Quadratic Equations

Given,

⇒ 4x² - 9 = 0

⇒ 4x² = 9

⇒ x² = $\frac{9}{4}$

⇒ x = $\sqrt{\frac{9}{4}} = \pm \frac{3}{2}$

Hence, Option 4 is the correct option.

Given,

⇒ (x - 3)(x + 5) = 0

⇒ (x - 3) or (x + 5) = 0

⇒ x - 3 = 0 or x + 5 = 0

⇒ x = 3 or x = -5.

Hence, Option 3 is the correct option.

Since, 4 is the root of the equation x² + kx - 4 = 0.

∴ It will satisfy the equation x² + kx - 4 = 0.

∴ 4² + 4k - 4 = 0

⇒ 16 + 4k - 4 = 0

⇒ 4k + 12 = 0

⇒ 4k = -12

⇒ k = -$\frac{12}{4}$ = -3.

Hence, Option 2 is the correct option.

Given,

x = 2 satisfies the equation 2x² - 3x + k = 0.

∴ 2(2)² - 3(2) + k = 0

⇒ 2(4) - 6 + k = 0

⇒ 8 - 6 + k = 0

⇒ k + 2 = 0

⇒ k = -2.

Hence, Option 1 is the correct option.

Given,

⇒ x² - 7x = 0

⇒ x(x - 7) = 0

⇒ x = 0 or x - 7 = 0

⇒ x = 0 or x = 7.

Hence, Option 4 is the correct option.

Since, $\sqrt{\frac{2}{3}}$ is a solution of equation 3x² + mx + 2 = 0.

$$\begin{aligned}\Rightarrow 3.\Big(\sqrt{\frac{2}{3}}\Big)^2 + m\Big(\sqrt{\frac{2}{3}}\Big) + 2 = 0 \\[1em] \Rightarrow 3 \times \frac{2}{3} + 2 + m\Big(\sqrt{\frac{2}{3}}\Big) = 0 \\[1em] \Rightarrow 2 + 2 + m\Big(\sqrt{\frac{2}{3}}\Big) = 0 \\[1em] \Rightarrow 4 + m\Big(\sqrt{\frac{2}{3}}\Big) = 0 \\[1em] \Rightarrow m\Big(\sqrt{\frac{2}{3}}\Big) = -4 \\[1em] \Rightarrow m = -4 \times \sqrt{\frac{3}{2}} \\[1em] \Rightarrow m = -2\sqrt{2}\sqrt{3} \\[1em] \Rightarrow m = -2\sqrt{6}.\end{aligned}$$

Hence, value of m is $-2\sqrt{6}.$

Since, $\frac{2}{3}$ is a solution of equation mx² + nx + 6 = 0.

Substituting $\frac{2}{3}$ in mx² + nx + 6 = 0,

$$\begin{aligned}\Rightarrow m\Big(\frac{2}{3}\Big)^2 + n \times \frac{2}{3} + 6 = 0 \\[1em] \Rightarrow m \times \frac{4}{9} + \frac{2n}{3} + 6 = 0 \\[1em] \Rightarrow \frac{4m}{9} + \frac{2n}{3} + 6 = 0 \\[1em] \Rightarrow \frac{4m + 6n + 54}{9} = 0 \\[1em] \Rightarrow 4m + 6n + 54 = 0 \\[1em] \Rightarrow 2(2m + 3n + 27) = 0 \\[1em] \Rightarrow 2m + 3n + 27 = 0 \\[1em] \Rightarrow 2m + 3n = -27 .......(i)\end{aligned}$$

Since, 1 is a solution of equation mx² + nx + 6 = 0.

Substituting 1 in mx² + nx + 6 = 0,

$$\begin{aligned}\Rightarrow m(1)^2 + n(1) + 6 = 0 \\[1em] \Rightarrow m + n + 6 = 0 \\[1em] \Rightarrow m = -(n + 6) ........(ii)\end{aligned}$$

Sustituting above value of m in eq. 1 we get,

$$\begin{aligned}\Rightarrow 2.-(n + 6) + 3n = -27 \\[1em] \Rightarrow -2(n + 6) + 3n = -27 \\[1em] \Rightarrow -2n - 12 + 3n = -27 \\[1em] \Rightarrow n - 12 = -27 \\[1em] \Rightarrow n = -27 + 12 \\[1em] \Rightarrow n = -15.\end{aligned}$$

Substituting value of n in (ii) we get,

$$\begin{aligned}\Rightarrow m = -(-15 + 6) \\[1em] \Rightarrow m = -(-9) \\[1em] \Rightarrow m = 9.\end{aligned}$$

Hence, the value of m = 9 and n = -15.

Given,

⇒ x² - 6x - 7 = 0

⇒ x² - 7x + x - 7 = 0

⇒ x(x - 7) + 1(x - 7) = 0

⇒ (x + 1)(x - 7) = 0

⇒ x + 1 = 0 or x - 7 = 0

⇒ x = -1 or x = 7.

Hence, Option 3 is the correct option.

Given,

⇒ x(x + 8) + 12 = 0

⇒ x² + 8x + 12 = 0

⇒ x² + 2x + 6x + 12 = 0

⇒ x(x + 2) + 6(x + 2) = 0

⇒ (x + 6)(x + 2) = 0

⇒ x + 6 = 0 or x + 2 = 0

⇒ x = -6 or x = -2.

Hence, Option 2 is the correct option.

Given,

One root of equation (p - 3)x² + x + p = 0 is 2.

∴ x = 2 satisfies the equation.

∴ (p - 3)(2)² + 2 + p = 0

⇒ 4(p - 3) + 2 + p = 0

⇒ 4p - 12 + 2 + p = 0

⇒ 5p - 10 = 0

⇒ 5p = 10

⇒ p = $\frac{10}{5}$ = 2.

Hence, Option 2 is the correct option.

Given,

$$\begin{aligned}\Rightarrow x + \frac{1}{x} = 2.5 \\[1em] \Rightarrow \frac{x^2 + 1}{x} = \frac{25}{10} \\[1em] \Rightarrow 10(x^2 + 1) = 25x \\[1em] \Rightarrow 10x^2 + 10 = 25x \\[1em] \Rightarrow 10x^2 - 25x + 10 = 0 \\[1em] \Rightarrow 10x^2 - 20x - 5x + 10 = 0 \\[1em] \Rightarrow 10x(x - 2) - 5(x - 2) = 0 \\[1em] \Rightarrow (x - 2)(10x - 5) = 0 \\[1em] \Rightarrow x - 2 = 0 \text{ or } 10x - 5 =0 \\[1em] \Rightarrow x = 2 \text{ or } 10x = 5 \\[1em] \Rightarrow x = 2 \text{ or } x = \frac{5}{10} = \frac{1}{2}.\end{aligned}$$

Hence, Option 3 is the correct option.

For quadratic equation :

2x + $\frac{5}{x}$ = 5

If x = 0, $\frac{5}{x}$ will not be defined.

∴ x ≠ 0.

Hence, Option 1 is the correct option.

⇒ (2x - 3)² = 7²

⇒ (2x - 3)² - 7² = 0

As, (a² - b²) = (a + b)(a - b)

⇒ (2x - 3 + 7)(2x - 3 - 7) = 0

⇒ (2x + 4)(2x - 10) = 0

⇒ 2x + 4 = 0 or 2x - 10 = 0      [Using Zero-product rule]

⇒ 2x = -4 or 2x = 10

⇒ x = -2 or x = 5.

Hence, x = -2 or x = 5.

⇒ (x + 1)(2x + 8) = (x + 7)(x + 3)

⇒ 2x² + 8x + 2x + 8 = x² + 3x + 7x + 21

⇒ 2x² - x² + 10x + 8 = 10x + 21

⇒ x² + 10x - 10x = 21 - 8

⇒ x² = 13

⇒ x = $\pm \sqrt{13}$

Hence, x = $+ \sqrt{13}$ or x = $- \sqrt{13}$.

i.e., x = 3.61 or x = -3.61

4(2x - 3)² - (2x - 3) - 14 = 0

⇒ 4(4x² + 9 - 12x) - 2x + 3 - 14 = 0

⇒ 16x² + 36 - 48x - 2x - 11 = 0

⇒ 16x² - 50x + 25 = 0

⇒ 16x² - 40x - 10x + 25 = 0

⇒ 8x(2x - 5) - 5(2x - 5) = 0

⇒ (8x - 5)(2x - 5) = 0

⇒ 8x - 5 = 0 or 2x - 5 = 0      [Using Zero-product rule]

⇒ 8x = 5 or 2x = 5

⇒ x = $\frac{5}{8}$ or x = $\frac{5}{2}$.

Hence, x = $\frac{5}{8} \text{ or } x = \frac{5}{2}$.

2x² - 9x + 10 = 0

⇒ 2x² - 4x - 5x + 10 = 0

⇒ 2x(x - 2) - 5(x - 2) = 0

⇒ (2x - 5)(x - 2) = 0

⇒ 2x - 5 = 0 or x - 2 = 0      [Using Zero-product rule]

⇒ 2x = 5 or x = 2

⇒ x = $\frac{5}{2}$ or x = 2.

(i) Since, x ∈ N

Hence, value of x = 2.

(ii) Since, x ∈ Q

Hence, value of x = $\frac{5}{2}$ or 2.

Given,

$$\begin{aligned}\phantom {\Rightarrow} \frac{x - 3}{x + 3} + \frac{x + 3}{x - 3} = 2\frac{1}{2} \\[1em] \Rightarrow \frac{(x - 3)(x - 3) + (x + 3)(x + 3)}{(x + 3)(x - 3)} = \frac{5}{2} \\[1em] \Rightarrow \frac{x^2 - 3x - 3x + 9 + x^2 + 3x + 3x + 9}{x^2 + 3x - 3x - 9} = \frac{5}{2} \\[1em] \Rightarrow \frac{x^2 - 6x + 9 + x^2 + 6x + 9}{x^2 - 9} = \frac{5}{2} \\[1em] \Rightarrow \frac{2x^2 + 18}{x^2 - 9} = \frac{5}{2} \\[1em] \Rightarrow 2(2x^2 + 18) = 5(x^2 - 9) \\[1em] \Rightarrow 4x^2 + 36 = 5x^2 - 45 \\[1em] \Rightarrow 5x^2 - 4x^2 = 36 + 45 \\[1em] \Rightarrow x^2 = 81 \\[1em] \Rightarrow x = \sqrt{81} = \pm 9.\end{aligned}$$

Hence, value of x = +9 or -9.

