour 40 minutes less

1 hour 40 minutes = $\frac{100}{60}$ hours

Since, Time = $\frac{\text{Distance}}{\text{Speed}}$

$$\begin{aligned}\therefore \frac{400}{x} - \frac{400}{x + 12} = \frac{100}{60} \\[1em] \Rightarrow 100(\frac{4}{x} - \frac{4}{x + 12}) = \frac{100}{60} \\[1em] \Rightarrow \frac{4}{x} - \frac{4}{x + 12} = \frac{1}{60} \\[1em] \Rightarrow \frac{4(x + 12) - 4x}{x(x + 12)} = \frac{1}{60} \\[1em] \Rightarrow 60(4x + 48 - 4x) = x(x + 12) \\[1em] \Rightarrow 60 \times 48 = x^2 + 12x \\[1em] \Rightarrow x^2 + 12x - 2880 = 0 \\[1em] \Rightarrow x^2 + 60x - 48x - 2880 = 0 \\[1em] \Rightarrow x(x + 60) - 48(x + 60) = 0 \\[1em] \Rightarrow (x - 48)(x + 60) = 0 \\[1em] \Rightarrow x - 48 = 0 \text{ or } x + 60 = 0 \\[1em] x = 48 \text{ or } x = -60\end{aligned}$$

Since speed of train cannot be negative hence, x ≠ -60

Hence, the speed of express train is 48 km/h.

(i) Distance covered by plane = 400 km

Average speed of plane = x km/h

Time = $\frac{\text{ Distance}}{\text{Speed}} = \frac{400}{x} \text{ hrs}$

(ii) Distance covered by plane = 400 km

Average speed of plane = (x + 40) km/h

Time = $\frac{\text{ Distance}}{\text{Speed}} = \frac{400}{x + 40} \text{ hrs}$

According to question,

$$\begin{aligned}\Rightarrow \frac{400}{x} - \frac{400}{x + 40} = \frac{30}{60} \\[1em] \Rightarrow \frac{400(x + 40) - 400x}{x(x + 40)} = \frac{1}{2} \\[1em] \Rightarrow \frac{400x - 400x + 16000}{x(x + 40)} = \frac{1}{2} \\[1em] \Rightarrow 16000 \times 2 = x(x + 40) \\[1em] \Rightarrow 32000 = x^2 + 40x \\[1em] \Rightarrow x^2 + 40x - 32000 = 0 \\[1em] \Rightarrow x^2 + 200x - 160x - 32000 = 0 \\[1em] \Rightarrow x(x + 200) - 160(x + 200) = 0 \\[1em] \Rightarrow (x - 160)(x + 200) = 0 \\[1em] \Rightarrow x - 160 = 0 \text{ or } x + 200 = 0 \\[1em] x = 160 \text{ or } x = -200\end{aligned}$$

Since speed of aeroplane cannot be negative hence, x ≠ -200

∴ x = 160

Equation in x : $\frac{400}{x} - \frac{400}{x + 40} = \frac{1}{2}$
Hence, the speed of aerolane is 160 km/h.

(i) Speed of car = x km/h

Distance between point A and B by road = 216 km

Time taken = $\frac{\text{Distance}}{\text{Speed}} = \frac{216}{x}$ hours

(ii) Speed of train = (x + 16) km/h

Distance between point A and B by rail = 208 km

Time taken = $\frac{\text{Distance}}{\text{Speed}} = \frac{208}{x + 16}$ hours

(iii) Given,

Train takes 2 hours less than car to reach town B

$$\begin{aligned}\therefore \frac{216}{x} - \frac{208}{x + 16} = 2 \\[1em] \Rightarrow \frac{216(x + 16) - 208x}{x(x + 16)} = 2 \\[1em] \Rightarrow 216x -208x + 3456 = 2x(x + 16) \\[1em] \Rightarrow 8x + 3456 =2x^2 + 32x \\[1em] \Rightarrow 2x^2 + 32x -8x - 3456 = 0 \\[1em] \Rightarrow 2x^2 + 24x - 3456 = 0 \\[1em] \Rightarrow 2(x^2 + 12x - 1728) = 0 \\[1em] \Rightarrow x^2 + 12x - 1728 = 0 \\[1em] \Rightarrow x^2 + 48x - 36x - 1728 = 0 \\[1em] \Rightarrow x(x + 48) - 36(x + 48) = 0 \\[1em] \Rightarrow (x - 36)(x + 48) = 0 \\[1em] \Rightarrow x - 36 = 0 \text{ or } x + 48 = 0 \\[1em] \Rightarrow x = 36 \text{ or } x = -48\end{aligned}$$

Since speed of car cannot be negative hence x ≠ -48

∴ x = 36

Equation : $\frac{216}{x} - \frac{208}{x + 16} = 2$

(iv) Speed of train = (x + 16) = (36 + 16) = 52 km/h

Hence, speed of train is 52 km/h.

Let the speed of plane in still air be x km/h

Speed of wind = 30 km/h

∴ Speed of plane in wind = (x + 30) km/h and Speed of plane against wind = (x - 30) km/h.

40 minutes = $\frac{40}{60} \text{ hours } = \frac{2}{3} \text{ hours }$

According to question,

$$\begin{aligned}\Rightarrow \frac{3600}{x - 30} - \frac{3600}{x + 30} = \frac{2}{3} \\[1em] \Rightarrow \frac{3600(x + 30) - 3600(x - 30)}{(x - 30)(x + 30)} = \frac{2}{3} \\[1em] \Rightarrow \frac{3600x + 108000 - 3600x + 108000}{x^2 - 30x + 30x - 900} = \frac{2}{3} \\[1em] \Rightarrow 216000 \times 3 = 2(x^2 - 900) \text{ (On cross multiplying) }\\[1em] \Rightarrow 648000 = 2(x^2 - 900) \\[1em] \Rightarrow 324000 = x^2 - 900 \text{ (Dividing the complete equation by 2) } \\[1em] \Rightarrow x^2 - 900 - 324000 = 0 \\[1em] \Rightarrow x^2 - 324900 = 0 \\[1em] \Rightarrow x^2 - (570)^2 = 0 \\[1em] \Rightarrow (x - 570)(x + 570) = 0 \\[1em] \Rightarrow x - 570 = 0 \text{ or } x + 570 = 0 \\[1em] x = 570 \text{ or } x = -570 \\[1em]\end{aligned}$$

Since speed of aeroplane cannot be negative hence, x ≠ -570.

The speed of aeroplane in still air is 570 km/h.

Let the speed of bus while reaching picnic spot be x km/h

Since, the speed of bus decrease by 5 km/h due to rain hence, speed of bus in rain = (x - 5) km/h

∴ Time taken to reach picnic spot = $\frac{150}{x}$ and Time taken to reach back to school = $\frac{150}{x - 5}$

According to given,

$$\begin{aligned}\Rightarrow \frac{150}{x - 5} - \frac{150}{x} = 1 \\[1em] \Rightarrow \frac{150x - 150(x - 5)}{x(x - 5)} = 1 \\[1em] \Rightarrow \frac{150x - 150x + 750}{x^2 - 5x} = 1 \\[1em] \Rightarrow 750 = x^2 - 5x \text{ (On cross multiplying) } \\[1em] \Rightarrow x^2 - 5x - 750 = 0 \\[1em] \Rightarrow x^2 - 30x + 25x - 750 = 0 \\[1em] \Rightarrow x(x - 30) + 25(x - 30) = 0 \\[1em] \Rightarrow (x + 25)(x - 30) = 0 \\[1em] \Rightarrow x + 25 = 0 \text{ or } x - 30 = 0 \\[1em] x = -25 \text{ or } x = 30\end{aligned}$$

Since speed of bus cannot be negative hence, x ≠ -25.

∴ x = 30

If x = 30 , x - 5 = 25.

Time taken to return = $\frac{150}{x - 5}$ hours = 6 hours.

Hence, time taken on return trip is 6 hours.

Let the speed of boat in still water be x km/h

Speed of stream = 1.5 km/h

∴ Speed of boat upstream = (x - 1.5) km/h and Speed of boat downstream = (x + 1.5) km/h

According to given,

$$\begin{aligned}\Rightarrow \frac{10}{x - 1.5} + \frac{5}{x + 1.5} = 6 \\[1em] \Rightarrow \frac{10(x + 1.5) + 5(x - 1.5)}{(x + 1.5)(x - 1.5)} = 6 \\[1em] \Rightarrow \frac{10x + 15 + 5x - 7.5}{x^2 - 1.5x + 1.5x - 2.25} = 6 \\[1em] \Rightarrow \frac{15x + 7.5}{x^2 - 2.25} = 6 \\[1em] \Rightarrow 15x + 7.5 = 6(x^2 - 2.25) \\[1em] \Rightarrow 15x + 7.5 = 6x^2 - 13.50 \\[1em] \Rightarrow 6x^2 - 13.5 - 7.5 - 15x = 0 \\[1em] \Rightarrow 6x^2 - 15x - 21 = 0 \\[1em] \Rightarrow 3(2x^2 - 5x - 7) = 0 \\[1em] \Rightarrow 2x^2 - 5x - 7 = 0 \\[1em] \Rightarrow 2x^2 - 7x + 2x - 7 = 0 \\[1em] \Rightarrow x(2x - 7) + 1(2x - 7) = 0 \\[1em] \Rightarrow (x + 1)(2x - 7) = 0 \\[1em] \Rightarrow x = -1 \text{ or } 2x - 7 = 0 \\[1em] \Rightarrow x + 1 = 0 \text{ or } x = \frac{7}{2} \\[1em] \Rightarrow x = -1 \text{ or } x = 3.5\end{aligned}$$

Since speed of bus cannot be negative hence, x ≠ -1

Hence speed of boat in still water is 3.5 km/h.

The tank is filled by the two pipes together in $11\frac{1}{9}$ minutes i.e. in $\frac{100}{9}$ minutes,

∴ the part of tank filled in one minute = $\frac{9}{100}$

Let the time taken by two pipes to fill tank separately be x minutes and (x + 5) minutes.

∴ the part of tank filled by the first pipe in one minute = $\frac{1}{x}$ and

the part of tank filled by the second pipe in one minute = $\frac{1}{x + 5}$

According to the question

$$\begin{aligned}\frac{1}{x} + \frac{1}{x + 5} = \frac{9}{100} \\[1em] \Rightarrow \frac{x + 5 + x}{x(x + 5)} = \frac{9}{100} \\[1em] \Rightarrow \frac{2x + 5}{x^2 + 5x} = \frac{9}{100} \\[1em] \Rightarrow 100(2x + 5) = 9(x^2 + 5x) \\[1em] \Rightarrow 200x + 500 = 9x^2 + 45x \\[1em] \Rightarrow 9x^2 + 45x - 200x - 500 = 0 \\[1em] \Rightarrow 9x^2 - 155x - 500 = 0 \\[1em] \Rightarrow 9x^2 - 180x + 25x - 500 = 0 \\[1em] \Rightarrow 9x(x - 20) + 25(x - 20) = 0 \\[1em] \Rightarrow (9x + 25)(x - 20) = 0 \\[1em] \Rightarrow x = -\frac{25}{9} \text{ or } x = 20\end{aligned}$$

Since, time cannot be negative hence, x ≠ $-\frac{25}{9}$

∴ x = 20 , x + 5 = 25.