Given,

$$\begin{aligned}\Rightarrow \frac{4}{x + 2} - \frac{1}{x + 3} = \frac{4}{2x + 1} \\[1em] \Rightarrow \frac{4(x + 3) - (x + 2)}{(x + 2)(x + 3)} = \frac{4}{2x + 1} \\[1em] \Rightarrow \frac{4x + 12 - x - 2}{x^2 + 3x + 2x + 6} = \frac{4}{2x + 1} \\[1em] \Rightarrow \frac{3x + 10}{x^2 + 5x + 6} = \frac{4}{2x + 1} \\[1em] \Rightarrow (3x + 10)(2x + 1) = 4(x^2 + 5x + 6) \\[1em] \Rightarrow 6x^2 + 3x + 20x + 10 = 4x^2 + 20x + 24 \\[1em] \Rightarrow 6x^2 - 4x^2 + 23x - 20x + 10 -24 = 0 \\[1em] \Rightarrow 2x^2 + 3x - 14 = 0 \\[1em] \Rightarrow 2x^2 + 7x - 4x - 14 = 0 \\[1em] \Rightarrow x(2x + 7) - 2(2x + 7) = 0 \\[1em] \Rightarrow (x - 2)(2x + 7) = 0 \\[1em] \Rightarrow (x - 2) = 0 \text{ or } 2x + 7 = 0 \\[1em] \Rightarrow x = 2 \text{ or } 2x = -7 \\[1em] \Rightarrow x = 2 \text{ or } x = -\frac{7}{2}.\end{aligned}$$

Hence, value of x = 2 or -$\frac{7}{2}$.

Given,

$$\begin{aligned}\Rightarrow \frac{5(x + 6) - 3(x - 2)}{(x - 2)(x + 6)} = \frac{4}{x} \\[1em] \Rightarrow \frac{5x + 30 - 3x + 6}{x^2 + 6x - 2x - 12} = \frac{4}{x} \\[1em] \Rightarrow \frac{2x + 36}{x^2 + 4x - 12} = \frac{4}{x} \\[1em] \Rightarrow x(2x + 36) = 4(x^2 + 4x - 12) \\[1em] \Rightarrow 2x^2 + 36x = 4x^2 + 16x - 48 \\[1em] \Rightarrow 4x^2 - 2x^2 + 16x - 36x - 48 = 0 \\[1em] \Rightarrow 2x^2 - 20x - 48 = 0 \\[1em] \Rightarrow 2(x^2 - 10x - 24) = 0 \\[1em] \Rightarrow x^2 - 10x - 24 = 0 \\[1em] \Rightarrow x^2 - 12x + 2x - 24 = 0 \\[1em] \Rightarrow x(x - 12) + 2(x - 12) = 0 \\[1em] \Rightarrow (x + 2)(x - 12) = 0 \\[1em] \Rightarrow x = -2 \text{ or } x = 12.\end{aligned}$$

Hence, value of x = -2 or 12.

Given,

$$\begin{aligned}\Rightarrow \Big(1 + \frac{1}{x + 1}\Big)\Big(1 - \frac{1}{x - 1}\Big) = \frac{7}{8} \\[1em] \Rightarrow \frac{x + 1 + 1}{x + 1} \times \frac{x - 1 - 1}{x - 1} = \frac{7}{8} \\[1em] \Rightarrow \frac{x + 2}{x + 1} \times \frac{x - 2}{x - 1} = \frac{7}{8} \\[1em] \Rightarrow \frac{x^2 - 4}{x^2 - 1} = \frac{7}{8} \\[1em] \Rightarrow 8(x^2 - 4) = 7(x^2 - 1) \\[1em] \Rightarrow 8x^2 - 32 = 7x^2 - 7 \\[1em] \Rightarrow 8x^2 - 7x^2 = - 7 + 32 \\[1em] \Rightarrow x^2 = 25 \\[1em] \Rightarrow x = \sqrt{25} \\[1em] \Rightarrow x = \pm 5.\end{aligned}$$

Hence, value of x = -5 or +5.

(i) Since, {3, 5} is solution set.

It means 3 and 5 are roots of the equation,

∴ x = 3 or x = 5

⇒ x - 3 = 0 or x - 5 = 0

⇒ (x - 3)(x - 5) = 0

⇒ (x² - 5x - 3x + 15) = 0

⇒ x² - 8x + 15 = 0.

Hence, quadratic equation with solution set {3, 5} = x² - 8x + 15 = 0.

(ii) Since, {-2, 3} is solution set.

It means -2 and 3 are roots of the equation,

∴ x = -2 or x = 3

⇒ x + 2 = 0 or x - 3 = 0

⇒ (x + 2)(x - 3) = 0

⇒ (x² - 3x + 2x - 6) = 0

⇒ x² - x - 6 = 0.

Hence, quadratic equation with solution set {-2, 3} = x² - x - 6 = 0.

Given,

$$\begin{aligned}\Rightarrow \frac{x}{3} + \frac{3}{6 - x} = \frac{2(6 + x)}{15} \\[1em] \Rightarrow \frac{x(6 - x) + 9}{3(6 - x)} = \frac{12 + 2x}{15} \\[1em] \Rightarrow \frac{6x - x^2 + 9}{18 - 3x} = \frac{12 + 2x}{15} \\[1em] \Rightarrow 15(6x - x^2 + 9) = (12 + 2x)(18 - 3x) \\[1em] \Rightarrow 90x - 15x^2 + 135 = 216 - 36x + 36x - 6x^2 \\[1em] \Rightarrow 90x - 15x^2 + 135 = 216 - 6x^2 \\[1em] \Rightarrow 15x^2 - 6x^2 - 90x + 216 - 135 = 0 \\[1em] \Rightarrow 9x^2 - 90x + 81 = 0 \\[1em] \Rightarrow 9(x^2 - 10x + 9) = 0 \\[1em] \Rightarrow x^2 - 10x + 9 = 0 \\[1em] \Rightarrow x^2 - 9x - x + 9 = 0 \\[1em] \Rightarrow x(x - 9) - 1(x - 9) = 0 \\[1em] \Rightarrow (x - 1)(x - 9) = 0 \\[1em] \Rightarrow x = 1 \text{ or } x = 9.\end{aligned}$$

Hence, x = 1 or x = 9.

Given,

9x² + $\frac{3x}{4} + 2$ = 0

$\Rightarrow \frac{36x^2 + 3x + 8}{4} = 0$

⇒ 36x² + 3x + 8 = 0

Comparing 36x² + 3x + 8 = 0 with ax² + bx + c = 0 we get,

a = 36, b = 3 and c = 8

D = b² - 4ac = (3)² - 4(36)(8) = 9 - 1152 = -1143.

Since, D < 0 hence, roots are imaginary.

There is no possibility of real roots.

Given,

⇒ a + 7 = 0 and b + 10 = 0

⇒ a = -7 and b = -10

Substituting value of a and b in 12x² = ax - b,

⇒ 12x² = (-7)x - (-10)

⇒ 12x² = -7x + 10

⇒ 12x² + 7x - 10 = 0

⇒ 12x² + 15x - 8x - 10 = 0

⇒ 3x(4x + 5) - 2(4x + 5) = 0

⇒ (3x - 2)(4x + 5) = 0

⇒ 3x - 2 = 0 or 4x + 5 = 0      [Using Zero-product rule]

⇒ 3x = 2 or 4x = - 5

⇒ x = $\frac{2}{3} \text{ or } x = -\frac{5}{4}$.

Hence, x = $\frac{2}{3} \text{ or } -\frac{5}{4}$.

Substituting, 2x + 3 = y in 4(2x + 3)² - (2x + 3) - 14 = 0 we get,

⇒ 4y² - y - 14 = 0

⇒ 4y² - 8y + 7y - 14 = 0

⇒ 4y(y - 2) + 7(y - 2) = 0

⇒ (4y + 7)(y - 2) = 0

⇒ 4y + 7 = 0 or y - 2 = 0

⇒ 4y = -7 or y = 2

⇒ y = $-\frac{7}{4}$ or y = 2.

∴ 2x + 3 = $-\frac{7}{4}$ or 2x + 3 = 2

⇒ 2x = $-\frac{7}{4} - 3$ or 2x = 2 - 3

⇒ 2x = $\frac{-7 - 12}{4}$ or 2x = -1

⇒ 2x = $-\frac{19}{4} \text{ or } x = -\frac{1}{2}$

⇒ x = $-\frac{19}{8} \text{ or } x = -\frac{1}{2}$

Hence, x = $-\frac{19}{8} \text{ or } x = -\frac{1}{2}$.

Given,

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

Hence, x = a + b or -(a + b).

Given,

$$\begin{aligned}\Rightarrow \Big(\frac{1200}{x} + 2\Big)(x - 10) - 1200 = 60 \\[1em] \Rightarrow \frac{(1200 + 2x)(x - 10)}{x} = 60 + 1200 \\[1em] \Rightarrow \frac{1200x - 12000 + 2x^2 - 20x }{x} = 1260 \\[1em] \Rightarrow 2x^2 + 1180x - 12000 = 1260x \\[1em] \Rightarrow 2x^2 + 1180x -1260x - 12000 = 0 \\[1em] \Rightarrow 2x^2 - 80x - 12000 = 0 \\[1em] \Rightarrow 2(x^2 - 40x - 6000) = 0 \\[1em] \Rightarrow x^2 - 40x - 6000 = 0 \\[1em] \Rightarrow x^2 - 100x + 60x - 6000 = 0 \\[1em] \Rightarrow x(x - 100) + 60(x - 100) = 0 \\[1em] \Rightarrow (x + 60)(x - 100) = 0 \\[1em] \Rightarrow (x + 60) = 0 \text{ or } (x - 100) = 0 \\[1em] \Rightarrow x = -60 \text{ or } x = 100.\end{aligned}$$

Hence, x = -60 or 100.

Given,

$$\begin{aligned}\Rightarrow \frac{14}{x + 3} - 1 = \frac{5}{x + 1} \\[1em] \Rightarrow \frac{14 - (x + 3)}{x + 3} = \frac{5}{x + 1} \\[1em] \Rightarrow \frac{14 - x - 3}{x + 3} = \frac{5}{x + 1} \\[1em] \Rightarrow \frac{11 - x}{x + 3} = \frac{5}{x + 1} \\[1em] \Rightarrow (11 - x)(x + 1) = 5(x + 3) \\[1em] \Rightarrow 11x + 11 - x^2 - x = 5x + 15 \\[1em] \Rightarrow -x^2 + 10x + 11 = 5x + 15 \\[1em] \Rightarrow -x^2 + 10x - 5x + 11 - 15 = 0 \\[1em] \Rightarrow -x^2 + 5x - 4 = 0 \\[1em] \Rightarrow x^2 - 5x + 4 = 0 \\[1em] \Rightarrow x^2 − 4x − x + 4 = 0 \\[1em] \Rightarrow x(x − 4) −1(x − 4) = 0 \\[1em] \Rightarrow (x − 1)(x − 4) = 0 \\[1em] \Rightarrow x = 1 \text{ or } x = 4.\end{aligned}$$

Hence, x = 1, 4.