Time taken by each pipe is 20 minutes and 25 minutes.

Number of children = x

Money recieved by each = $\frac{480}{x}$

If number of children were 20 more then money received by each child = $\frac{480}{x + 20}$

Given, if there are 20 more children then each will get ₹12 less

$$\begin{aligned}\therefore \frac{480}{x} - \frac{480}{x + 20} = 12 \\[1em] \Rightarrow \frac{40}{x} - \frac{40}{x + 20} = 1 \text{ (Dividing the equation by 12) } \\[1em] \Rightarrow \frac{40(x + 20) - 40x}{x(x + 20)} = 1 \\[1em] \Rightarrow \frac{40x + 800 - 40x}{x^2 + 20x} = 1 \\[1em] \Rightarrow 800 = x^2 + 20x \\[1em] \Rightarrow x^2 + 20x - 800 = 0 \\[1em] \Rightarrow x^2 + 40x - 20x - 800 = 0 \\[1em] \Rightarrow x(x + 40) - 20(x + 40) = 0 \\[1em] \Rightarrow (x - 20)(x + 40) = 0 \\[1em] \Rightarrow x - 20 = 0 \text{ or } x + 40 = 0 \\[1em] x = 20 \text{ or } x = -40\end{aligned}$$

Since, number of children cannot be negative hence, x ≠ -40

∴ x = 20

Hence, number of children are 20.

Let the number of children be x

₹7500 is divided equally among x children so , each received = ₹$\frac{7500}{x}$

Given, if there were 20 less children, each would have received ₹ 100 more.

If there are 20 less children then each would receive = ₹$\frac{7500}{x - 20}$

$$\begin{aligned}\therefore \frac{7500}{x - 20} - \frac{7500}{x} = 100 \\[1em] \Rightarrow \frac{75}{x - 20} - \frac{75}{x} = 1 \text{ (Dividing the equation by 100) } \\[1em] \Rightarrow \frac{75x - 75(x - 20)}{x(x - 20)} = 1 \\[1em] \Rightarrow \frac{75x - 75x + 1500}{x(x - 20)} = 1 \\[1em] \Rightarrow \frac{1500}{x(x - 20)} = 1 \\[1em] \Rightarrow 1500 = x^2 - 20x \text{ (On cross multiplication) }\\[1em] \Rightarrow x^2 - 20x - 1500 = 0 \\[1em] \Rightarrow x^2 - 50x + 30x - 1500 = 0 \\[1em] \Rightarrow x(x - 50) + 30(x - 50) = 0 \\[1em] \Rightarrow (x + 30)(x - 50) = 0 \\[1em] \Rightarrow x + 30 = 0 \text{ or } x - 50 = 0 \\[1em] x = -30 \text{ or } x = 50\end{aligned}$$

Since, number of children cannot be negative hence, x ≠ -30

∴ x = 50

Hence, number of children are 50.

2x articles cost ₹(5x + 54)

So, cost of each article = ₹$\Big(\frac{5x + 54}{2x}\Big)$

Similar (x + 2) articles cost ₹(10x - 4)

So, cost of each article = ₹$\Big(\frac{10x - 4}{x + 2}\Big)$

$$\begin{aligned}\therefore \frac{5x + 54}{2x} = \frac{10x - 4}{x + 2} \\[1em] \Rightarrow (5x + 54)(x + 2) = 2x(10x - 4) \\[1em] \Rightarrow 5x^2 + 10x + 54x + 108 = 20x^2 - 8x \\[1em] \Rightarrow 5x^2 - 20x^2 + 64x + 8x + 108 = 0 \\[1em] \Rightarrow -15x^2 + 72x + 108 = 0 \\[1em] \Rightarrow 15x^2 - 72x - 108 = 0 \text{ (On multiplying equation by -1) }\\[1em] \Rightarrow 3(5x^2 - 24x - 36) = 0 \\[1em] \Rightarrow 5x^2 - 24x - 36 = 0 \\[1em] \Rightarrow 5x^2 - 30x + 6x - 36 = 0 \\[1em] \Rightarrow 5x(x - 6) + 6(x - 6) = 0 \\[1em] \Rightarrow (5x + 6)(x - 6) = 0 \\[1em] \Rightarrow (5x + 6) = 0 \text{ or } x - 6 = 0 \\[1em] x = -\frac{6}{5} \text{ or } x = 6\end{aligned}$$

Value of x cannot be negative and in fraction as that will make number of articles in fraction which is not possible hence, x ≠ -$\frac{6}{5}$

∴ x = 6

Hence , value of x is 6.

(i) x articles cost ₹600.

So, cost of each article = $\frac{600}{x}$

(ii) Given,

If the cost per article were ₹5 more, the number of articles that can be bought for ₹600 would be four less.

$$\begin{aligned}\therefore \frac{600}{x - 4} - \frac{600}{x} = 5 \\[1em] \Rightarrow \frac{600x - 600(x - 4)}{x(x - 4)} = 5 \\[1em] \Rightarrow 600x - 600x + 2400 = 5x(x - 4) \text{ (On cross multiplication) } \\[1em] \Rightarrow 5x^2 - 20x - 2400 = 0 \\[1em] \Rightarrow 5(x^2 - 4x - 480) = 0 \\[1em] \Rightarrow x^2 - 4x - 480 = 0 \\[1em] \Rightarrow x^2 - 24x + 20x - 480 = 0 \\[1em] \Rightarrow x(x - 24) + 20(x - 24) = 0 \\[1em] \Rightarrow (x - 24)(x + 20) = 0 \\[1em] x = 24 \text{ or } x = -20\end{aligned}$$

Since, no of articles cannot be negative hence, x ≠ -20

∴ x = 24

Hence, the number of articles that trader buys are 24.

Let the cost of each book be ₹x

Number of books that can be bought for ₹960 = $\frac{960}{x}$

If the cost of book is ₹8 less then number of books that can be bought for ₹960 = $\frac{960}{x - 8}$

Given, number of books would be 4 more if price would be ₹8 less

$$\begin{aligned}\therefore \frac{960}{x - 8} - \frac{960}{x} = 4 \\[1em] \Rightarrow \frac{960x - 960(x - 8)}{x(x - 8)} = 4 \\[1em] \Rightarrow \frac{960x - 960x + 7680}{x^2 - 8x} = 4 \\[1em] \Rightarrow 7680 = 4(x^2 - 8x) \\[1em] \Rightarrow 4x^2 - 32x - 7680 = 0 \\[1em] \Rightarrow 4(x^2 - 8x - 1920) = 0 \\[1em] \Rightarrow x^2 - 8x - 1920 = 0 \\[1em] \Rightarrow x^2 - 48x + 40x - 1920 = 0 \\[1em] \Rightarrow x(x - 48) + 40(x - 48) = 0 \\[1em] \Rightarrow (x - 48)(x + 40) = 0 \\[1em] \Rightarrow x - 48 = 0 \text{ or } x + 40 = 0 \\[1em] x = 48 \text{ or } x = -40\end{aligned}$$

Since, cost of book cannot be negative hence, x ≠ -40

∴ x = 48

Hence, the cost of each book is ₹48.

Let the length of original cloth be x metres

Since, cost of total piece = ₹300

So, cost of each metre = $\frac{300}{x}$

Given, new length is 5 metre more i.e. (x + 5) metres and cost of each metre is ₹2 less i.e. $\frac{300}{x} - 2$

Since total cost remains unchanged

$$\begin{aligned}\therefore (x + 5) \times \big(\frac{300}{x} - 2 \big) = 300 \\[1em] \Rightarrow (x + 5)\big(\frac{300 - 2x}{x}\big) = 300 \\[1em] \Rightarrow (x + 5)(300 - 2x) = 300x \text{ (On cross multiplication) } \\[1em] \Rightarrow 300x - 2x^2 + 1500 - 10x - 300x = 0 \\[1em] \Rightarrow -2x^2 + 1500 - 10x = 0 \\[1em] \Rightarrow -2(x^2 + 5x - 750) = 0 \\[1em] \Rightarrow x^2 + 5x - 750 = 0 \\[1em] \Rightarrow x^2 + 30x - 25x - 750 = 0 \\[1em] \Rightarrow x(x + 30) - 25(x + 30) = 0 \\[1em] \Rightarrow (x + 30)(x - 25) = 0 \\[1em] \Rightarrow x + 30 = 0 \text{ or } x - 25 = 0 \\[1em] x = -30 \text{ or } x = 25\end{aligned}$$

Since, length cannot be negative hence, x ≠ -30

∴ x = 25 , $\frac{300}{x}$ = 12

Hence, the length of original cloth is 25 metre and cost per metre is ₹12.

Let the number of people be x

Total bill = ₹4800

∴ Bill amount each person has to pay = $\frac{4800}{x}$

If there were 4 more people then bill each has to pay = $\frac{4800}{x + 4}$

According to question,

$$\begin{aligned}\Rightarrow \frac{4800}{x} - \frac{4800}{x + 4} = 200 \\[1em] \Rightarrow \frac{4800(x + 4) - 4800x}{x(x + 4)} = 200 \\[1em] \Rightarrow 4800x - 4800x + 19200 = 200x(x + 4) \\[1em] \Rightarrow 19200 = 200x^2 + 800x \\[1em] \Rightarrow 200x^2 + 800x - 19200 = 0 \\[1em] \Rightarrow x^2 + 4x - 96 = 0 \text{ (On Dividing the equation by 200) } \\[1em] \Rightarrow x^2 + 12x - 8x - 96 = 0 \\[1em] \Rightarrow x(x + 12) - 8(x + 12) = 0 \\[1em] \Rightarrow (x - 8)(x + 12) = 0 \\[1em] x = 8 \text{ or } x = -12\end{aligned}$$

Since, number of people cannot be negative hence, x ≠ -12

∴ x = 8

Hence, the number of people staying overnight are 8.

Let number of days for which person plans trip be x

Total expense for trip = ₹3000

∴ Expense of each day = ₹$\Big(\frac{3000}{x}\Big)$

Days extended = 5 , so total days = (x + 5)

Daily expense to be cut down is ₹20,

∴ New Expense of each day = ₹$\Big(\frac{3000}{x} - 20\Big)$

According to question,

$$\begin{aligned}\Rightarrow (x + 5)\Big(\frac{3000}{x} - 20\Big) = 3000 \\[1em] \Rightarrow (x + 5)\Big(\frac{3000 - 20x}{x}\Big) = 3000 \\[1em] \Rightarrow (x + 5)(3000 - 20x) = 3000x \text{ (Cross multiplying) } \\[1em] \Rightarrow 3000x - 20x^2 + 15000 - 100x = 3000x \\[1em] \Rightarrow -20x^2 + 3000x - 3000x -100x + 15000 = 0 \\[1em] \Rightarrow -20x^2 - 100x + 15000 = 0 \\[1em] \Rightarrow -20(x^2 + 5x - 750) = 0 \\[1em] \Rightarrow x^2 + 5x - 750 = 0 \\[1em] \Rightarrow x^2 + 30x - 25x - 750 = 0 \\[1em] \Rightarrow x(x + 30) - 25(x + 30) = 0 \\[1em] \Rightarrow (x - 25)(x + 30) = 0 \\[1em] \Rightarrow x - 25 = 0 \text{ or } x + 30 = 0 \\[1em] x = 25 \text{ or } x = -30\end{aligned}$$

Since, number of days cannot be negative hence, x ≠ -30

∴ x = 25

Hence, the number of days of tour programme are 25.

Let the age of Vivek be x years.

Since sum of ages of vivek and amit is 47, so amit's age = (47 - x) years

Product of ages = 550

∴ x(47 - x) = 550

$$\begin{aligned}\Rightarrow 47x - x^2 = 550 \\[1em] \Rightarrow x^2 - 47x + 550 = 0 \\[1em] \Rightarrow x^2 - 25x - 22x + 550 = 0 \\[1em] \Rightarrow x(x - 25) - 22(x - 25) = 0 \\[1em] \Rightarrow (x - 22)(x - 25) = 0 \\[1em] \Rightarrow x - 22 = 0 \text{ or } x - 25 = 0 \\[1em] x = 22 \text{ or } x = 25.\end{aligned}$$

Vivek's age = 25 years and Amit's age = 22 years.

Let paul's present age be x years

Father's age = 2x²

After 10 years,

Paul's age = (x + 10) years

Father's age = (2x² + 10)

According to question,

$$\begin{aligned}\Rightarrow 2x^2 + 10 = 4(x + 10) \\[1em] \Rightarrow 2x^2 + 10 = 4x + 40 \\[1em] \Rightarrow 2x^2 - 4x + 10 - 40 = 0 \\[1em] \Rightarrow 2x^2 - 4x - 30 = 0 \\[1em] \Rightarrow 2(x^2 - 2x - 15)= 0 \\[1em] \Rightarrow x^2 - 2x - 15 = 0 \\[1em] \Rightarrow x^2 - 5x + 3x - 15 = 0 \\[1em] \Rightarrow x(x - 5) + 3( x - 5) = 0 \\[1em] \Rightarrow (x + 3)(x - 5) = 0\\[1em] \Rightarrow x + 3 = 0 \text{ or } x - 5 = 0 \\[1em] x = -3 \text{ or } x = 5\end{aligned}$$

Since, age cannot be negative hence x ≠ -3

∴ x = 5 , 2x² = 50

Paul's age is 5 years , while his father's age is 50 years.

Let son's present age be x years

Man's age = 2x²

After 8 years,

Son's age = (x + 8) years

Man's age = (2x² + 8)

According to question,

$$\begin{aligned}\Rightarrow 2x^2 + 8 = 3(x + 8) + 4 \\[1em] \Rightarrow 2x^2 + 8 = 3x + 24 + 4 \\[1em] \Rightarrow 2x^2 + 8 = 3x + 28 \\[1em] \Rightarrow 2x^2 - 3x + 8 - 28 = 0 \\[1em] \Rightarrow 2x^2 - 3x - 20 = 0 \\[1em] \Rightarrow 2x^2 - 8x + 5x - 20 = 0 \\[1em] \Rightarrow 2x(x - 4) + 5(x - 4) = 0 \\[1em] \Rightarrow (x - 4)(2x + 5) = 0 \\[1em] \Rightarrow x - 4 = 0 \text{ or } 2x + 5 = 0 \\[1em] x = 4 \text{ or } x = -\frac{5}{2}\end{aligned}$$

Since, age cannot be negative hence x ≠ $-\frac{5}{2}$

∴ x = 4 , 2x² = 32

The present age of son is 4 years while the age of man is 32 years.

Let present age of daughter be x years

Two years before daughter's age = (x - 2) years

Man's age two years before = 3(x - 2)²

So, present age of man = 3(x - 2)² + 2

Three year's later

Daughter's age = (x + 3) years

Man's age = (3(x - 2)² + 2 + 3) years

According to question,

$$\begin{aligned}\Rightarrow (3(x - 2)^2 + 2 + 3) = 4(x + 3) \\[1em] \Rightarrow 3(x^2 + 4 - 4x) + 5 = 4x + 12 \\[1em] \Rightarrow 3x^2 + 12 - 12x + 5 = 4x + 12 \\[1em] \Rightarrow 3x^2 - 12x - 4x + 17 - 12 = 0 \\[1em] \Rightarrow 3x^2 - 16x + 5 = 0 \\[1em] \Rightarrow 3x^2 - 15x - x + 5 = 0 \\[1em] \Rightarrow 3x(x - 5) - 1(x - 5) = 0 \\[1em] \Rightarrow (3x - 1)(x - 5) = 0 \\[1em] \Rightarrow 3x - 1 = 0 \text{ or } x - 5 = 0 \\[1em] x = \frac{1}{3} \text{ or } x = 5\end{aligned}$$

Since, age cannot be in fraction so x ≠ $\frac{1}{3}$

∴ x = 5, 3(x - 2)² + 2 = 3(5 - 2)² + 2 = 3(3)² + 2 = 29

The present age of daughter is 5 years and of man is 29 years.

Let the lengths of the two sides other than hypotenuse be x cm and y cm

According to question,

Hypotenuse = x + 2 (in terms of 1st side)

Hypotenuse = 2y + 1 (in terms of 2nd side)

∴ x + 2 = 2y + 1
x = 2y + 1 - 2
x = 2y - 1

Hypotenuse = 2y + 1

As the given triangle is right-angled, by using Pythagoras theorem, we get:

$x^2 + y^2 = (2y + 1)^2$

Putting value of x = 2y - 1 in above equation

$$\begin{aligned}\Rightarrow (2y - 1)^2 + y^2 = (2y + 1)^2 \\[1em] \Rightarrow 4y^2 + 1 - 4y + y^2 = 4y^2 + 1 + 4y \\[1em] \Rightarrow 4y^2 - 4y^2 + y^2 + 1 - 1 - 4y -4y = 0 \\[1em] \Rightarrow y^2 - 8y = 0 \\[1em] \Rightarrow y(y - 8) = 0 \\[1em] \Rightarrow y = 0 \text{ or } y - 8 = 0 \\[1em] y = 0 \text{ or } y = 8\end{aligned}$$

y ≠ 0 , as that will make value of x negative and length cannot be negative.

∴ y = 8 , x = 2y - 1 = 15 , hypotenuse = 2y + 1 = 17

Perimeter = 8 + 15 + 17 = 40 cm

$$\begin{aligned}\text{Area} = \frac{1}{2} \times \text{first side } \times \text{ second side} \\[1em] = \frac{1}{2} \times 8 \times 15 \text{ cm}^2 \\[1em] = 60 \text{ cm}^2\end{aligned}$$

Hence, the value of first side = 8cm, second side = 15 cm , Hypotenuse = 17cm , Perimeter = 40cm , Area = 60 cm².

Let the sides of the smaller and bigger square be x cm and y cm, respectively.

Area of smaller square = x² cm²

Area of larger square = y² cm²

According to question,

$$\begin{aligned}y^2 - 2x^2 = 14 \\[1em] \Rightarrow y^2 = 14 + 2x^2 \qquad \text{[...Eq 1]}\end{aligned}$$

and

$3x^2 + 2y^2 = 203 \qquad \text{[...Eq 2]}$

Putting value of y² from Equation 1 in Equation 2, we get:

$$\begin{aligned}\Rightarrow 3x^2 + 2(14 + 2x^2) = 203 \\[1em] \Rightarrow 3x^2 + 28 + 4x^2 = 203 \\[1em] \Rightarrow 7x^2 = 203 - 28 \\[1em] \Rightarrow 7x^2 = 175 \\[1em] \Rightarrow x^2 = 25 \\[1em] \Rightarrow x = \sqrt{25} \\[1em] x = 5 \text{ or } x = -5\end{aligned}$$

Since, side of a square cannot be negative hence, x ≠ -5.

∴ x = 5
y² = 14 + 2² = 64
⇒ y = 8

Hence, the length of larger square's side is 8cm and smaller square's is 5cm.

Option 1:

$$\begin{aligned}(x + 2)^2 = 2(x + 3) \\[0.5em] \Rightarrow x^2 + 4 + 4x = 2x + 6 \\[0.5em] \Rightarrow x^2 + 4x - 2x + 4 - 6 = 0 \\[0.5em] \Rightarrow x^2 + 2x - 2 = 0\end{aligned}$$

It is a quadratic equation as highest power of x is 2.

Option 2:

$$\begin{aligned}x^2 + 3x = (-1)(1 - 3x) \\[0.5em] \Rightarrow x^2 + 3x = -1 + 3x \\[0.5em] \Rightarrow x^2 + 3x - 3x + 1 = 0 \\[0.5em] x^2 + 1 = 0\end{aligned}$$

It is a quadratic equation as the highest power of x is 2.

Option 3:

$$\begin{aligned}(x + 2)(x - 1) = x^2 - 2x - 3 \\[0.5em] \Rightarrow x^2 - x + 2x - 2 = x^2 - 2x - 3 \\[0.5em] \Rightarrow x^2 - x^2 - x + 2x + 2x - 2 + 3 = 0 \\[0.5em] 3x + 1 = 0\end{aligned}$$

It is not a quadratic equation as highest power of x is not 2.

Option 4:

$$\begin{aligned}x^3 - x^2 + 2x + 1 = (x + 1)^3 \\[0.5em] \Rightarrow x^3 - x^2 + 2x + 1 = x^3 + 1 + 3x(x + 1) \\[0.5em] \Rightarrow x^3 - x^2 + 2x + 1 = x^3 + 1 + 3x^2 + 3x \\[0.5em] \Rightarrow x^3 - x^3 - x^2 - 3x^2 + 2x - 3x + 1 - 1 = 0 \\[0.5em] \Rightarrow -4x^2 - x = 0 \\[0.5em] 4x^2 + x = 0\end{aligned}$$

It is a quadratic equation as highest power of x is 2.

∴ Option 3 is the correct option.

Given,

3 is a root of the quadratic equation x² - px + 3 = 0

∴ 3² - 3p + 3 = 0

⇒ 9 - 3p + 3 = 0

⇒ 12 - 3p = 0

⇒ 3p = 12

⇒ p = $\frac{12}{3}$ = 4.

Hence, Option 1 is the correct option.

Given,

$$\begin{aligned}x^2 - 3x - 10 = 0 \\[0.5em] \Rightarrow x^2 - 5x + 2x - 10 = 0 \\[0.5em] \Rightarrow x(x - 5) + 2(x - 5) = 0 \\[0.5em] \Rightarrow (x + 2)(x - 5) = 0 \\[0.5em] \Rightarrow x + 2 = 0 \text{ or } x - 5 = 0 \\[0.5em] x = -2 \text{ or } x = 5 \\[0.5em]\end{aligned}$$

∴ Option 2 is the correct option

Irrational roots occur in conjugate pair .

Hence if one root is $\frac{3 - \sqrt{5}}{2}$ then the other root is =

$\frac{3 + \sqrt{5}}{2}$

∴ Option 3 is the correct option

Given ,

2x² - 5x + (k + 3) = 0 has equal roots

Comparing equation with ax² + bx + c = 0
a= 2 , b = -5 , c = k + 3

Since, equation has equal roots

∴ b² - 4ac = 0

$$\begin{aligned}\Rightarrow (-5)^2 - 4 \times 2 \times (k + 3) = 0 \\[0.5em] \Rightarrow 25 - 8(k + 3) = 0 \\[0.5em] \Rightarrow 25 - 8k - 24 = 0 \\[0.5em] \Rightarrow 1 - 8k = 0 \\[0.5em] \Rightarrow 8k = 1 \\[0.5em] k = \frac{1}{8}\end{aligned}$$

∴ Option 3 is the correct option.