Given,

⇒ 2x² + ax - a² = 0

⇒ 2x² + 2ax - ax - a² = 0

⇒ 2x(x + a) - a(x + a) = 0

⇒ (2x - a)(x + a) = 0

⇒ (2x - a) = 0 or (x + a) = 0

⇒ x = $\frac{a}{2}$ or x = -a

Hence, x = $\frac{a}{2}$ or -a.

Given,

$$\begin{aligned}\Rightarrow \sqrt{2x + 9} + x = 13 \\[1em] \Rightarrow \sqrt{2x + 9} = 13 - x \\[1em] \text{Squaring both sides, we get :} \\[1em] \Rightarrow (\sqrt{2x + 9})^2 = (13 - x)^2 \\[1em] \Rightarrow 2x + 9 = 169 - 26x + x^2 \\[1em] \Rightarrow x^2 - 26x - 2x + 169 - 9 = 0 \\[1em] \Rightarrow x^2 - 28x + 160 = 0 \\[1em] \Rightarrow x^2 - 20x - 8x + 160 = 0 \\[1em] \Rightarrow x(x - 20) - 8(x - 20) = 0 \\[1em] \Rightarrow (x - 20)(x - 8) = 0 \\[1em] \Rightarrow x = 20 \text{ or } x = 8\end{aligned}$$

Substituting value of x = 20, in L.H.S of equation $\sqrt{2x + 9} + x = 13$, we get:

$$\begin{aligned}\Rightarrow \sqrt{2(20) + 9} + 20 \\[1em] \Rightarrow \sqrt{40 + 9} + 20 \\[1em] \Rightarrow \sqrt{49} + 20 \\[1em] \Rightarrow 7 + 20 = 27\end{aligned}$$

L.H.S ≠ R.H.S

x = 20 is not valid

Substituting value of x = 8, in L.H.S of equation $\sqrt{2x + 9} + x = 13$, we get:

$$\begin{aligned}\Rightarrow \sqrt{2(8) + 9} + 8 \\[1em] \Rightarrow \sqrt{16 + 9} + 8 \\[1em] \Rightarrow \sqrt{25} + 8 \\[1em] \Rightarrow 5 + 8 \\[1em] \Rightarrow 13\end{aligned}$$

L.H.S = R.H.S

x = 8 is valid

Hence, x = 8.

$$\begin{aligned}\Rightarrow \frac{2800}{x - 100} - \frac{2800}{x} = \frac{1}{2} \\[1em] \Rightarrow \frac{2800x - 2800(x - 100)}{x(x - 100)} = \frac{1}{2} \\[1em] \Rightarrow \frac{2800x - 2800x + 280000}{x^2 - 100x} = \frac{1}{2} \\[1em] \Rightarrow 2(280000) = x^2 - 100x \\[1em] \Rightarrow 560000 = x^2 - 100x \\[1em] \Rightarrow x^2 - 100x - 560000 = 0 \\[1em] \Rightarrow x^2 - 800x + 700x - 560000 = 0 \\[1em] \Rightarrow x(x - 800) + 700(x - 800) = 0 \\[1em] \Rightarrow (x + 700)(x - 800) = 0 \\[1em] \Rightarrow x = -700 \text { or } x = 800\end{aligned}$$

Hence, x = 800.

$$\begin{aligned}\Rightarrow 3x^2 - 2\sqrt{6}x + 2 = 0 \\[1em] \Rightarrow 3x^2 - \sqrt{6}x - \sqrt{6}x + 2 = 0 \\[1em] \Rightarrow \sqrt{3}x (\sqrt{3}x - \sqrt2) - \sqrt2(\sqrt{3}x - \sqrt2) = 0 \\[1em] \Rightarrow (\sqrt{3}x - \sqrt2) (\sqrt{3}x - \sqrt2) = 0 \\[1em] \Rightarrow (\sqrt{3}x - \sqrt2)^2 = 0 \\[1em] \Rightarrow (\sqrt{3}x - \sqrt2) = 0 \\[1em] \Rightarrow \sqrt{3}x = \sqrt2 \\[1em] \Rightarrow x = \frac{\sqrt2}{\sqrt{3}} \\[1em] \Rightarrow x = \frac{\sqrt2 \times \sqrt3}{\sqrt{3} \times \sqrt3} \\[1em] \Rightarrow x = \frac{\sqrt6}{3}.\end{aligned}$$

Hence, x = $\frac{\sqrt6}{3}$.

Comparing equation x² - 3x + 2 = 0 with ax² + bx + c = 0, we get :

a = 1, b = -3 and c = 2.

By formula,

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

Substituting values we get :

$$\begin{aligned}x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \times 1 \times 2}}{2 \times 1} \\[1em] = \frac{3 \pm \sqrt{9 - 8}}{2} \\[1em] = \frac{3 \pm \sqrt{1}}{2} \\[1em] = \frac{3 \pm 1}{2} \\[1em] = \frac{3 + 1}{2} \text{ or } \frac{3 - 1}{2} \\[1em] = \frac{4}{2} \text{ or } \frac{2}{2} \\[1em] = 2.0 \text{ or } 1.0\end{aligned}$$

Hence, Option 2 is the correct option.

Comparing equation x² - 4x - 5 = 0 with ax² + bx + c = 0, we get :

a = 1, b = -4 and c = -5.

By formula,

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

Substituting values we get :

$$\begin{aligned}x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \times 1 \times -5}}{2 \times 1} \\[1em] = \frac{4 \pm \sqrt{16 + 20}}{2} \\[1em] = \frac{4 \pm \sqrt{36}}{2} \\[1em] = \frac{4 \pm 6}{2} \\[1em] = \frac{4 + 6}{2} \text{ or } \frac{4 - 6}{2} \\[1em] = \frac{10}{2} \text{ or } \frac{-2}{2} \\[1em] = 5.00 \text{ or } -1.00\end{aligned}$$

Hence, Option 1 is the correct option.

Comparing equation x² - 8x - 9 = 0 with ax² + bx + c = 0, we get :

a = 1, b = -8 and c = -9.

By formula,

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

Substituting values we get :

$$\begin{aligned}x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4 \times 1 \times -9}}{2 \times 1} \\[1em] = \frac{8 \pm \sqrt{64 + 36}}{2} \\[1em] = \frac{8 \pm \sqrt{100}}{2} \\[1em] = \frac{8 \pm 10}{2} \\[1em] = \frac{8 + 10}{2} \text{ or } \frac{8 - 10}{2} \\[1em] = \frac{18}{2} \text{ or } \frac{-2}{2} \\[1em] = 9 \text{ or } -1\end{aligned}$$

Hence, Option 2 is the correct option.

Given,

⇒ x² - 2x - 3 = 0

⇒ x² - 3x + x - 3 = 0

⇒ x(x - 3) + 1(x - 3) = 0

⇒ (x + 1)(x - 3) = 0

⇒ x + 1 = 0 or x - 3 = 0

⇒ x = -1 or x = 3.

Rounding off to two significant figures, we get :

⇒ x = -1.0 or x = 3.0

Hence, Option 3 is the correct option.

Comparing equation x² - 6x - 16 = 0 with ax² + bx + c = 0, we get :

a = 1, b = -6 and c = -16.

By formula,

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

Substituting values we get :

$$\begin{aligned}x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4 \times 1 \times -16}}{2 \times 1} \\[1em] = \frac{6 \pm \sqrt{36 + 64}}{2} \\[1em] = \frac{6 \pm \sqrt{100}}{2} \\[1em] = \frac{6 \pm 10}{2} \\[1em] = \frac{6 + 10}{2} \text{ or } \frac{6 - 10}{2} \\[1em] = \frac{16}{2} \text{ or } \frac{-4}{2} \\[1em] = 8 \text{ or } -2.\end{aligned}$$

Substituting x = 8 in L.H.S. of equation x² - 6x - 16 = 0, we get :

⇒ 8² - 6(8) - 16

⇒ 64 - 48 - 16

⇒ 0.

Substituting x = -2 in L.H.S. of equation x² - 6x - 16 = 0, we get :

⇒ (-2)² - 6(-2) - 16

⇒ 4 + 12 - 16

⇒ 0.

Since, L.H.S. = R.H.S., thus x = -2 is a solution of the equation.

Thus 8 and -2 are the solution of the equation x² - 6x - 16 = 0.

Hence, Option 3 is the correct option.

Comparing x² - 8x + 5 = 0 with ax² + bx + c = 0 we get,

a = 1, b = -8 and c = 5.

We know that,

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

Substituting values of a, b and c in above equation we get,

$$\begin{aligned}\Rightarrow x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4.(1).(5)}}{2(1)} \\[1em] = \frac{8 \pm \sqrt{64 - 20}}{2} \\[1em] = \frac{8 \pm \sqrt{44}}{2} \\[1em] = \frac{8 \pm 2\sqrt{11}}{2} \\[1em] = 4 \pm \sqrt{11} \\[1em] = 4 + \sqrt{11} \text{ and } 4 - \sqrt{11} \\[1em] = 4 + 3.3 \text{ and } 4 - 3.3 \\[1em] = 7.3 \text{ and } 0.7\end{aligned}$$

Hence, x = 7.3 and 0.7

Comparing 5x² + 10x - 3 = 0 with ax² + bx + c = 0 we get,

a = 5, b = 10 and c = -3.

We know that,

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

Substituting values of a, b and c in above equation we get,

$$\begin{aligned}\Rightarrow x = \frac{-(10) \pm \sqrt{(10)^2 - 4.(5).(-3)}}{2(5)} \\[1em] = \frac{-10 \pm \sqrt{100 + 60}}{10} \\[1em] = \frac{-10 \pm \sqrt{160}}{10} \\[1em] = \frac{-10 \pm 4\sqrt{10}}{10} \\[1em] = \frac{-10 \pm 12.8}{10} \\[1em] = \frac{-10 + 12.8}{10} \text{ and } \frac{-10 - 12.8}{10} \\[1em] = \frac{2.8}{10} \text{ and } \frac{-22.8}{10} \\[1em] = 0.28 \text{ and } -2.28 \\[1em] \approx 0.3 \text{ and } -2.3\end{aligned}$$

Hence, x = 0.3 and -2.3

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

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

We know that,

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

Substituting values of a, b and c in above equation we get,

$$\begin{aligned}\Rightarrow x = \frac{-(-10) \pm \sqrt{(-10)^2 - 4.(2).(5)}}{2(2)} \\[1em] = \frac{10 \pm \sqrt{100 - 40}}{4} \\[1em] = \frac{10 \pm \sqrt{60}}{4} \\[1em] = \frac{10 \pm 2\sqrt{15}}{4} \\[1em] = \frac{10 \pm 7.74}{4} \\[1em] = \frac{10 + 7.74}{4} \text{ and } \frac{10 - 7.74}{4} \\[1em] = \frac{17.74}{4} \text{ and } \frac{2.26}{4} \\[1em] = 4.44 \text{ and } 0.56\end{aligned}$$

Hence, x = 4.44 and 0.56

Given,

$$\begin{aligned}\Rightarrow 4x + \frac{6}{x} + 13 = 0 \\[1em] \Rightarrow \frac{4x^2 + 6 + 13x}{x} = 0 \\[1em] \Rightarrow 4x^2 + 13x + 6 = 0\end{aligned}$$

Comparing 4x² + 13x + 6 = 0 with ax² + bx + c = 0 we get,

a = 4, b = 13 and c = 6.

We know that,

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

Substituting values of a, b and c in above equation we get,

$$\begin{aligned}\Rightarrow x = \frac{-(13) \pm \sqrt{(-13)^2 - 4.(4).(6)}}{2(4)} \\[1em] = \frac{-13 \pm \sqrt{169 - 96}}{8} \\[1em] = \frac{-13 \pm \sqrt{73}}{8} \\[1em] = \frac{-13 \pm 8.54}{8} \\[1em] = \frac{-13 + 8.54}{8} \text{ and } \frac{-13 - 8.54}{8} \\[1em] = \frac{-4.46}{8} \text{ and } \frac{-21.54}{8} \\[1em] = -0.56 \text{ and } -2.69\end{aligned}$$

Hence, x = -0.56 and -2.69

Comparing 4x² - 5x - 3 = 0 with ax² + bx + c = 0 we get,

a = 4, b = -5 and c = -3.

We know that,

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

Substituting values of a, b and c in above equation we get,

$$\begin{aligned}\Rightarrow x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4.(4).(-3)}}{2(4)} \\[1em] = \frac{5 \pm \sqrt{25 + 48}}{8} \\[1em] = \frac{5 \pm \sqrt{73}}{8} \\[1em] = \frac{5 \pm 8.54}{8} \\[1em] = \frac{5 + 8.54}{8} \text{ and } \frac{5 - 8.54}{8} \\[1em] = \frac{13.54}{8} \text{ and } \frac{-3.54}{8} \\[1em] = 1.69 \text{ and } -0.44\end{aligned}$$

Hence, x = 1.69 and -0.44

Comparing 3x² - 12x - 1 = 0 with ax² + bx + c = 0 we get,

a = 3, b = -12 and c = -1.

We know that,

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

Substituting values of a, b and c in above equation we get,

$$\begin{aligned}\Rightarrow x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4.(3).(-1)}}{2(3)} \\[1em] = \frac{12 \pm \sqrt{144 + 12}}{6} \\[1em] = \frac{12 \pm \sqrt{156}}{6} \\[1em] = \frac{12 \pm 12.49}{6} \\[1em] = \frac{12 + 12.49}{6} \text{ or } \frac{12 - 12.49}{6} \\[1em] = \frac{24.49}{6} \text{ or } \frac{-0.49}{6} \\[1em] = 4.082 \text{ or } -0.082\end{aligned}$$

Hence, x = 4.082 or -0.082

Comparing x² - 16x + 6 = 0 with ax² + bx + c = 0 we get,

a = 1, b = -16 and c = 6.

We know that,

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

Substituting values of a, b and c in above equation we get,

$$\begin{aligned}\Rightarrow x = \frac{-(-16) \pm \sqrt{(-16)^2 - 4.(1).(6)}}{2(1)} \\[1em] = \frac{16 \pm \sqrt{256 - 24}}{2} \\[1em] = \frac{16 \pm \sqrt{232}}{2} \\[1em] = \frac{16 \pm 15.232}{2} \\[1em] = \frac{16 + 15.232}{2} \text{ or } \frac{16 - 15.232}{2} \\[1em] = \frac{31.232}{2} \text{ or } \frac{0.768}{2} \\[1em] = 15.616 \text{ or } 0.384\end{aligned}$$

Hence, x = 15.616 or 0.384

Comparing 2x² + 11x + 4 = 0 with ax² + bx + c = 0 we get,

a = 2, b = 11 and c = 4.

We know that,

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

Substituting values of a, b and c in above equation we get,

$$\begin{aligned}\Rightarrow x = \frac{-(11) \pm \sqrt{(11)^2 - 4.(2).(4)}}{2(2)} \\[1em] = \frac{-11 \pm \sqrt{121 - 32}}{4} \\[1em] = \frac{-11 \pm \sqrt{89}}{4} \\[1em] = \frac{-11 \pm 9.434}{4} \\[1em] = \frac{-11 + 9.434}{4} \text{ or } \frac{-11 - 9.434}{4} \\[1em] = \frac{-1.566}{4} \text{ or } \frac{-20.434}{4} \\[1em] = -0.392 \text{ or } -5.109\end{aligned}$$

Hence, x = -0.392 or -5.109

Comparing 5x² - 3x - 4 = 0 with ax² + bx + c = 0 we get,

a = 5, b = -3 and c = -4.

We know that,

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

Substituting values of a, b and c in above equation we get,

$$\begin{aligned}\Rightarrow x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4.(5).(-4)}}{2(5)} \\[1em] = \frac{3 \pm \sqrt{9 + 80}}{10} \\[1em] = \frac{3 \pm \sqrt{89}}{10} \\[1em] = \frac{3 \pm 9.434}{10} \\[1em] = \frac{3 + 9.434}{10} \text{ or } \frac{3 - 9.434}{10} \\[1em] = \frac{12.434}{10} \text{ or } \frac{-6.434}{10} \\[1em] = 1.2434 \text{ or } -0.6434 \\[1em] \approx 1.24 \text{ or } -0.643\end{aligned}$$

Hence, x = 1.24 or -0.643

Given,

⇒ (x - 1)² - 3x + 4 = 0

⇒ x² + 1 - 2x - 3x + 4 = 0

⇒ x² - 5x + 5 = 0

Comparing x² - 5x + 5 = 0 with ax² + bx + c = 0 we get,

a = 1, b = -5 and c = 5.

We know that,

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

Substituting values of a, b and c in above equation we get,

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

Hence, x = 3.6 or 1.4

Given,

x = 3 is a solution of the quadratic equation (k + 2)x² - kx + 6 = 0.

Substituting x = 3 in equation (k + 2)x² - kx + 6 = 0,

⇒ (k + 2)(3)² - k(3) + 6 = 0

⇒ (k + 2)(9) - 3k + 6 = 0

⇒ 9k + 18 - 3k + 6 = 0

⇒ 6k + 24 = 0

⇒ 6k = -24

⇒ k = $\frac{-24}{6}$

⇒ k = -4.

Substituting k = −4 in equation (k + 2)x² - kx + 6 = 0 :

⇒ (-4 + 2)x² - (-4)x + 6 = 0

⇒ -2x² + 4x + 6 = 0

⇒ -2(x² - 2x - 3) = 0

⇒ x² - 2x - 3 = 0

⇒ x² - 3x + x - 3 = 0

⇒ x(x - 3) + 1(x - 3) = 0

⇒ (x + 1)(x - 3) = 0

⇒ (x + 1) = 0 or (x - 3) = 0

⇒ x = -1 or x = 3.

Hence, option 4 is the correct option.

Comparing equation 2x² - 3x + 1 = 0, with ax² + bx + c = 0, we get :

a = 2, b = -3 and c = 1.

By formula,

D = b² - 4ac

= (-3)² - 4 × 2 × 1

= 9 - 8

= 1; which is positive.

Since, a, b and c are real numbers; a ≠ 0 and b² - 4ac > 0

∴ Roots are real and unequal.

Hence, Option 1 is the correct option.

Comparing equation x² - 5x + 6 = 0, with ax² + bx + c = 0, we get :

a = 1, b = -5 and c = 6.

By formula,

D = b² - 4ac

= (-5)² - 4 × 1 × 6

= 25 - 24

= 1; which is positive.

Since, a, b and c are real numbers; a ≠ 0 and b² - 4ac > 0

∴ Roots are real and distinct.

Hence, Option 1 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.

Comparing equation x² - px + 4 = 0, with ax² + bx + c = 0, we get :

a = 1, b = -p and c = 4.

Since, roots are equal.

∴ D = 0

∴ b² - 4ac = 0

⇒ (-p)² - 4 × 1 × 4 = 0

⇒ p² - 16 = 0

⇒ p² - 4² = 0

⇒ (p - 4)(p + 4) = 0

⇒ (p - 4) = 0 or (p + 4) = 0

⇒ p = 4 or p = -4.

Hence, Option 4 is the correct option.

Comparing equation x² + 10x - 3 = 0, with ax² + bx + c = 0, we get :

a = 1, b = 10 and c = -3.

By formula,

D = b² - 4ac

= (10)² - 4 × 1 × -3

= 100 + 12

= 112; which is positive.

Comparing equation 2x² - 5x + 9 = 0, with ax² + bx + c = 0, we get :

a = 2, b = -5 and c = 9.

By formula,

D = b² - 4ac

= (-5)² - 4 × 2 × 9

= 25 - 72

= -47; which is negative.

∴ Roots are imaginary.

Hence, Option 2 is the correct option.

Since, 1 is the root of equation 3x² - mx + 4 = 0.

∴ x = 1, will satisfy the equation 3x² - mx + 4 = 0.

⇒ 3(1)² - m(1) + 4 = 0

⇒ 3 - m + 4 = 0

⇒ 7 - m = 0

⇒ m = 7.

Hence, Option 1 is the correct option.

Comparing 7x² - 9x + 2 = 0 with ax² + bx + c = 0 we get,

a = 7, b = -9 and c = 2.