Given ,

2x² - kx + k = 0 has equal roots

Comparing equation with ax² + bx + c = 0
a= 2 , b = -k , c = k

Since, equation has equal roots

∴ b² - 4ac = 0

$$\begin{aligned}\Rightarrow (-k)^2 - 4 \times 2 \times k = 0 \\[0.5em] \Rightarrow k^2 - 8k = 0 \\[0.5em] \Rightarrow k(k - 8) = 0 \\[0.5em] \Rightarrow k = 0 \text{ or } k - 8 = 0 \\[0.5em] k = 0 \text{ or } k = 8 \\[0.5em]\end{aligned}$$

∴ Option 4 is the correct option.

Given ,

3x² - kx + 2k = 0 has equal roots

Comparing equation with ax² + bx + c = 0
a= 3 , b = -k , c = 2k

Since, equation has equal roots

∴ b² - 4ac = 0

$$\begin{aligned}\Rightarrow (-k)^2 - 4 \times 3 \times 2k = 0 \\[0.5em] \Rightarrow k^2 - 24k = 0 \\[0.5em] \Rightarrow k(k - 24) = 0 \\[0.5em] \Rightarrow k = 0 \text{ or } k - 24 = 0 \\[0.5em] k = 0 \text{ or } k = 24 \\[0.5em]\end{aligned}$$

∴ Option 4 is the correct option.

Given ,

(k + 1)x² - 2(k - 1)x + 1 = 0 has equal roots

Comparing equation with ax² + bx + c = 0
a= (k + 1) , b = -2(k - 1) , c = 1

Since, equation has equal roots

∴ b² - 4ac = 0

$$\begin{aligned}\Rightarrow (-2(k - 1))^2 - 4 \times (k + 1) \times 1 = 0 \\[0.5em] \Rightarrow 4(k^2 + 1 - 2k) - 4(k + 1) = 0 \\[0.5em] \Rightarrow 4k^2 + 4 - 8k - 4k - 4 = 0 \\[0.5em] \Rightarrow 4k^2 - 12k = 0 \\[0.5em] \Rightarrow 4k(k - 3) = 0 \\[0.5em] \Rightarrow 4k = 0 \text{ or } k - 3 = 0 \\[0.5em] k = 0 \text{ or } k = 3.\end{aligned}$$

∴ Option 2 is the correct option.

Given ,

2x² - 6x + p = 0 has real and different roots

Comparing equation with ax² + bx + c = 0
a= 2 , b = -6 , c = p

Since, equation has real and different roots

∴ b² - 4ac > 0

$$\begin{aligned}\Rightarrow (-6)^2 - 4 \times 2 \times p \gt 0 \\[0.5em] \Rightarrow (-6)^2 - 8p \gt 0 \\[0.5em] \Rightarrow 36 - 8p \gt 0 \\[0.5em] \Rightarrow 8p \lt 36 \\[0.5em] \Rightarrow p \lt \frac{36}{8} \\[0.5em] p \lt \frac{9}{2}\end{aligned}$$

∴ Option 1 is the correct option.

In order to find nature of roots we need to find the value of, b² - 4ac

Given,

$2x^2 - \sqrt{5}x + 1 = 0$

Comparing equation with ax² + bx + c = 0
a= 2 , b = -$\sqrt{5}$ , c = 1

Putting values in b² - 4ac

$$\begin{aligned}(-\sqrt{5})^2 - 4 \times 2 \times 1 \\[0.5em] \Rightarrow 5 - 8 \\[0.5em] \Rightarrow -3\end{aligned}$$

Since, b² - 4ac = -3 < 0 , hence there are no real roots

∴ Option 3 is the correct option.

Given,

Roots of equation x² - 6x + k = 0 are real and distinct.

∴ D > 0

⇒ b² - 4ac > 0

⇒ (-6)² - 4 × 1 × k > 0

⇒ 36 - 4k > 0

⇒ 4k < 36

⇒ k < $\frac{36}{4}$

⇒ k < 9.

Hence, Option 4 is the correct option.

By formula,

D = b² - 4ac

For equation, px² - qx + r = 0

D = (-q)² - 4 × p × r

We know that,

Roots of a quadratic equation are real and equal if discriminant = 0.

⇒ q² - 4pr = 0

⇒ q² = 4pr.

Hence, Option 2 is the correct option.

Given,

⇒ x² + kx + 6 = (x - 2)(x - 3)

⇒ x² + kx + 6 = x² - 3x - 2x + 6

⇒ x² + kx + 6 = x² - 5x + 6

⇒ x² - x² + kx + 6 - 6 = -5x

⇒ kx = -5x

⇒ k = -5.

Hence, Option 1 is the correct option.

Solving,

⇒ x² - 1 = 0

⇒ (x + 1)(x - 1) = 0

⇒ x + 1 = 0 or x - 1 = 0

⇒ x = -1 or x = 1.

Hence, Option 4 is the correct option.

The quadratic equation: ax² + bx + c = 0.

The expression b² - 4ac is called the discriminant (D).

When,

1. D > 0; two distinct real roots

2. D = 0; real and equal roots

3. D < 0; then roots are imaginary

thus, assertion (A) is false but reason(R) is true.

Hence, option 2 is the correct option.

For, the quadratic equation: ax² + bx + c = 0. The equation has real roots if

b² - 4ac ≥ 0

So, reason (R) is true.

Comparing equation 4x² + 12x + 15 = 0, with ax² + bx + c = 0, we get :

a = 4, b = 12, c = 15

D = b² - 4ac

= 12² - 4 x 4 x 15

= 144 - 240 = -96.

Since, D < 0, so the equation has no real roots.

So, assertion (A) is true and reason (R) correctly explains assertion (A).

Thus, both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).

Hence, option 3 is the correct option.

We know that,

The quadratic equation ax² + bx + c = 0 has real roots if 'discriminant' = b² - 4ac ≥ 0.

So, reason (R) is false.

Given, 9x² + 6x - k = 0

Comparing above equation with ax² + bx + c = 0, we get :

a = 9, b = 6 and c = -k

If the equation has real roots, then D ≥ 0

⇒ b² - 4ac ≥ 0

⇒ 6² - 4 x 9 x (-k) ≥ 0

⇒ 36 + 36k ≥ 0

⇒ 36k ≥ -36

⇒ k ≥ -$\frac{36}{36}$

⇒ k ≥ -1

So, assertion (A) is true.

Thus, Assertion (A) is true, but Reason (R) is false.

Hence, option 1 is the correct option.

Given,

Polynomial : 2x² - 3x + 5

Discriminant (D) = b² - 4ac

= (-3)² - 4 x 2 x 5

= 9 - 40

= -31.

So, reason (R) is true.

Since the discriminant is negative, this quadratic has no real roots and cannot be factorized into linear factors with real coefficients.

So, assertion (A) is true and reason (R) correctly explains assertion (A).

Thus, both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).

Hence, option 3 is the correct option.

Given,

Equation : k²x² - 2kx + 1 = 0

Comparing above equation with ax² + bx + c = 0, we get :

a = k², b = -2k and c = 1

Discriminant (D) = b² - 4ac

= (-2k)² - 4 x k² x 1

= 4k² - 4k²

= 0.

So, reason (R) is true.

Since, D = 0 this means the equation has one repeated real root.

So, assertion (A) is true and reason (R) correctly explains assertion (A).

Thus, both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).

Hence, option 3 is the correct option.

Given,

$$\begin{aligned}\Rightarrow x^2 + 6x - 16 = 0 \\[1em] \Rightarrow x^2 + 8x - 2x - 16 = 0 \\[1em] \Rightarrow x(x + 8) - 2(x + 8) = 0 \\[1em] \Rightarrow (x - 2)(x + 8) = 0 \\[1em] \Rightarrow x - 2 = 0 \text{ or } x + 8 = 0 \\[1em] \Rightarrow x = 2 \text{ or } x = -8\end{aligned}$$

Hence, roots of given equation are 2, -8.

Given,

$$\begin{aligned}\Rightarrow 3x^2 + 11x + 10 = 0 \\[1em] \Rightarrow 3x^2 + 6x + 5x + 10 = 0 \\[1em] \Rightarrow 3x(x + 2) + 5(x + 2) = 0 \\[1em] \Rightarrow (x + 2)(3x + 5) = 0 \\[1em] \Rightarrow x + 2 = 0 \text{ or } 3x + 5 = 0 \\[1em] \Rightarrow x = -2 \text{ or } x = -\frac{5}{3}\end{aligned}$$

Hence, roots of given equation are -2, -$\frac{5}{3}$.

Given,

$$\begin{aligned}\Rightarrow 2x^2 + ax - a^2 = 0 \\[1em] \Rightarrow 2x^2 + 2ax - ax - a^2 = 0 \\[1em] \Rightarrow 2x(x + a) - a(x + a) = 0 \\[1em] \Rightarrow (x + a)(2x - a) = 0 \\[1em] \Rightarrow x + a = 0 \text{ or } 2x - a = 0 \\[1em] \Rightarrow x = -a \text{ or } x = \frac{a}{2}\end{aligned}$$

Hence, roots of given equation are $-a, \frac{a}{2}$.

Given,

$$\begin{aligned}\Rightarrow \sqrt{3}x^2 + 10x + 7\sqrt{3} = 0 \\[1em] \Rightarrow \sqrt{3}x^2 + 7x + 3x + 7\sqrt{3} = 0 \\[1em] \Rightarrow x(\sqrt{3}x + 7) + \sqrt{3}(\sqrt{3}x + 7) = 0 \\[1em] \Rightarrow (x + \sqrt{3})(\sqrt{3}x + 7) = 0 \\[1em] \Rightarrow x + \sqrt{3} = 0 \text{ or } \sqrt{3}x + 7 = 0 \\[1em] \Rightarrow x = -\sqrt{3} \text{ or } x = -\frac{7}{\sqrt{3}} \\[1em] \Rightarrow x = -\sqrt{3} \text{ or } x = -\frac{7 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} \\[1em] \Rightarrow x = -\sqrt{3} \text{ or } x = - \frac{7\sqrt{3}}{3}\end{aligned}$$

Hence, roots of given equation are $-\sqrt{3}, -\frac{7\sqrt{3}}{3}$.

Given,

$$\begin{aligned}\Rightarrow x(x + 1) + (x + 2)(x + 3) = 42 \\[1em] \Rightarrow x^2 + x + x^2 + 3x + 2x + 6 = 42 \\[1em] \Rightarrow 2x^2 + 6x + 6 = 42 \\[1em] \Rightarrow 2x^2 + 6x + 6 - 42 = 0 \\[1em] \Rightarrow 2x^2 + 6x - 36 = 0 \\[1em] \Rightarrow 2(x^2 + 3x - 18) = 0 \\[1em] \Rightarrow x^2 + 3x - 18 = 0 \\[1em] \Rightarrow x^2 + 6x - 3x - 18 = 0 \\[1em] \Rightarrow x(x + 6) - 3(x + 6) = 0 \\[1em] \Rightarrow (x - 3)(x + 6) = 0 \\[1em] \Rightarrow x - 3 = 0 \text{ or } x + 6 = 0 \\[1em] x = 3 \text{ or } x = -6\end{aligned}$$

Hence, roots of given equation are 3 , -6.

Given,

$$\begin{aligned}\frac{6}{x} - \frac{2}{x - 1} = \frac{1}{x - 2} \\[1em] \Rightarrow \frac{6(x - 1) - 2x}{x(x - 1)} = \frac{1}{x - 2} \\[1em] \Rightarrow \frac{6x - 6 - 2x}{x^2 - x} = \frac{1}{x - 2} \\[1em] \Rightarrow \frac{4x - 6}{x^2 - x} = \frac{1}{x - 2} \\[1em] \Rightarrow (4x - 6)(x - 2) = x^2 - x \\[1em] \Rightarrow (4x^2 - 8x - 6x + 12) = x^2 - x \\[1em] \Rightarrow 4x^2 - x^2 - 14x + x + 12 = 0 \\[1em] \Rightarrow 3x^2 - 13x + 12 = 0 \\[1em] \Rightarrow 3x^2 - 9x - 4x + 12 = 0 \\[1em] \Rightarrow 3x(x - 3) - 4(x - 3) = 0 \\[1em] \Rightarrow (3x - 4)(x - 3) = 0 \\[1em] \Rightarrow 3x - 4 = 0 \text{ or } x - 3 = 0 \\[1em] x = \frac{4}{3} \text{ or } x = 3.\end{aligned}$$

Hence, roots of given equation are $3 ,\frac{4}{3}$

Given,

$$\begin{aligned}\Rightarrow \sqrt{x + 15} = x + 3 \\[1em] \Rightarrow x + 15 = (x + 3)^2 \text{ (On squaring both sides) } \\[1em] \Rightarrow x + 15 = x^2 + 9 + 6x \\[1em] \Rightarrow x^2 + 6x - x + 9 - 15 = 0 \\[1em] \Rightarrow x^2 + 5x - 6 = 0 \\[1em] \Rightarrow x^2 + 6x - x - 6 = 0 \\[1em] \Rightarrow x(x + 6) - 1(x + 6) = 0 \\[1em] \Rightarrow (x - 1)(x + 6) = 0 \\[1em] \Rightarrow x - 1 = 0 \text{ or } x + 6 = 0 \\[1em] \Rightarrow x = 1 \text{ or } x = -6\end{aligned}$$

Since we squared the equation, so roots need to be checked

Putting x = -6 in equation

$$\begin{aligned}\Rightarrow \sqrt{-6 + 15} = -6 + 3 \\[1em] \Rightarrow \sqrt{ 9 } = -3 \\[1em]\end{aligned}$$

L.H.S. = 3 and R.H.S. = -3

Since, L.H.S. ≠ R.H.S. hence, x = -6 is not root of the equation

Putting x = 1 in equation

$$\begin{aligned}\Rightarrow \sqrt{1 + 15} = 1 + 3 \\[1em] \Rightarrow \sqrt{ 16 } = 4 \\[1em]\end{aligned}$$

L.H.S. = R.H.S. = 4

Hence, root of the equation is 1.

Given,

$$\begin{aligned}\Rightarrow \sqrt{3x^2 - 2x - 1} = 2x - 2 \\[1em] \Rightarrow 3x^2 - 2x - 1 = (2x - 2)^2 \text{ (On squaring both sides) } \\[1em] \Rightarrow 3x^2 - 2x - 1 = 4x^2 + 4 - 8x \\[1em] \Rightarrow 3x^2 - 4x^2 - 2x + 8x - 1 - 4 = 0 \\[1em] \Rightarrow -x^2 + 6x - 5 = 0 \\[1em] \Rightarrow x^2 - 6x + 5 = 0 \text{ (On multiplying equation by -1) }\\[1em] \Rightarrow x^2 - 5x - x + 5 = 0 \\[1em] \Rightarrow x(x - 5) - 1(x - 5) = 0 \\[1em] \Rightarrow (x - 1)(x - 5) = 0 \\[1em] \Rightarrow x - 1 = 0 \text{ or } x - 5 = 0 \\[1em] x = 1 \text{ or } x = 5\end{aligned}$$

Since we squared the equation, so roots need to be checked

Putting x = 1 in equation

$$\begin{aligned}\Rightarrow \sqrt{3(1)^2 - 2(1) - 1} = 2(1) - 2 \\[1em] \Rightarrow \sqrt{ 0 } = 0 \\[1em]\end{aligned}$$

L.H.S. = R.H.S. = 0

Putting x = 5 in equation

$$\begin{aligned}\Rightarrow \sqrt{3(5)^2 - 2(5) - 1} = 2(5) - 2 \\[1em] \Rightarrow \sqrt{ 64 } = 8 \\[1em]\end{aligned}$$

L.H.S. = R.H.S. = 8

Hence, roots of the equation are 1, 5.

The given equation is 2x² - 3x - 1 = 0

Comparing it with ax² + bx + c = 0
a = 2, b = -3, c = -1

By using the formula , x = $\frac{-b ± \sqrt{b^2 - 4ac}}{2a}$ , we obtain

$$\begin{aligned}\Rightarrow \frac{-(-3) ± \sqrt{(-3)^2 - 4 \times 2 \times -1 }}{2 \times 2} \\[1em] \Rightarrow \frac{3 ± \sqrt{9 + 8}}{4} \\[1em] \Rightarrow \frac{3 + \sqrt{17}}{4} \text{ or } \frac{3 - \sqrt{17}}{4} \\[1em]\end{aligned}$$

Hence, roots of the equation are $\frac{3 + \sqrt{17}}{4} , \frac{3 - \sqrt{17}}{4}$.

The given equation is $x \Big(3x + \frac{1}{2} \Big)$ = 6

$$\begin{aligned}\Rightarrow 3x^2 + \frac{1}{2}x = 6 \\[1em] \Rightarrow 6x^2 + x = 12 \\[1em] \Rightarrow 6x^2 + x - 12 = 0\end{aligned}$$

Comparing it with ax² + bx + c = 0
a= 6, b = 1, c = -12

By using the formula , x = $\frac{-b ± \sqrt{b^2 - 4ac}}{2a}$ , we obtain

$$\begin{aligned}\Rightarrow \frac{-(1) ± \sqrt{(1)^2 - 4 \times 6 \times -12 }}{2 \times 6} \\[1em] \Rightarrow \frac{-1 ± \sqrt{1 + 288}}{12} \\[1em] \Rightarrow \frac{-1 + \sqrt{289}}{12} \text{ or } \frac{-1 - \sqrt{289}}{12} \\[1em] \Rightarrow \frac{-1 + 17}{12} \text{ or } \frac{-1 - 17}{12} \\[1em] \Rightarrow \frac{16}{12} \text{ or } -\frac{18}{12} \\[1em] \Rightarrow \frac{4}{3} \text{ or } -\frac{3}{2}\end{aligned}$$

Hence, roots of the equation are $\frac{4}{3} , -\frac{3}{2}$.

Given,

$$\begin{aligned}\frac{2x + 5}{3x + 4} = \frac{x + 1}{x + 3} \\[1em] \Rightarrow (2x + 5)(x + 3) = (x + 1)(3x + 4) \\[1em] \Rightarrow (2x^2 + 6x + 5x + 15) = (3x^2 + 4x + 3x + 4) \\[1em] \Rightarrow 2x^2 - 3x^2 + 11x - 7x + 15 - 4 = 0 \\[1em] \Rightarrow -x^2 + 4x + 11 = 0 \\[1em] \Rightarrow x^2 - 4x - 11 = 0 \text{ ( On multiplying equation by -1) }\end{aligned}$$

Comparing it with ax² + bx + c = 0
a= 1, b = -4, c = -11

By using the formula , x = $\frac{-b ± \sqrt{b^2 - 4ac}}{2a}$ , we obtain

$$\begin{aligned}\Rightarrow \frac{-(-4) ± \sqrt{(-4)^2 - 4 \times 1 \times -11 }}{2 \times 1} \\[1em] \Rightarrow \frac{4 ± \sqrt{16 + 44}}{2} \\[1em] \Rightarrow \frac{4 + \sqrt{60}}{2} \text{ or } \frac{4 - \sqrt{60}}{2} \\[1em] \Rightarrow \frac{4 + 2\sqrt{15}}{2} \text{ or } \frac{4 - 2\sqrt{15}}{2} \\[1em] \Rightarrow 2 + \sqrt{15} \text{ or } 2 - \sqrt{15}\end{aligned}$$

Hence, roots of the equation are $2 + \sqrt{15} , 2 - \sqrt{15}$.

The given equation is $\frac{2}{x + 2} - \frac{1}{x + 1} = \frac{4}{x + 4} - \frac{3}{x + 3}$

$$\begin{aligned}\Rightarrow \frac{2(x + 1) - (x + 2)}{(x + 2)(x + 1)} = \frac{4(x + 3) - 3(x + 4)}{(x + 4)(x + 3)} \\[1em] \Rightarrow \frac{2x + 2 - x - 2}{x^2 + x + 2x + 2} = \frac{4x + 12 - 3x - 12}{x^2 + 3x + 4x + 12} \\[1em] \Rightarrow \frac{x}{x^2 + 3x + 2} = \frac{x}{x^2 + 7x + 12} \\[1em] \Rightarrow x(x^2 + 7x + 12) = x(x^2 + 3x + 2) \\[1em] \Rightarrow (x^3 + 7x^2 + 12x) = (x^3 + 3x^2 + 2x) \\[1em] \Rightarrow x^3 + 7x^2 + 12x - x^3 - 3x^2 - 2x = 0 \\[1em] \Rightarrow 4x^2 + 10x = 0 \\[1em] \Rightarrow 2x(2x + 5) = 0 \\[1em] \Rightarrow 2x = 0 \text{ or } 2x + 5 = 0 \\[1em] \Rightarrow x = 0 \text{ or } x = -\frac{5}{2}\end{aligned}$$

Hence, roots of the equation are $0 ,-\frac{5}{2}$.

The given equation is $\frac{3x - 4}{7} + \frac{7}{3x - 4} = \frac{5}{2}$

$$\begin{aligned}\Rightarrow \frac{(3x - 4)^2 + 7^2}{7(3x - 4)} = \frac{5}{2} \text{ (On taking L.C.M. ) }\\[1em] \Rightarrow \frac{9x^2 + 16 - 24x + 49}{21x - 28} = \frac{5}{2} \\[1em] \Rightarrow 2(9x^2 - 24x + 65) = 5(21x - 28) \\[1em] \Rightarrow 18x^2 - 48x + 130 = 105x - 140 \\[1em] \Rightarrow 18x^2 - 48x - 105x + 130 + 140 = 0 \\[1em] \Rightarrow 18x^2 - 153x + 270 = 0 \\[1em] \Rightarrow 9(2x^2 - 17x + 30) = 0 \\[1em] \Rightarrow 2x^2 - 17x + 30 = 0\end{aligned}$$

Comparing it with ax² + bx + c = 0
a= 2, b = -17, c = 30

By using the formula , x = $\frac{-b ± \sqrt{b^2 - 4ac}}{2a}$ , we obtain

$$\begin{aligned}\Rightarrow \frac{-(-17) ± \sqrt{(-17)^2 - 4 \times 2 \times 30}}{2 \times 2} \\[1em] \Rightarrow \frac{17 ± \sqrt{289 - 240}}{4} \\[1em] \Rightarrow \frac{17 + \sqrt{49}}{4} \text{ or } \frac{17 - \sqrt{49}}{4} \\[1em] \Rightarrow \frac{17 + 7}{4} \text{ or } \frac{17 - 7}{4} \\[1em] \Rightarrow \frac{24}{4} \text{ or } \frac{10}{4} \\[1em] 6 \text{ or } \frac{5}{2}\end{aligned}$$

Hence, roots of the given equation are $6, \frac{5}{2}$.