We know that,

Discriminant = D = b² - 4ac = (-9)² - 4(7)(2)

= 81 - 56 = 25; which is positive.

Since, a, b and c are real numbers; a ≠ 0 and b² - 4ac > 0

∴ The roots are real and unequal.

Comparing 6x² - 13x + 4 = 0 with ax² + bx + c = 0 we get,

a = 6, b = -13 and c = 4.

We know that,

Discriminant = D = b² - 4ac = (-13)² - 4(6)(4)

= 169 - 96 = 73; which is positive.

Since, a, b and c are real numbers; a ≠ 0 and b² - 4ac > 0

∴ The roots are real and unequal.

Comparing 25x² - 10x + 1 = 0 with ax² + bx + c = 0 we get,

a = 25, b = -10 and c = 1.

We know that,

Discriminant = D = b² - 4ac = (-10)² - 4(25)(1)

= 100 - 100 = 0;

Since, a, b and c are real numbers; a ≠ 0 and b² - 4ac = 0

∴ The roots are real and equal.

Comparing x² + $2\sqrt{3}$x - 9 = 0 with ax² + bx + c = 0 we get,

a = 1, b = $2\sqrt{3}$ and c = -9.

We know that,

Discriminant = D = b² - 4ac = $(2\sqrt{3})$² - 4(1)(-9)

= 12 + 36 = 48; which is positive.

Since, a, b and c are real numbers; a ≠ 0 and b² - 4ac > 0

∴ The roots are real and unequal.

Comparing 3x² - 12x + (n - 5) = 0 with ax² + bx + c = 0 we get,

a = 3, b = -12 and c = (n - 5).

Since equations have equal roots,

∴ D = 0

⇒ (-12)² - 4.(3).(n - 5) = 0

⇒ 144 - 12(n - 5) = 0

⇒ 144 - 12n + 60 = 0

⇒ 204 - 12n = 0

⇒ 12n = 204

⇒ n = $\frac{204}{12}$

⇒ n = 17

Hence, n = 17.

Comparing (m - 2)x² - (5 + m)x + 16 = 0 with ax² + bx + c = 0 we get,

a = (m - 2), b = -(5 + m) and c = 16.

Since equations have equal roots,

∴ D = 0

⇒ (-(5 + m))² - 4.(m - 2).(16) = 0

⇒ 25 + m² + 10m - 64(m - 2) = 0

⇒ 25 + m² + 10m - 64m + 128 = 0

⇒ m² - 54m + 153 = 0

⇒ m² - 51m - 3m + 153 = 0

⇒ m(m - 51) - 3(m - 51) = 0

⇒ (m - 3)(m - 51) = 0

⇒ (m - 3) = 0 or (m - 51) = 0

⇒ m = 3 or m = 51.

Hence, m = 3 or 51.

Comparing 3x² - 6x + k = 0 with ax² + bx + c = 0 we get,

a = 3, b = -6 and c = k.

Since equations have distinct and real roots,

∴ D > 0

⇒ (-6)² - 4.(3).(k) > 0

⇒ 36 - 12k > 0

⇒ 12k < 36

⇒ k < 3.

Hence, k < 3.

Since, 2 is the root hence, it satisfies the equation 3x² - p(x + 1) = 0.

⇒ 3(2)² - p(2 + 1) = 0

⇒ 3(4) - 3p = 0

⇒ 3p = 12

⇒ p = 4.

Substituting value of p in px² - qx + 9 = 0

⇒ 4x² - qx + 9 = 0

Comparing 4x² - qx + 9 = 0 with ax² + bx + c = 0 we get,

a = 4, b = -q and c = 9.

Since equation has equal roots,

∴ D = 0

⇒ (-q)² - 4.(4).(9) = 0

⇒ q² - 144 = 0

⇒ q² = 144

⇒ q = 12 or -12.

Hence, p = 4 and q = 12 or -12.

Given,

$$\begin{aligned}\Rightarrow \frac{1}{x + 4} - \frac{1}{x - 7} = \frac{11}{30} \\[1em] \Rightarrow \frac{(x - 7) - (x + 4)}{(x + 4)(x - 7)} = \frac{11}{30} \\[1em] \Rightarrow \frac{x - 7 - x - 4}{(x + 4)(x - 7)} = \frac{11}{30} \\[1em] \Rightarrow \frac{-11}{(x + 4)(x - 7)} = \frac{11}{30} \\[1em] \Rightarrow -11 \times 30 = 11(x + 4)(x - 7) \\[1em] \Rightarrow \frac{-11 \times 30}{11} = x^2 - 7x + 4x - 28 \\[1em] \Rightarrow -30 = x^2 - 3x - 28 \\[1em] \Rightarrow x^2 - 3x + 2 = 0\end{aligned}$$

Now we have,

x² - 3x + 2 = 0

Comparing x² - 3x + 2 = 0 with ax² + bx + c = 0 we get,

a = 1, b = -3 and c = 2.

We know that,

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

Substituting values of a, b and c in above equation we get,

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

Hence, x = 1 or 2.

Given,

x² = 4x

x² - 4x = 0

Comparing x² - 4x = 0 with ax² + bx + c = 0 we get,

a = 1, b = -4 and c = 0.

We know that,

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

Substituting values of a, b and c in above equation we get,

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

Hence, x = 0 or 4.

Given,

3y + $\frac{5}{16y}$ = 2

$\frac{48y^2 + 5}{16y}$ = 2

48y² + 5 = 32y

48y² - 32y + 5 = 0

Comparing 48y² - 32y + 5 = 0 with ax² + bx + c = 0 we get,

a = 48, b = -32 and c = 5.

We know that,

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

Substituting values of a, b and c in above equation we get,

$$\begin{aligned}\Rightarrow x = \frac{-(-32) \pm \sqrt{(-32)^2 - 4(48)(5)}}{2(48)} \\[1em] = \frac{32 \pm \sqrt{1024 - 960}}{96} \\[1em] = \frac{32 \pm \sqrt{64}}{96} \\[1em] = \frac{32 \pm 8}{96} \\[1em] = \frac{32 + 8}{96} \text{ or } \frac{32 - 8}{96} \\[1em] = \frac{40}{96} \text{ or } \frac{24}{96} \\[1em] = \frac{5}{12} \text{ or } \frac{1}{4}.\end{aligned}$$

Hence, x = $\frac{5}{12} \text{ or } \frac{1}{4}$.

(i) kx(x - 2) + 6 = 0

Given,

kx(x - 2) + 6 = 0

kx² - 2kx + 6 = 0

Comparing kx² - 2kx + 6 = 0 with ax² + bx + c = 0 we get,

a = k, b = -2k and c = 6.

Since roots are real and equal,

We know that,

∴ D = 0

⇒ b² - 4ac = 0

⇒ (-2k)² - 4(k)(6) = 0

⇒ 4k² - 24k = 0

⇒ 4k(k - 6) = 0

⇒ 4k = 0 or (k - 6) = 0

⇒ k = 0 or k = 6

If k = 0, substituting in L.H.S. of kx² - 2kx + 6 = 0 we get,

= (0)x² - 2(0)x + 6

= 6

L.H.S. ≠ R.H.S.

Thus, k ≠ 0.

Substituting k = 0 in kx² - 2kx + 6 = 0 we get,

⇒ (6)x² - 2(6)x + 6 = 0

⇒ 6x² - 12x + 6 = 0

⇒ 6x² - 6x - 6x + 6 = 0

⇒ 6x(x - 1) - 6(x - 1) = 0

⇒ (6x - 6)(x - 1) = 0

⇒ (6x - 6) = 0 or (x - 1) = 0

⇒ x = $\frac{6}{6}$ or x = 1

⇒ x = 1 equation has real and equal roots

∴ k = 6

Hence, k = 6.

(ii) (k + 4)x² + (k + 1)x + 1 = 0

Given,

(k + 4)x² + (k + 1)x + 1 = 0

Comparing (k + 4)x² + (k + 1)x + 1 = 0 with ax² + bx + c = 0 we get,

a = k + 4, b = k + 1 and c = 1.

Since roots are real and equal,

We know that,

∴ D = 0

⇒ b² - 4ac = 0

⇒ (k + 1)² - 4(k + 4)(1) = 0

⇒ k² + 2k + 1 - 4k - 16 = 0

⇒ k² - 2k - 15 = 0

⇒ k² - 5k + 3k - 15 = 0

⇒ k(k - 5) + 3(k - 5) = 0

⇒ (k + 3)(k - 5) = 0

⇒ k + 3 = 0 or k - 5 = 0

⇒ k = -3 or k = 5.

If k = 5, substituting in (k + 4)x² + (k + 1)x + 1 = 0 we get,

⇒ (5 + 4)x² + (5 + 1)x + 1 = 0

⇒ 9x² + 6x + 1 = 0

⇒ (3x + 1)² = 0

⇒ 3x + 1 = 0

⇒ x = $\frac{-1}{3}$

Both roots are the same.

If k = -3, substituting in (k + 4)x² + (k + 1)x + 1 = 0 we get,

⇒ (-3 + 4)x² + (-3 + 1)x + 1 = 0

⇒ x² - 2x + 1 = 0

⇒ (x - 1)² = 0

⇒ x - 1 = 0

⇒ x = 1

Both roots are the same.

Hence, k = -3 or 5.

Given,

⇒ x⁴ - 5x² + 4 = 0

⇒ x⁴ - 4x² - x² + 4 = 0

⇒ x²(x² - 4) - 1(x² - 4) = 0

⇒ (x² - 1)(x² - 4) = 0

⇒ x² - 1 = 0 or x² - 4 = 0

⇒ x² = 1 or x² = 4

⇒ x = $\sqrt{1}$ or x = $\sqrt{4}$

⇒ x = ± 1 or x = ± 2.

Hence, Option 2 is the correct option.

Given,

$$\begin{aligned}\Rightarrow \frac{1}{x} + \frac{1}{x - 5} = \frac{3}{10} \\[1em] \Rightarrow \frac{x - 5 + x}{x(x - 5)} = \frac{3}{10} \\[1em] \Rightarrow \frac{2x - 5}{x^2 - 5x} = \frac{3}{10} \\[1em] \Rightarrow 10(2x - 5) = 3(x^2 - 5x) \\[1em] \Rightarrow 20x - 50 = 3x^2 - 15x \\[1em] \Rightarrow 3x^2 - 15x - 20x + 50 = 0 \\[1em] \Rightarrow 3x^2 - 35x + 50 = 0 \\[1em] \Rightarrow 3x^2 - 30x - 5x + 50 = 0 \\[1em] \Rightarrow 3x(x - 10) - 5(x - 10) = 0 \\[1em] \Rightarrow (3x - 5)(x - 10) = 0 \\[1em] \Rightarrow 3x - 5 = 0 \text{ or } x - 10 = 0 \\[1em] \Rightarrow 3x = 5 \text{ or } x = 10 \\[1em] \Rightarrow x = \frac{5}{3} \text{ or } x = 10.\end{aligned}$$

Hence, Option 2 is the correct option.