The given equation is $\frac{4}{x} - 3 = \frac{5}{2x + 3}$

$$\begin{aligned}\Rightarrow \frac{4 - 3x}{x} = \frac{5}{2x + 3} \text{ (On taking L.C.M.) } \\[1em] \Rightarrow (4 - 3x)(2x + 3) = 5x \text{ (On Cross multiplication) } \\[1em] \Rightarrow 8x + 12 - 6x^2 - 9x = 5x \\[1em] \Rightarrow 6x^2 + 5x + 9x - 8x - 12 = 0 \\[1em] \Rightarrow 6x^2 + 6x - 12 = 0 \\[1em] \Rightarrow 6(x^2 + x - 2) = 0 \\[1em] x^2 + x - 2 = 0\end{aligned}$$

Comparing it with ax² + bx + c = 0
a= 1, b = 1, c = -2

By using the formula , x = $\frac{-b ± \sqrt{b^2 - 4ac}}{2a}$ , we obtain

$$\begin{aligned}\Rightarrow \frac{-(1) ± \sqrt{(1)^2 - 4 \times 1 \times -2}}{2 \times 1} \\[1em] \Rightarrow \frac{-1 ± \sqrt{1 + 8}}{2} \\[1em] \Rightarrow \frac{-1 + \sqrt{9}}{2} \text{ or } \frac{-1 - \sqrt{9}}{2} \\[1em] \Rightarrow \frac{-1 + 3}{2} \text{ or } \frac{-1 - 3}{2} \\[1em] \Rightarrow \frac{2}{2} \text{ or } \frac{-4}{2} \\[1em] 1 \text{ or } -2\end{aligned}$$

Hence, roots of the given equation are 1, -2.

The given equation is x² + (4 - 3a)x - 12a = 0

Comparing it with ax² + bx + c = 0
a= 1, b = (4 - 3a), c = -12a

By using the formula , x = $\frac{-b ± \sqrt{b^2 - 4ac}}{2a}$ , we obtain

$$\begin{aligned}\Rightarrow \frac{-(4 - 3a) ± \sqrt{(4 - 3a)^2 - 4 \times 1 \times -12a}}{2 \times 1} \\[1em] \Rightarrow \frac{(3a - 4) ± \sqrt{16 + 9a^2 - 24a + 48a}}{2} \\[1em] \Rightarrow \frac{3a - 4 ± \sqrt{16 + 9a^2 + 24a}}{2} \\[1em] \Rightarrow \frac{3a - 4 ± \sqrt{(4 + 3a)^2}}{2}\\[1em] \Rightarrow \frac{3a - 4 + |4 + 3a|}{2} \text{ or } \frac{3a - 4 - |4 + 3a|}{2} \\[1em] \Rightarrow \frac{6a}{2} \text{ or } -\frac{8}{2} \\[1em] 3a \text{ or } -4\end{aligned}$$

Hence, roots of the given equation are 3a, -4.

The given equation is 10ax² - 6x + 15ax - 9 = 0

Comparing it with ax² + bx + c = 0
a= 10a, b = (15a - 6), c = -9

By using the formula , x = $\frac{-b ± \sqrt{b^2 - 4ac}}{2a}$ , we obtain

$$\begin{aligned}\Rightarrow \frac{-(15a - 6) ± \sqrt{(15a - 6)^2 - 4 \times 10a \times -9}}{2 \times 10a} \\[1em] \Rightarrow \frac{(6 - 15a) ± \sqrt{225a^2 + 36 - 180a + 360a}}{20a} \\[1em] \Rightarrow \frac{6 - 15a ± \sqrt{225a^2 + 36 + 180a}}{20a} \\[1em] \Rightarrow \frac{6 - 15a ± \sqrt{(15a + 6)^2}}{20a}\\[1em] \Rightarrow \frac{6 - 15a + |15a + 6|}{20a} \text{ or } \frac{6 - 15a - |15a + 6|}{20a} \\[1em] \Rightarrow \frac{12}{20a} \text{ or } -\frac{30a}{20a} \\[1em] \frac{3}{5a} \text{ or } -\frac{3}{2}\end{aligned}$$

Hence, roots of the given equation are $\frac{3}{5a}, -\frac{3}{2}$.

Given,

$$\begin{aligned}(x - 1)^2 - 3x + 4 = 0 \\[0.5em] \Rightarrow x^2 + 1 - 2x - 3x + 4 = 0 \\[0.5em] \Rightarrow x^2 - 5x + 5 = 0\end{aligned}$$

Comparing it with ax² + bx + c = 0
a= 1, b = -5, c = 5

By using the formula , x = $\frac{-b ± \sqrt{b^2 - 4ac}}{2a}$ , we obtain

$$\begin{aligned}\Rightarrow \frac{-(-5) ± \sqrt{(-5)^2 - 4 \times 1 \times 5}}{2 \times 1} \\[1em] \Rightarrow \frac{5 ± \sqrt{25 - 20}}{2} \\[1em] \Rightarrow \frac{5 + \sqrt{5}}{2} \text{ or } \frac{5 - \sqrt{5}}{2} \\[1em] \Rightarrow \frac{5 + 2.2}{2} \text{ or } \frac{5 - 2.2}{2} \\[1em] \Rightarrow \frac{7.2}{2} \text{ or } \frac{2.8}{2} \\[1em] 3.6 \text{ or } 1.4\end{aligned}$$

Hence, roots of the given equation are 3.6, 1.4.

(i) The given equation is 3x² - 7x + 8 = 0

Comparing it with ax² + bx + c = 0
a= 3, b = -7, c = 8

$$\begin{aligned}\therefore \text{Discriminant }= b^2 - 4ac \\[0.5em] = (-7)^2 - 4 \times 3 \times 8 \\[0.5em] = 49 - 96 \\[0.5em] = -47 \lt 0\end{aligned}$$

Since, Discriminant < 0 , hence equation has no real roots.

(ii) The given equation is $x^2 - \frac{1}{2}x - 4 = 0$

Comparing it with ax² + bx + c = 0
a= 1, b = $-\frac{1}{2}$, c = -4

$$\begin{aligned}\therefore \text{Discriminant } = b^2 - 4ac \\[1em] = (-\frac{1}{2})^2 - 4 \times 1 \times -4 \\[1em] = \frac{1}{4} + 16 \\[1em] = \frac{65}{4} \gt 0\end{aligned}$$

Since, Discriminant > 0 , hence equation has two distinct and real roots.

By using the formula , $x = \frac{-b ± \sqrt{b^2 - 4ac}}{2a}$ , we obtain

$$\begin{aligned}\Rightarrow \frac{-(-\frac{1}{2}) ± \sqrt{(-\frac{1}{2})^2 - 4 \times 1 \times -4}}{2 \times 1} \\[1em] \Rightarrow \frac{\frac{1}{2} ± \sqrt{\frac{1}{4} + 16}}{2} \\[1em] \Rightarrow \frac{\frac{1}{2} + \sqrt{\frac{65}{4}}}{2} \text{ or } \frac{\frac{1}{2} - \sqrt{\frac{65}{4}}}{2} \\[1em] \Rightarrow \frac{\frac{1}{2} + \frac{\sqrt{65}}{2}}{2} \text{ or } \frac{\frac{1}{2} - \frac{\sqrt{65}}{2}}{2} \\[1em] \Rightarrow \frac{1 + \sqrt{65}}{4} \text{ or } \frac{1 - \sqrt{65}}{4}\end{aligned}$$

Hence, roots of the given equation are $\frac{1 + \sqrt{65}}{4} , \frac{1 - \sqrt{65}}{4}$.

(iii) $5x^2 - 6\sqrt{5}x + 9 = 0$

The given equation is $5x^2 - 6\sqrt{5}x + 9 = 0$

Comparing it with ax² + bx + c = 0
a= 5, b = -6√5, c = 9

$$\begin{aligned}\therefore \text{Discriminant } = b^2 - 4ac \\[1em] = (-6\sqrt{5})^2 - 4 \times 5 \times 9 \\[1em] = 180 - 180 \\[1em] = 0\end{aligned}$$

Since, Discriminant = 0, hence equation has two equal and real roots.

By using the formula , x = $\frac{-b ± \sqrt{b^2 - 4ac}}{2a}$ , we obtain

$$\begin{aligned}\Rightarrow \frac{-(-6\sqrt{5}) ± \sqrt{(-6\sqrt{5})^2 - 4 \times 5 \times 9}}{2 \times 5} \\[1em] \Rightarrow \frac{6\sqrt{5} ± \sqrt{180 - 180}}{10} \\[1em] \Rightarrow \frac{6\sqrt{5} + \sqrt{0}}{10} \text{ or } \frac{6\sqrt{5} - \sqrt{0}}{10} \\[1em] \Rightarrow \frac{6\sqrt{5}}{10} \text{ or } \frac{6\sqrt{5}}{10} \\[1em] \Rightarrow \frac{3\sqrt{5}}{5} \text{ or } \frac{3\sqrt{5}}{5} \\[1em] \frac{3}{\sqrt{5}} \text{ or } \frac{3}{\sqrt{5}}\end{aligned}$$

Hence, roots of the given equation are $\frac{3}{\sqrt{5}}, \frac{3}{\sqrt{5}}$.

(iv) $\sqrt{3}x^2 - 2x - \sqrt{3} = 0$

The given equation is $\sqrt{3}x^2 - 2x - \sqrt{3} = 0$.

Comparing it with ax² + bx + c = 0
a= $\sqrt{3}$, b = -2, c = -$\sqrt{3}$

$$\begin{aligned}\therefore \text{Discriminant } = b^2 - 4ac \\[1em] = (-2)^2 - 4 \times \sqrt{3} \times -\sqrt{3} \\[1em] = 4 + 12 \\[1em] = 16\end{aligned}$$

Since, Discriminant > 0 , hence equation has two distinct and real roots.

By using the formula , x = $\frac{-b ± \sqrt{b^2 - 4ac}}{2a}$ , we obtain

$$\begin{aligned}\Rightarrow \frac{-(-2) ± \sqrt{(-2)^2 - 4 \times \sqrt{3} \times -\sqrt{3}}}{2 \times \sqrt{3}} \\[1em] \Rightarrow \frac{2 ± \sqrt{4 + 12}}{2\sqrt{3}} \\[1em] \Rightarrow \frac{2 + \sqrt{16}}{2\sqrt{3}} \text{ or } \frac{2 - \sqrt{16}}{2\sqrt{3}} \\[1em] \Rightarrow \frac{2 + 4}{2\sqrt{3}} \text{ or } \frac{2- 4}{2\sqrt{3}} \\[1em] \Rightarrow \frac{6}{2\sqrt{3}} \text{ or } -\frac{2}{2\sqrt{3}} \\[1em] \sqrt{3} \text{ or } -\frac{1}{\sqrt{3}}\end{aligned}$$

Hence, roots of the given equation are $\sqrt{3}, -\frac{1}{\sqrt{3}}$.