Given,

$\frac{1}{x - 3} - \frac{1}{x + 5}$ = 1

So, in above equation.

x - 3 and x + 5 cannot be equal to zero.

⇒ x - 3 ≠ 0

⇒ x ≠ 3

⇒ x + 5 ≠ 0

⇒ x ≠ -5.

Hence, Option 3 is the correct option.

Let x² = y,

⇒ 2x⁴ - 5x² + 3 = 0

⇒ 2y² - 5y + 3 = 0

⇒ 2y² - 2y - 3y + 3 = 0

⇒ 2y(y - 1) - 3(y - 1) = 0

⇒ (2y - 3)(y - 1) = 0

⇒ 2y - 3 = 0 or y - 1 = 0      [Zero product rule]

⇒ 2y = 3 or y = 1

⇒ y = $\frac{3}{2}$ or y = 1.

∴ x² = $\frac{3}{2}$ or x² = 1

⇒ x = $\pm \sqrt{\frac{3}{2}} = \pm 1.22$ or x = $\sqrt{1} = \pm 1$

Hence, x = +1.22, -1.22, +1, -1.

Let x² = y,

⇒ x⁴ - 2x² - 3 = 0

⇒ y² - 2y - 3 = 0

⇒ y² - 3y + y - 3 = 0

⇒ y(y - 3) + 1(y - 3) = 0

⇒ (y + 1)(y - 3) = 0

⇒ y + 1 = 0 or y - 3 = 0      [Zero product rule]

⇒ y = -1 or y = 3

∴ x² = -1 or x² = 3

Since, square of a number cannot be negative,

∴ x² = 3

⇒ x = $\sqrt{3} = \pm 1.73$

Hence, x = +1.73, -1.73.

Let x² - x = a

Substituting value in (x² - x)² + 5(x² - x) + 4 = 0 we get,

⇒ a² + 5a + 4 = 0

⇒ a² + 4a + a + 4 = 0

⇒ a(a + 4) + 1(a + 4) = 0

⇒ (a + 1)(a + 4) = 0

⇒ a + 1 = 0 or a + 4 = 0      [Zero product rule]

⇒ a = -1 or a = -4

∴ x² - x = -1 and x² - x = -4

Solving, x² - x = -1

⇒ x² - x = -1

⇒ x² - x + 1 = 0

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

a = 1, b = -1, c = 1

Discriminant = D = b² - 4ac = (-1)² - 4(1)(1) = 1 - 4 = -3.

-3 < 0

∴ No real solution.

Solving, x² - x = -4

⇒ x² - x + 4 = 0

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

a = 1, b = -1, c = 4

Discriminant = D = b² - 4ac = (-1)² - 4(1)(4) = 1 - 16 = -15.

-15 < 0

∴ No real solution.

Hence, there is no real solution.

Let x² - 3x = a

Substituting value in (x² - 3x)² - 16(x² - 3x) - 36 = 0 we get,

⇒ a² - 16a - 36 = 0

⇒ a² - 18a + 2a - 36 = 0

⇒ a(a - 18) + 2(a - 18) = 0

⇒ (a + 2)(a - 18) = 0

⇒ a + 2 = 0 or a - 18 = 0      [Zero product rule]

⇒ a = -2 or a = 18

∴ x² - 3x = -2 and x² - 3x = 18

Solving, x² - 3x = -2

⇒ x² - 3x = -2

⇒ x² - 3x + 2 = 0

⇒ x² - 2x - x + 2 = 0

⇒ x(x - 2) - 1(x - 2) = 0

⇒ (x - 1)(x - 2) = 0

⇒ x - 1 = 0 or x - 2 = 0      [Zero product rule]

⇒ x = 1 or x = 2.

Solving, x² - 3x = 18

⇒ x² - 3x = 18

⇒ x² - 3x - 18 = 0

⇒ x² - 6x + 3x - 18 = 0

⇒ x(x - 6) + 3(x - 6) = 0

⇒ (x + 3)(x - 6) = 0

⇒ x + 3 = 0 or x - 6 = 0      [Zero product rule]

⇒ x = -3 or x = 6.

Hence, x = 1, 2, -3, 6.

Let $\sqrt{\frac{x}{x - 3}} =$ a .......(i)

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

Substituting value of a = $\frac{1}{2}$ in (i) we get,

$\Rightarrow \sqrt{\frac{x}{x - 3}} = \frac{1}{2}$

Squaring both sides we get,

$$\begin{aligned}\Rightarrow \frac{x}{x - 3} = \frac{1}{4} \\[1em] \Rightarrow 4x = x - 3 \\[1em] \Rightarrow 4x - x = -3 \\[1em] \Rightarrow 3x = -3 \\[1em] \Rightarrow x = -1.\end{aligned}$$

Substituting value of a = 2 in (i) we get,

$\Rightarrow \sqrt{\frac{x}{x - 3}} = 2$

Squaring both sides we get,

$$\begin{aligned}\Rightarrow \frac{x}{x - 3} = 4 \\[1em] \Rightarrow x = 4(x - 3) \\[1em] \Rightarrow x = 4x - 12 \\[1em] \Rightarrow 4x - x = 12 \\[1em] \Rightarrow 3x = 12 \\[1em] \Rightarrow x = 4.\end{aligned}$$

Hence, x = -1, 4.

Let $\frac{2x - 3}{x - 1} =$ a .......(i)

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

Substituting value of a = -1 in (i) we get,

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

Substituting value of a = 4 in (i) we get,

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

Hence, x = $1\frac{1}{3}, \frac{1}{2}$.

Let $\Big(\frac{3x + 1}{x + 1}\Big) = a$ .......(i)

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

Substituting value of a = $\frac{1}{2}$ in (i) we get,

$$\begin{aligned}\Rightarrow \Big(\frac{3x + 1}{x + 1}\Big) = \frac{1}{2} \\[1em] \Rightarrow 2(3x + 1) = x + 1 \\[1em] \Rightarrow 6x + 2 = x + 1 \\[1em] \Rightarrow 6x - x = 1 - 2 \\[1em] \Rightarrow 5x = -1 \\[1em] \Rightarrow x = -\frac{1}{5}\end{aligned}$$

Substituting value of a = 2 in (i) we get,

$$\begin{aligned}\Rightarrow \Big(\frac{3x + 1}{x + 1}\Big) = 2 \\[1em] \Rightarrow 3x + 1 = 2(x + 1) \\[1em] \Rightarrow 3x + 1 = 2x + 2 \\[1em] \Rightarrow 3x - 2x = 2 - 1 \\[1em] \Rightarrow x = 1.\end{aligned}$$

Hence, x = 1, $-\frac{1}{5}$.

Let $x + \frac{1}{x} = a$ ........(i)

Squaring, both sides we get,

$$\begin{aligned}\Rightarrow x^2 + \frac{1}{x^2} + 2 = a^2 \\[1em] \Rightarrow x^2 + \frac{1}{x^2} = a^2 - 2 .......(ii)\end{aligned}$$

Substituting the values from equations (i) and (ii) we get,

⇒ 9(a² - 2) - 9a - 52 = 0

⇒ 9a² - 18 - 9a - 52 = 0

⇒ 9a² - 9a - 70 = 0

⇒ 9a² - 30a + 21a - 70 = 0

⇒ 3a(3a - 10) + 7(3a - 10) = 0

⇒ (3a + 7)(3a - 10) = 0

⇒ (3a + 7) = 0 or 3a - 10 = 0      [Zero product rule]

⇒ 3a = -7 or 3a = 10

⇒ $a = -\frac{7}{3} \text{ or } a = \frac{10}{3}$

Considering a = $\frac{10}{3}$ we get,

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

Considering a = $-\frac{7}{3}$ we get,

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

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

a = 3, b = 7 and c = 3.

We know that,

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

$$\begin{aligned}\Rightarrow x = \frac{-7 \pm \sqrt{(7)^2 - 4(3)(3)}}{2(3)} \\[1em] = \frac{-7 \pm \sqrt{49 - 36}}{6} \\[1em] = \frac{-7 \pm \sqrt{13}}{6}.\end{aligned}$$

Hence, x = 3, $\frac{1}{3}, \frac{-7 \pm \sqrt{13}}{6}.$

Let $x - \frac{1}{x} = a$ ........(i)

Squaring, both sides we get,

$$\begin{aligned}\Rightarrow x^2 + \frac{1}{x^2} - 2 = a^2 \\[1em] \Rightarrow x^2 + \frac{1}{x^2} = a^2 + 2 .......(ii)\end{aligned}$$

Substituting the values from equations (i) and (ii) we get,

$(x^2 + \frac{1}{x^2}) - 3(x - \frac{1}{x}) - 2 = 0$

⇒ a² + 2 - 3a - 2 = 0

⇒ a² - 3a = 0

⇒ a(a - 3) = 0

⇒ a = 0 or a - 3 = 0

⇒ a = 0 or a = 3.

Considering a = 0 we get,

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

Considering a = 3 we get,

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

Comparing x² - 3x - 1 = 0 with ax² + bx + c = 0 we get,

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

We know that,

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

$$\begin{aligned}\Rightarrow x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-1)}}{2(1)} \\[1em] = \frac{3 \pm \sqrt{9 + 4}}{2} \\[1em] = \frac{3 \pm \sqrt{13}}{2}.\end{aligned}$$

Hence, x = 1, -1, $\frac{3 \pm \sqrt{13}}{2}$.

Let x² + 5x = a we get,

⇒ (a + 4)(a + 6) = 120

⇒ a² + 6a + 4a + 24 = 120

⇒ a² + 10a + 24 - 120 = 0

⇒ a² + 10a - 96 = 0

⇒ a² + 16a - 6a - 96 = 0

⇒ a(a + 16) - 6(a + 16) = 0

⇒ (a - 6)(a + 16) = 0

⇒ (a - 6) = 0 or (a + 16) = 0

⇒ a = 6 or a = -16.