The given equation is (4 - k)x² + 2(k + 2)x + (8k + 1) = 0

Comparing it with ax² + bx + c = 0
a= (4 - k), b = (2k + 4), c = (8k + 1)

Given,

Equation has real and equal roots

∴ b² - 4ac = 0

$$\begin{aligned}\Rightarrow (2k + 4)^2 - 4 \times (4 - k) \times (8k + 1) = 0 \\[1em] \Rightarrow (4k^2 + 16 + 16k) - 4(32k + 4 - 8k^2 - k) = 0 \\[1em] \Rightarrow 4k^2 + 16 + 16k - 128k - 16 + 32k^2 + 4k = 0 \\[1em] \Rightarrow 4k^2 + 32k^2 + 16k - 128k + 4k + 16 - 16 = 0 \\[1em] \Rightarrow 36k^2 - 108k = 0 \\[1em] \Rightarrow 36k(k - 3) = 0 \\[1em] \Rightarrow 36k = 0 \text{ or } k - 3 = 0 \\[1em] k = 0 \text{ or } k = 3\end{aligned}$$

Hence , the value of k are 0, 3.

The given equation is 3x² - 5x - 2m = 0

Comparing it with ax² + bx + c = 0
a= 3, b = -5, c = -2m

Given,

Equation has two real and distinct roots

∴ b² - 4ac > 0

$$\begin{aligned}\Rightarrow (-5)^2 - 4 \times 3 \times -2m \gt 0 \\[1em] \Rightarrow 25 + 24m \gt 0 \\[1em] \Rightarrow 24m \gt -25 \\[1em] m \gt -\frac{25}{24}\end{aligned}$$

Hence, the required value is m > $-\frac{25}{24}$

(i) Given,

$$\begin{aligned}3kx^2 = 4(kx - 1) \\[0.5em] \Rightarrow 3kx^2 - 4(kx - 1) = 0 \\[0.5em] \Rightarrow 3kx^2 - 4kx + 4 = 0 \\[0.5em]\end{aligned}$$

Comparing it with ax² + bx + c = 0
a= 3k, b = -4k, c = 4

Given,

Equation has equal roots

∴ b² - 4ac = 0

$$\begin{aligned}\Rightarrow (-4k)^2 - 4 \times 3k \times 4 = 0 \\[0.5em] \Rightarrow 16k^2 - 48k = 0 \\[0.5em] \Rightarrow 16k(k - 3) = 0 \\[0.5em] \Rightarrow 16k = 0 \text{ or } k - 3 = 0 \\[0.5em] k = 0 \text{ or } k = 3\end{aligned}$$

But k cannot be equal to 0 as that will make a = 3k = 0 , which will make roots equal to infinity.

∴ k = 3

Hence equation is 9x² - 12x + 4 = 0

$$\begin{aligned}\Rightarrow 9x^2 - 6x - 6x + 4 = 0 \\[0.5em] \Rightarrow 3x(3x - 2) - 2(3x - 2) = 0 \\[0.5em] \Rightarrow (3x - 2)(3x - 2) = 0 \\[0.5em] \Rightarrow 3x - 2 = 0 \text{ or } 3x - 2 = 0 \\[0.5em] x = \frac{2}{3} \text{ or } x = \frac{2}{3}\end{aligned}$$

Hence the value of k is 3 and the roots are $\frac{2}{3}, \frac{2}{3}.$

(ii) The given equation is (k + 4)x² +(k + 1)x + 1 = 0

Comparing it with ax² + bx + c = 0
a = (k + 4), b = (k + 1), c = 1

Given,

Equation has equal roots

∴ b² - 4ac = 0

$$\begin{aligned}\Rightarrow (k + 1)^2 - 4 \times (k + 4) \times 1 = 0 \\[1em] \Rightarrow (k^2 + 1 + 2k) - 4(k + 4) = 0 \\[1em] \Rightarrow k^2 + 1 + 2k - 4k - 16 = 0 \\[1em] \Rightarrow k^2 - 2k - 15 = 0 \\[1em] \Rightarrow k^2 - 5k + 3k - 15 = 0 \\[1em] \Rightarrow k(k - 5) + 3(k - 5) = 0 \\[1em] \Rightarrow (k + 3)(k - 5) = 0 \\[1em] \Rightarrow k + 3 = 0 \text{ or } k - 5 = 0 \\[1em] k = -3 \text{ or } k = 5\end{aligned}$$

∴ When k = -3 , equation is x² - 2x + 1 = 0

$$\begin{aligned}\Rightarrow x^2 - x - x + 1 = 0 \\[1em] \Rightarrow x(x - 1) - 1(x - 1) = 0 \\[1em] \Rightarrow (x - 1)(x - 1) = 0 \\[1em] \Rightarrow x - 1 = 0 \text{ or } x - 1 = 0 \\[1em] x = 1 \text{ or } x = 1\end{aligned}$$

∴ When k = 5 , equation is 9x² + 6x + 1 = 0

$$\begin{aligned}\Rightarrow 9x^2 + 3x + 3x + 1 = 0 \\[1em] \Rightarrow 3x(3x + 1) + 1(3x + 1) = 0 \\[1em] \Rightarrow (3x + 1)(3x + 1) = 0 \\[1em] \Rightarrow 3x + 1 = 0 \text{ or } 3x + 1 = 0 \\[1em] x = -\frac{1}{3} \text{ or } -\frac{1}{3}\end{aligned}$$

k = -3, 5
When k = -3 , roots are 1, 1
When k = 5 , roots are $-\frac{1}{3}, -\frac{1}{3}$

Let first number be x

Since difference between two numbers is 3 hence, the other number is (x + 3).

Given , sum of the squares of number = 117

∴ x² + (x + 3)² = 117

$$\begin{aligned}\Rightarrow x^2 + x^2 + 9 + 6x = 117 \\[1em] \Rightarrow 2x^2 + 6x + 9 - 117 = 0 \\[1em] \Rightarrow 2x^2 + 6x - 108 = 0 \\[1em] \Rightarrow 2(x^2 + 3x - 54) = 0 \\[1em] \Rightarrow x^2 + 3x - 54 = 0 \\[1em] \Rightarrow x^2 + 9x - 6x - 54 = 0 \\[1em] \Rightarrow x(x + 9) - 6(x + 9) = 0 \\[1em] \Rightarrow (x - 6)(x + 9) = 0 \\[1em] \Rightarrow x - 6 = 0 \text{ or } x + 9 = 0 \\[1em] x = 6 \text{ or } x = -9\end{aligned}$$

Since, numbers are natural hence, x ≠ -9.

∴ x = 6 , x + 3 = 9.

Hence, the required numbers are 6, 9.

Let the larger number be x , so the smaller number is 16 - x.

Given, twice the square of the larger part exceeds the square of the smaller part by 164

$$\begin{aligned}\therefore 2x^2 - (16 - x)^2 = 164 \\[1em] \Rightarrow 2x^2 - (256 + x^2 - 32x) = 164 \\[1em] \Rightarrow 2x^2 - 256 - x^2 + 32x - 164 = 0 \\[1em] \Rightarrow x^2 + 32x - 420 = 0 \\[1em] \Rightarrow x^2 + 42x - 10x - 420 = 0 \\[1em] \Rightarrow x(x + 42) - 10(x + 42) = 0 \\[1em] \Rightarrow (x - 10)(x + 42) = 0 \\[1em] \Rightarrow x - 10 = 0 \text{ or } x + 42 = 0 \\[1em] x = 10 \text{ or } x = -42\end{aligned}$$

Since, numbers are natural hence, x ≠ -42.

∴ x = 10 , 16 - x = 6.

Hence, the required numbers are 10, 6.

Since, the numbers are in the ratio 3 : 4, hence the numbers be 3x and 4x.

Given , difference between their squares = 175

∴ (4x)² - (3x)² = 175

$$\begin{aligned}\Rightarrow 16x^2 - 9x^2 = 175 \\[1em] \Rightarrow 7x^2 = 175 \\[1em] \Rightarrow x^2 = \frac{175}{7} \\[1em] \Rightarrow x^2 = 25 \\[1em] \Rightarrow x^2 - 25 = 0 \\[1em] \Rightarrow (x - 5)(x + 5) = 0 \\[1em] \Rightarrow x - 5 = 0 \text{ or } x + 5 = 0 \\[1em] x = 5 \text{ or } x = -5\end{aligned}$$

Since, numbers are natural hence, x ≠ -5.

∴ x = 5, 3x = 15 , 4x = 20.

Hence, the required numbers are 15, 20.

Area of a square = (side)²

∴ Area of first square = x² and Area of second square = (x + 4)²

Given, sum of areas of two squares is = 656 cm²

∴ x² + (x + 4)² = 656

$$\begin{aligned}\Rightarrow x^2 + x^2 + 16 + 8x = 656 \\[1em] \Rightarrow 2x^2 + 8x + 16 - 656 = 0 \\[1em] \Rightarrow 2x^2 + 8x - 640 = 0 \\[1em] \Rightarrow 2(x^2 + 4x - 320) = 0 \\[1em] \Rightarrow x^2 + 4x - 320 = 0 \\[1em] \Rightarrow x^2 + 20x - 16x - 320 = 0 \\[1em] \Rightarrow x(x + 20) - 16(x + 20) = 0 \\[1em] \Rightarrow (x + 20)(x - 16) = 0 \\[1em] \Rightarrow x + 20 = 0 \text{ or } x - 16 = 0 \\[1em] x = -20 \text{ or } x = 16\end{aligned}$$

Since, length cannot be negative hence x ≠ -20.

∴ x = 16 , x + 4 = 20.

Hence, the sides of two squares are 16 cm and 20 cm.

Let the value of breadth be x metre

So, length = (x + 12) metre

Perimeter = 2(Length + Breadth) = 2(x + x + 12) = 2(2x + 12) = (4x + 24) metre

Area = Length $\times$ Breadth = x(x + 12) = (x² + 12x) metre²

Given, area is equal to 4 times the perimeter

∴ x² + 12x = 4(4x + 24)

$$\begin{aligned}\Rightarrow x^2 + 12x = 16x + 96 \\[1em] \Rightarrow x^2 + 12x - 16x - 96 = 0 \\[1em] \Rightarrow x^2 - 4x - 96 = 0 \\[1em] \Rightarrow x^2 - 12x + 8x - 96 = 0 \\[1em] \Rightarrow x(x - 12) + 8(x - 12) = 0 \\[1em] \Rightarrow (x - 12)(x + 8) = 0 \\[1em] \Rightarrow x - 12 = 0 \text{ or } x + 8 = 0 \\[1em] x = 12 \text{ or } x = -8 \\[1em]\end{aligned}$$

Since, breadth cannot be negative hence, x ≠ -8

∴ x = 12 , x + 12 = 24

Hence, the length of the garden is 24m and breadth is 12m.