Considering a = 6 we get,

⇒ x² + 5x = 6

⇒ x² + 5x - 6 = 0

⇒ x² + 6x - x - 6 = 0

⇒ x(x + 6) - 1(x + 6) = 0

⇒ (x - 1)(x + 6) = 0

⇒ (x - 1) = 0 or (x + 6) = 0

⇒ x = 1 or x = -6.

Considering a = -16 we get,

⇒ x² + 5x = -16

⇒ x² + 5x + 16 = 0

In this case,

D = b² - 4ac = 5² - 4(1)(16) = 25 - 64 = -39 < 0.

It means roots are imaginary in this case.

Hence, x = 1, -6.

Solving,

$$\begin{aligned}\Rightarrow 3\Big(\frac{3x - 1}{2x + 3}\Big) - 2\Big(\frac{2x + 3}{3x - 1}\Big) = 5 \\[1em] \Rightarrow \frac{3(3x - 1)^2 - 2(2x + 3)^2}{(2x + 3)(3x - 1)} = 5 \\[1em] \Rightarrow 3(3x - 1)^2 - 2(2x + 3)^2 = 5(2x + 3)(3x - 1) \\[1em] \Rightarrow 3(9x^2 - 6x + 1) - 2(4x^2 + 12x + 9) = 5(6x^2 - 2x + 9x - 3) \\[1em] \Rightarrow 27x^2 - 18x + 3 - 8x^2 - 24x - 18 = 5(6x^2 + 7x - 3) \\[1em] \Rightarrow 19x^2 - 42x - 15 = 30x^2 + 35x - 15 \\[1em] \Rightarrow 30x^2 - 19x^2 + 35x + 42x - 15 + 15 = 0 \\[1em] \Rightarrow 11x^2 + 77x = 0 \\[1em] \Rightarrow 11x(x + 7) = 0 \\[1em] \Rightarrow 11x = 0 \text{ or } x + 7 = 0 \\[1em] \Rightarrow x = 0 \text{ or } x = -7.\end{aligned}$$

Hence, x = 0 \text{ or } x = -7.

Given,

5^x+1 + 5^2-x = 5³ + 1

⇒ 5^x.5¹ + 5².5^-x = 125 + 1

⇒ 5^x.5¹ + $\frac{5^2}{5^x}$ = 126

Let 5^x be y.

⇒ 5y + $\frac{25}{\text{y}}$ = 126

⇒ 5y² + 25 = 126y

⇒ 5y² - 126y + 25 = 0

⇒ 5y² - 125y - y + 25 = 0

⇒ 5y(y - 25) - (y - 25) = 0

⇒ (y - 25)(5y - 1) = 0

⇒ (y - 25) = 0 or (5y - 1) = 0

⇒ y = 25 or y = $\frac{1}{5}$

⇒ y = 5² or y = 5^-1

Substituting the value of y,

⇒ 5^x = 5² or 5^x = 5^-1

⇒ x = 2 or -1

Hence, the value of x = 2 or -1.

Given,

Equation : (k + 2)x² - 2x + 1 = 0

Comparing (k + 2)x² - 2x + 1 = 0 with ax² + bx + c = 0 we get,

a = (k + 2), b = -2 and c = 1.

Since, roots are real,

∴ D ≥ 0

⇒ b² - 4ac ≥ 0

⇒ (-2)² - 4(k + 2)(1) ≥ 0

⇒ 4 - 4(k + 2) ≥ 0

⇒ 4 - 4k - 8 ≥ 0

⇒ -4k - 4 ≥ 0

⇒ 4k ≤ -4

⇒ k ≤ -1.

Thus, greatest value of k = -1.

Hence, option 3 is the correct option.

Given,

x² - 4x + k = 0

Comparing x² - 4x + k = 0 with ax² + bx + c = 0 we get,

a = 1, b = -4 and c = k.

Since roots are distinct and real,

∴ D > 0

⇒ b² - 4ac > 0

⇒ (-4)² - 4(1)(k) > 0

⇒ 16 - 4k > 0

⇒ 16 > 4k

⇒ $\frac{16}{4}$ > k

⇒ k < 4

Since k ∈ N,

Possible natural numbers less than 4 are: 1, 2, 3

The greatest value is: 3

Hence, Option 2 is the correct option.

Comparing equation kx² + kx + 1 = 0, with ax² + bx + c = 0 , we get :

a = k, b = k and c = 1.

Since, quadratic equation has real and equal roots.

∴ D = 0

∴ b² - 4ac = 0

⇒ k² - 4 × k × 1 = 0

⇒ k² - 4k = 0

⇒ k(k - 4) = 0

⇒ k = 0 or (k - 4) = 0.

⇒ k = 0 or k = 4.

Hence, Option 4 is the correct option.

Given,

⇒ x² - 4x = 5

⇒ x² - 4x - 5 = 0

⇒ x² - 5x + x - 5 = 0

⇒ x(x - 5) + 1(x - 5) = 0

⇒ (x + 1)(x - 5) = 0

⇒ (x + 1) = 0 or (x - 5) = 0

⇒ x = -1 or x = 5.

Hence, Option 3 is the correct option.

Given,

⇒ x² - 7x = 0

⇒ x(x - 7) = 0

⇒ x = 0 or x - 7 = 0

⇒ x = 0 or x = 7.

Hence, Option 4 is the correct option.

Given,

x = 1 is a root of the equation $\sqrt{x} + kx - 2$ = 0.

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

Hence, Option 1 is the correct option.

A is true, R is false.

Reason

Given,

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

The quadratic equation mentioned in the reason is $x^2 - 2x - 15 = 0$ whereas we see that the correct quadratic equation is $x^2 + 2x - 15 = 0$
∴ Reason (R) is false.

Our solution shows that one of the roots is 3.
∴ Assertion (A) is true.

Hence, option 1 is correct.

Both A and R are true and R is correct reason for A.

Reason

Given, 2x² + 5x - 3 = 0

As we know that the roots of equation ax² + bx + c = 0 are real and unequal if b² - 4ac > 0.

⇒ b² - 4ac = 5² - 4 x 2 x (-3)

= 25 + 24 = 49 > 0

So, Assertion (A) is true.

And, Reason (R) is also true and it clearly explain assertion as a positive discriminant (b² - 4ac > 0) guarantees that the roots are real and unequal.

Hence, option 3 is correct.

Both the statements are true.

Reason

We are given that one root of the quadratic equation is 3 + $\sqrt{2}$. If the quadratic equation has real coefficients, the conjugate of a root that involves a surd (i.e., a square root or irrational number) must also be a root of the quadratic equation.

Therefore, 3 - $\sqrt{2}$ is other root of the given quadratic equation.

So, statement (1) is true.

The property of conjugate roots holds for quadratic equations with real coefficients, meaning if one root involves a surd, the other root will be its conjugate. In this case, since 3 + $\sqrt{2}$ is a surd, the other root must be 3 - $\sqrt{2}$.

So, statement (2) is true.

Hence, option 1 is correct.

Given,

⇒ p - 15 = 0

⇒ p = 15.

Substituting value of p in 2x² + px + 25 = 0 we get,

⇒ 2x² + 15x + 25 = 0

⇒ 2x² + 10x + 5x + 25 = 0

⇒ 2x(x + 5) + 5(x + 5) = 0

⇒ (2x + 5)(x + 5) = 0

⇒ 2x + 5 = 0 or x + 5 = 0

⇒ 2x = -5 or x = -5

⇒ x = -$\frac{5}{2}$ or x = -5.

Hence, x = -5 or $-\frac{5}{2}$.

Given,

$$\begin{aligned}\Rightarrow \frac{1}{p} + \frac{1}{q} + \frac{1}{x} = \frac{1}{x + p + q} \\[1em] \Big(\frac{1}{p} + \frac{1}{q}\Big) + \Big(\frac{1}{x} - \frac{1}{x + p + q}\Big) = 0 \\[1em] \Big(\frac{q + p}{pq}\Big) + \Big(\frac{x + p + q - x}{x(x + p + q)}\Big) = 0 \\[1em] \Big(\frac{q + p}{pq}\Big) + \Big(\frac{p + q}{x(x + p + q)}\Big) = 0 \\[1em] (p + q)\Big(\frac{1}{pq} + \frac{1}{x(x + p + q)}\Big) = 0 \\[1em] (p + q)\Big(\frac{x(x + p + q) + pq}{xpq(x + p + q)} \Big) = 0 \\[1em] (p + q)\Big(\frac{x^2 + x(p + q) + pq}{xpq(x + p + q)}\Big) = 0 \\[1em] \therefore \Big(\frac{x^2 + x(p + q) + pq}{xpq(x + p + q)}\Big) = 0 \\[1em] x^2 + x(p + q) + pq = 0 \\[1em]\end{aligned}$$

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

a = 1, b = (p + q), c = pq

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

Substituting value in above equation we get,

$$\begin{aligned}\Rightarrow x = \frac{-(p + q) \pm \sqrt{(p + q)^2 - 4.(1).pq}}{2(1)} \\[1em] = \frac{-(p + q) \pm \sqrt{p^2 + q^2 + 2pq - 4pq}}{2} \\[1em] = \frac{-(p + q) \pm \sqrt{(p - q)^2}}{2} \\[1em] = \frac{-(p + q) \pm (p - q)}{2} \\[1em] = \frac{-(p + q) + (p - q)}{2} \text{ or } \frac{-(p + q) - (p - q)}{2} \\[1em] = \frac{-p - q + p - q}{2} \text{ or } \frac{-p - q - p + q}{2} \\[1em] = \frac{-2q}{2} \text{ or } \frac{-2p}{2} \\[1em] = -q \text{ or } -p.\end{aligned}$$

Hence, x = -q or -p.

Given,

⇒ 8x² - 14x + 3 = 0

⇒ 8x² - 12x - 2x + 3 = 0

⇒ 4x(2x - 3) - 1(2x - 3) = 0

⇒ (4x - 1)(2x - 3) = 0

⇒ 4x - 1 = 0 or 2x - 3 = 0

⇒ 4x = 1 or 2x = 3

⇒ x = $\frac{1}{4}$ or x = $\frac{3}{2}$.

(i) Since, there is no integer in the solution,

Hence, no solution.

(ii) Here, x ∈ Q

Hence, x = $\frac{1}{4}, \frac{3}{2}$.

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

a = 1, b = 1, c = -(a + 2)(a + 1)

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

Substituting values we get,

$$\begin{aligned}\Rightarrow \frac{-1 \pm \sqrt{(1)^2 - 4(1)(-(a + 2)(a + 1))}}{2(1)} \\[1em] = \frac{-1 \pm \sqrt{1 + 4(a + 2)(a + 1)}}{2} \\[1em] = \frac{-1 \pm \sqrt{1 + 4(a^2 + a + 2a + 2)}}{2} \\[1em] = \frac{-1 \pm \sqrt{1 + 4(a^2 + 3a + 2)}}{2} \\[1em] = \frac{-1 \pm \sqrt{4a^2 + 12a + 8 + 1}}{2} \\[1em] = \frac{-1 \pm \sqrt{4a^2 + 12a + 9}}{2} \\[1em] = \frac{-1 \pm \sqrt{(2a + 3)^2}}{2} \\[1em] = \frac{-1 \pm 2a + 3}{2} \\[1em] = \frac{-1 + (2a + 3)}{2} \text{ or } \frac{-1 - (2a + 3)}{2} \\[1em] = \frac{2a + 2}{2} \text{ or } \frac{-2a - 4}{2} \\[1em] = (a + 1) \text{ or } -(a + 2).\end{aligned}$$

Hence, x = (a + 1) or -(a + 2).

Solving,

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

Comparing 3x² - 6x + 2 = 0 with ax² + bx + c = 0 we get,

a = 3, b = -6 and c = 2.

We know that,

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

$$\begin{aligned}x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(3)(2)}}{2(3)} \\[1em] x = \frac{6 \pm \sqrt{36 - 24}}{6} \\[1em] x = \frac{6 \pm \sqrt{12}}{6} \\[1em] x = \frac{6 \pm \sqrt{4 \times 3}}{6} \\[1em] x = \frac{6 \pm 2\sqrt{3}}{6} \\[1em] x = \frac{2(3 \pm \sqrt{3})}{6} \\[1em] x = \frac{3 + \sqrt{3}}{3}, \frac{3 - \sqrt{3}}{3} \\[1em] \therefore m = \frac{3 + \sqrt{3}}{3} \text{ and } n = \frac{3 - \sqrt{3}}{3}.\end{aligned}$$

Solving m × n,

$$\begin{aligned}m × n = \Big(\frac{3 + \sqrt{3}}{3}\Big)\Big(\frac{3 - \sqrt{3}}{3}\Big) \\[1em] = \frac{(3)^2 - (\sqrt{3})^2}{9} \\[1em] = \frac{9 - 3}{9} \\[1em] = \frac{6}{9} \\[1em] = \frac{2}{3}.\end{aligned}$$

Hence, m × n = $\frac{2}{3}$.

Given, $\frac{3}{4}$ is root of 8x² + mx + 15 = 0

$$\begin{aligned}\therefore 8\Big(\frac{3}{4}\Big)^2 + m \times \frac{3}{4} + 15 = 0 \\[1em] \Rightarrow 8 \times \frac{9}{16} + \frac{3m}{4} + 15 = 0 \\[1em] \Rightarrow \frac{9}{2} + \frac{3m}{4} + 15 = 0 \\[1em] \Rightarrow \frac{3m}{4} + \frac{30 + 9}{2} = 0 \\[1em] \Rightarrow \frac{3m}{4} = -\frac{39}{2} \\[1em] \Rightarrow m = -\frac{39}{2} \times \frac{4}{3} \\[1em] \Rightarrow m = -26.\end{aligned}$$

Substituting value of m in equation,

⇒ 8x² + mx + 15 = 0

⇒ 8x² - 26x + 15 = 0

⇒ 8x² - 20x - 6x + 15 = 0

⇒ 4x(2x - 5) - 3(2x - 5) = 0

⇒ (4x - 3)(2x - 5) = 0

⇒ 4x - 3 = 0 or 2x - 5 = 0

⇒ 4x = 3 or 2x = 5

⇒ x = $\frac{3}{4}$ or x = $\frac{5}{2}$.

Hence, m = -26 and other root = $\frac{5}{2}$.

Substituting x = -3 in x² + (3 - 2a)x - 6a = 0,

⇒ (-3)² + (3 - 2a)(-3) - 6a = 0

⇒ 9 - 9 + 6a - 6a = 0

⇒ 0 = 0.

Hence, -3 is one root of the quadratic equation.

We know that,

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

$$\begin{aligned}= \frac{-(3 - 2a) \pm \sqrt{(3 - 2a)^2 - 4(1)(-6a)}}{2} \\[1em] = \frac{2a - 3 \pm \sqrt{9 + 4a^2 - 12a + 24a}}{2} \\[1em] = \frac{2a - 3 \pm \sqrt{4a^2 + 12a + 9}}{2} \\[1em] = \frac{2a - 3 \pm \sqrt{(2a + 3)^2}}{2} \\[1em] = \frac{(2a - 3) + (2a + 3)}{2} \text{ or } \frac{(2a - 3) - (2a + 3)}{2} \\[1em] = \frac{4a}{2} \text{ or } \frac{-6}{2} \\[1em] = 2a \text{ or } -3.\end{aligned}$$

Hence, the other root is 2a.

Given, m + 5 = 0 and n - 1 = 0

∴ m = -5 and n = 1.

Substituting values of m and n in 2x² - mx - 25n = 0 we get,

⇒ 2x² - (-5)x - 25(1) = 0

⇒ 2x² + 5x - 25 = 0

⇒ 2x² + 10x - 5x - 25 = 0

⇒ 2x(x + 5) - 5(x + 5) = 0

⇒ (2x - 5)(x + 5) = 0

⇒ (2x - 5) = 0 or x + 5 = 0

⇒ x = $\frac{5}{2}$ or x = -5.

Hence, x = $\frac{5}{2}$ or -5.

Let (a + b)x = y

⇒ (a + b)²x² - (a + b)x - 6 = 0

⇒ y² - y - 6 = 0

⇒ y² - 3y + 2y - 6 = 0

⇒ y(y - 3) + 2(y - 3) = 0

⇒ (y + 2)(y - 3) = 0

⇒ (y + 2) = 0 or y - 3 = 0

⇒ y = -2 or y = 3.

∴ (a + b)x = -2 or (a + b)x = 3

⇒ x = $-\frac{2}{a + b} \text{ or } \frac{3}{a + b}$.

Hence, x = $-\frac{2}{a + b} \text{ or } \frac{3}{a + b}.$

Since, equation has equal roots, D = 0.

∴ b² - 4ac = 0

⇒ (2(m - 1))² - 4(1)(m + 5) = 0

⇒ (2m - 2)² - (4m + 20) = 0

⇒ 4m² + 4 - 8m - 4m - 20 = 0

⇒ 4m² - 12m - 16 = 0

⇒ 4(m² - 3m - 4) = 0

⇒ m² - 3m - 4 = 0

⇒ m² - 4m + m - 4 = 0

⇒ m(m - 4) + 1(m - 4) = 0

⇒ (m + 1)(m - 4) = 0

⇒ m + 1 = 0 or m - 4 = 0

⇒ m = -1 or m = 4.

Hence, m = -1, 4.

Since equations has real roots, D ≥ 0

∴ b² - 4ac ≥ 0

⇒ 8² - 4(4)(-k) ≥ 0

⇒ 64 + 16k ≥ 0

⇒ 16k ≥ -64

Dividing both sides by 16 we get,

⇒ k ≥ -4

Hence, k ≥ -4.

Since, -2 is a root of the equation 3x² + 7x + p = 1,

∴ 3(-2)² + 7(-2) + p = 1

⇒ 3(4) - 14 + p = 1

⇒ 12 - 14 + p = 1

⇒ -2 + p = 1

⇒ p = 3.

Substituting value of p in x² + k(4x + k - 1) + p = 0 we get,

⇒ x² + k(4x + k - 1) + 3 = 0

⇒ x² + 4kx + k² - k + 3 = 0

Since, roots are equal, D = 0.

∴ b² - 4ac = 0

⇒ (4k)² - 4(1)(k² - k + 3) = 0

⇒ 16k² - 4k² + 4k - 12 = 0

⇒ 12k² + 4k - 12 = 0

⇒ 4(3k² + k - 3) = 0

⇒ 3k² + k - 3 = 0

$$\begin{aligned}\Rightarrow k = \frac{-1 \pm \sqrt{(1)^2 - 4(3)(-3)}}{2(3)} \\[1em] = \frac{-1 \pm \sqrt{1 + 36}}{6} \\[1em] = \frac{-1 \pm \sqrt{37}}{6}.\end{aligned}$$

Hence, p = 3 and k = $\frac{-1 \pm \sqrt{37}}{6}$.

(i) Let the usual speed of the plane be x km/hr.

Distance from Vizag to Colombo = 1800 km

Time taken at usual speed = $\frac{\text{Distance}}{\text{Speed}} = \frac{1800}{x}$ hr

Increased speed = x + 300

Time taken at increased speed = $\frac{\text{Distance}}{\text{Speed}} = \frac{1800}{x + 300}$ hr

Since flight delayed by 60 minutes = 1 hour,

$\frac{1800}{x} - \frac{1800}{x + 300} = 1$

Solving,

$$\begin{aligned}\Rightarrow \frac{1800(x + 300) - 1800x}{x(x + 300)} = 1 \\[1em] \Rightarrow \frac{1800x + 540000 - 1800x}{x(x + 300)} = 1 \\[1em] \Rightarrow 1800x + 540000 - 1800x = x(x + 300) \\[1em] \Rightarrow 540000 = x^2 + 300x \\[1em] \Rightarrow x^2 + 300x - 540000 = 0\end{aligned}$$

Hence, the required equation is x² + 300x - 540000 = 0.

(ii) Comparing x² + 300x - 540000 = 0 with ax² + bx + c = 0 we get,

a = 1, b = 300 and c = -540000.

We know that,

D = b² - 4ac

= (300)² - 4(1)(-540000)

= 90000 + 2160000

= 2250000

∴ D > 0

Hence, roots are real and unequal.

(iii) Comparing x² + 300x - 540000 = 0 with ax² + bx + c = 0 we get,

a = 1, b = 300 and c = -540000.

We know that,

x = $\frac{-b \pm \sqrt{D}}{2a}$

Substituting values of a, b and c in above equation we get,

$$\begin{aligned}\Rightarrow x = \frac{-(300) \pm \sqrt{2250000}}{2(1)} \\[1em] = \frac{-300 \pm 1500}{2} \\[1em] = \frac{-300 + 1500}{2} \text{ or } \frac{-300 - 1500}{2} \\[1em] = \frac{1200}{2} \text{ or } \frac{-1800}{2} \\[1em] = 600 \text{ or } -900.\end{aligned}$$

Speed cannot be negative,

∴ x = 600 km/h

Hence, usual speed of the plane = 600 km/h.