Let x metres be the length of the side opposite to unfenced side, then length of each of two others sides = $\frac{1}{2}(30 -x).$

According to question,

$$\begin{aligned}\Rightarrow x \times \frac{1}{2}(30 - x) = 100 \\[1em] \Rightarrow 15x - \frac{x^2}{2} = 100 \\[1em] \Rightarrow \frac{30x - x^2}{2} = 100 \text{ (On taking L.C.M.) }\\[1em] \Rightarrow 30x - x^2 = 200 \text { (On cross multiplying) } \\[1em] \Rightarrow x^2 - 30x + 200 = 0 \\[1em] \Rightarrow x^2 - 20x - 10x + 200 = 0 \\[1em] \Rightarrow x(x - 20) - 10(x - 20) = 0 \\[1em] \Rightarrow (x - 10)(x - 20) = 0 \\[1em] \Rightarrow x - 10 = 0 \text{ or } x - 20 = 0 \\[1em] x = 10 \text{ or } x = 20\end{aligned}$$

∴ if x = 10, $\frac{1}{2}(30 - x) :$

$$\begin{aligned}= \frac{1}{2}(30 - 10) \\[1em] = \frac{1}{2} \times 20 \\[1em] = 10\end{aligned}$$

∴ if x = 20, $\frac{1}{2}(30 - x) :$

$$\begin{aligned}= \frac{1}{2}(30 - 20) \\[1em] = \frac{1}{2} \times 10 \\[1em] = 5\end{aligned}$$

Hence, the dimensions of garden are 10m x 10m or 20m x 5m.

Let the shortest side be x metre

According to question,

Third side = (x + 1)m

Hypotenuse = (2x - 1)m

For right angled triangle,

(Hypotenuse)² = (Perpendicular)² + (Base)²

$$\begin{aligned}\therefore (2x - 1)^2 = (x + 1)^2 + x^2 \\[1em] \Rightarrow 4x^2 + 1 - 4x = x^2 + 1 + 2x + x^2 \\[1em] \Rightarrow 4x^2 + 1 - 4x = 2x^2 + 2x + 1 \\[1em] \Rightarrow 4x^2 - 2x^2 - 4x - 2x + 1 - 1 = 0 \\[1em] \Rightarrow 2x^2 - 6x = 0 \\[1em] \Rightarrow 2x(x - 3) = 0 \\[1em] \Rightarrow 2x = 0 \text{ or } x - 3 = 0 \\[1em] x = 0 \text{ or } x = 3\end{aligned}$$

Since, side's length cannot be equal to 0 , hence x ≠ 0.

∴ x = 3, (x + 1) = 4, (2x - 1) = 5

Hence, the sides of triangle are
Shortest side = 3 m
Hypotenuse = 5 m
Third side = 4 m

Sum of length of other two sides + Hypotenuse = 112

or, Sum of length of other two sides = 112 - Hypotenuse

∴ Sum of length of other two sides = 112 - 50 = 62 cm

Let the length of perpendicular = x cm

So, the length of base = (62 - x) cm

For right angled triangle,

(Perpendicular)² + (Base)² = (Hypotenuse)²

$$\begin{aligned}\therefore x^2 + (62 - x)^2 = (50)^2 \\[1em] \Rightarrow x^2 + 3844 + x^2 - 124x = 2500 \\[1em] \Rightarrow 2x^2 - 124x + 3844 - 2500 = 0 \\[1em] \Rightarrow 2x^2 - 124x + 1344 = 0 \\[1em] \Rightarrow 2(x^2 - 62x + 672) = 0 \\[1em] \Rightarrow x^2 - 62x + 672 = 0 \\[1em] \Rightarrow x^2 - 48x - 14x + 672 = 0 \\[1em] \Rightarrow x(x - 48) - 14(x - 48) = 0 \\[1em] \Rightarrow (x - 48)(x - 14) = 0 \\[1em] \Rightarrow x - 48 = 0 \text{ or } x - 14 = 0 \\[1em] x = 48 \text{ or } x = 14\end{aligned}$$

∴ The length of other two sides are 48 cm and 14 cm.

$$\begin{aligned}\text{Area} = \frac{1}{2} \times \text{Perpendicular} \times \text{Base} \\[1em] = \frac{1}{2} \times 48 \times 14 \\[1em] = 336\end{aligned}$$

Hence, the area of triangle is 336 cm^2.

Let the speed of the stream be x km/h.

Speed of boat in -

Still water = 11 km/h

Upstream = (11 - x) km/h

Downstream = (11 + x) km/h

Given,

Boat can go 12 km upstream and return downstream to original point in 2 hours 45 minutes.

Converting 2 hours 45 minutes to hours:

2 hours 45 minutes = ((2 x 60) + 45) minutes = 165 minutes.

165 minutes = $\frac{165}{60}$ hours

$$\begin{aligned}\therefore \frac{12}{11 - x} + \frac{12}{11 + x} = \frac{165}{60} \\[1em] \Rightarrow \frac{12(11 + x) + 12(11 - x)}{(11 + x)(11 - x)} = \frac{165}{60} \\[1em] \Rightarrow \frac{132 + 12x + 132 - 12x}{121 - x^2} = \frac{165}{60} \\[1em] \Rightarrow \frac{264}{121 - x^2} = \frac{11}{4}\\[1em] \Rightarrow 264 \times 4 = 11(121 - x^2) \\[1em] \Rightarrow 1056 = 1331 - 11x^2 \\[1em] \Rightarrow 11x^2 = 1331 - 1056 \\[1em] \Rightarrow 11x^2 = 275 \\[1em] \Rightarrow x^2 = \frac{275}{11} \\[1em] \Rightarrow x^2 = 25 \\[1em] \Rightarrow x^2 - 25 = 0 \\[1em] \Rightarrow (x - 5)(x + 5) = 0 \\[1em] \Rightarrow x = 5, -5\end{aligned}$$

Since speed cannot be negative hence, x ≠ -5.

Hence, speed of stream is 5 km/h.

In first case,

Amount spent = ₹2800

Price of each plant = ₹x

No. of plants = $\frac{2800}{x}$

In second case,

Amount spent = ₹2730

Price of plant is reduced by ₹1 , so new price = ₹(x - 1)

Given, in this case 10 more plants were received hence, no. of plants = $\frac{2800}{x} + 10$

According to question,

$$\begin{aligned}\Rightarrow \big(\frac{2800}{x} + 10\big)(x - 1) = 2730 \\[1em] \Rightarrow \big(\frac{2800 + 10x}{x}\big)(x - 1) = 2730 \\[1em] \Rightarrow \frac{(2800 + 10x)(x - 1)}{x} = 2730 \\[1em] \Rightarrow \frac{2800x - 2800 + 10x^2 - 10x}{x} = 2730 \\[1em] \Rightarrow 10x^2 + 2790x - 2800 = 2730x \text{ (On cross multiplication) } \\[1em] \Rightarrow 10x^2 + 60x - 2800 = 0 \\[1em] \Rightarrow x^2 + 6x - 280 = 0 \text{ (On dividing by 10) } \\[1em] \Rightarrow x^2 + 20x - 14x - 280 = 0 \\[1em] \Rightarrow x(x + 20) - 14(x + 20) = 0 \\[1em] \Rightarrow (x - 14)(x + 20) = 0 \\[1em] x = 14 \text{ or } x = -20\end{aligned}$$

Since price cannot be negative hence , x ≠ -20

Hence, the value of x is 14.

Let the present age of Mr. Pratap be x years.

After 40 years his age will be (x + 40) years

32 years before his age was (x - 32) years

According to question ,

$$\begin{aligned}\Rightarrow (x - 32)^2 = (x + 40) \\[1em] \Rightarrow x^2 + 1024 - 64x = x + 40 \\[1em] \Rightarrow x^2 - 64x - x + 1024 - 40 = 0 \\[1em] \Rightarrow x^2 - 65x + 984 = 0 \\[1em] \Rightarrow x^2 - 24x - 41x + 984 = 0 \\[1em] \Rightarrow x(x - 24) - 41(x - 24) = 0 \\[1em] \Rightarrow (x - 41)(x - 24) = 0 \\[1em] \Rightarrow x - 41 = 0 \text{ or } x - 24 = 0 \\[1em] x = 41 \text{ or } x = 24\end{aligned}$$

But x ≠ 24 as it is less than 32.

∴ x = 41.

The present age of Mr. Pratap is 41 years.

Let no. of people be x.

Total expense = ₹ 18,000

Expense per person = ₹ $\frac{18000}{x}$

Given,

If three more people join them, then the share of each reduces by ₹ 3,000.

No, of people now = x + 3

Expense per person = ₹ $\frac{18000}{x + 3}$

According to question,

$$\begin{aligned}\Rightarrow \frac{18000}{x} - \frac{18000}{x + 3} = 3000 \\[1em] \Rightarrow \frac{18000(x + 3) - 18000x}{x(x + 3)} = 3000 \\[1em] \Rightarrow \frac{18000x + 54000 - 18000x}{x^2 + 3x} = 3000 \\[1em] \Rightarrow \frac{54000}{x^2 + 3x} = 3000 \\[1em] \Rightarrow x^2 + 3x = \frac{54000}{3000} \\[1em] \Rightarrow x^2 + 3x = 18 \\[1em] \Rightarrow x^2 + 3x - 18 = 0 \\[1em] \Rightarrow x^2 + 6x - 3x - 18 = 0 \\[1em] \Rightarrow x(x + 6) - 3(x + 6) = 0 \\[1em] \Rightarrow (x - 3)(x + 6) = 0 \\[1em] \Rightarrow x - 3 = 0 \text{ or } x + 6 = 0 \\[1em] \Rightarrow x = 3 \text{ or } x = -6.\end{aligned}$$

Since, no. of people cannot be negative.

∴ x = 3.

Hence, original number of people = 3.

Given,

A car travels a distance of 72 km at a certain average speed of x km per hour and then travels a distance of 81 km at an average speed of 6 km per hour more than its original average speed.

Total time taken to complete the journey = 3 hours

$$\begin{aligned}\therefore \frac{72}{x} + \frac{81}{x + 6} = 3 \\[1em] \Rightarrow \frac{72(x + 6) + 81x}{x(x + 6)} = 3 \\[1em] \Rightarrow 72(x + 6) + 81x = 3x(x + 6) \\[1em] \Rightarrow 72x + 432 + 81x = 3x^2 + 18x \\[1em] \Rightarrow 3x^2 + 18x - 72x - 81x - 432 = 0 \\[1em] \Rightarrow 3x^2 - 135x - 432 = 0 \\[1em] \Rightarrow 3(x^2 - 45x - 144) = 0 \\[1em] \Rightarrow x^2 - 45x - 144 = 0 \\[1em] \Rightarrow x^2 - 48x + 3x - 144 = 0 \\[1em] \Rightarrow x(x - 48) + 3(x - 48) = 0 \\[1em] \Rightarrow (x + 3)(x - 48) = 0 \\[1em] \Rightarrow x + 3 = 0 \text{ or } x - 48 = 0 \\[1em] \Rightarrow x = -3 \text{ or } x = 48.\end{aligned}$$

As speed cannot be negative in this case,

Hence, the original speed = 48 km/hr.