Quadratic Equations

Given,

⇒ 6x² - x - 2 = 0

⇒ 6x² - 4x + 3x - 2 = 0

⇒ 2x(3x - 2) + 1(3x - 2) = 0

⇒ (2x + 1)(3x - 2) = 0

⇒ 2x + 1 = 0 or 3x - 2 = 0      [Using Zero-product rule]

⇒ 2x = -1 or 3x = 2

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

Hence, $-\frac{1}{2}, \frac{2}{3}$ are the solutions of equation 6x² - x - 2 = 0.

Given,

⇒ 9x² - 3x - 2 = 0

⇒ 9x² - 6x + 3x - 2 = 0

⇒ 3x(3x - 2) + 1(3x - 2) = 0

⇒ (3x + 1)(3x - 2) = 0

⇒ 3x + 1 = 0 or 3x - 2 = 0      [Using Zero-product rule]

⇒ 3x = -1 or 3x = 2

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

Hence, $\frac{-1}{3}, \frac{2}{3}$ are the solutions of the equation 9x² - 3x - 2 = 0.

Given,

⇒ 16x² = 25

⇒ 16x² - 25 = 0

⇒ (4x)² - (5)² = 0

⇒ (4x + 5)(4x - 5) = 0

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

⇒ 4x = -5 or 4x = 5

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

Hence, x = $\Big\lbrace\frac{5}{4}, \frac{-5}{4}\Big\rbrace$.

Given,

⇒ x² + 2x = 24

⇒ x² + 2x - 24 = 0

⇒ x² + 6x - 4x - 24 = 0

⇒ x(x + 6) - 4(x + 6) = 0

⇒ (x - 4)(x + 6) = 0

⇒ x - 4 = 0 or x + 6 = 0      [Using Zero-product rule]

⇒ x = 4 or x = -6.

Hence, x = {-6, 4}.

Given,

⇒ x² - x = 156

⇒ x² - x - 156 = 0

⇒ x² - 13x + 12x - 156 = 0

⇒ x(x - 13) + 12(x - 13) = 0

⇒ (x + 12)(x - 13) = 0

⇒ x + 12 = 0 or x - 13 = 0      [Using Zero-product rule]

⇒ x = -12 or x = 13.

Hence, x = {13, -12}.

Given,

⇒ x² - 11x = 42

⇒ x² - 11x - 42 = 0

⇒ x² - 14x + 3x - 42 = 0

⇒ x(x - 14) + 3(x - 14) = 0

⇒ (x + 3)(x - 14) = 0

⇒ x + 3 = 0 or x - 14 = 0      [Using Zero-product rule]

⇒ x = -3 or x = 14.

Hence, x = {14, -3}.

Given,

⇒ x² - 7x + 10 = 0

⇒ x² - 5x - 2x + 10 = 0

⇒ x(x - 5) - 2(x - 5) = 0

⇒ (x - 2)(x - 5) = 0

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

⇒ x = 2 or x = 5

Hence, x = {2, 5}.

Given,

⇒ x² + 18x = 40

⇒ x² + 18x - 40 = 0

⇒ x² + 20x - 2x - 40 = 0

⇒ x(x + 20) - 2(x + 20) = 0

⇒ (x - 2)(x + 20) = 0

⇒ x - 2 = 0 or x + 20 = 0      [Using Zero-product rule]

⇒ x = 2 or x = -20.

Hence, x = {-20, 2}.

Given,

⇒ x² + 17 = 18x

⇒ x² - 18x + 17 = 0

⇒ x² - 17x - x + 17 = 0

⇒ x(x - 17) - 1(x - 17) = 0

⇒ (x - 1)(x - 17) = 0

⇒ x - 1 = 0 or x - 17 = 0      [Using Zero-product rule]

⇒ x = 1 or x = 17.

Hence, x = {1, 17}.

Given,

⇒ 3x² = 5x

⇒ 3x² - 5x = 0

⇒ x(3x - 5) = 0

⇒ x = 0 or (3x - 5) = 0      [Using Zero-product rule]

⇒ x = 0 or 3x = 5

⇒ x = 0 or x = $\frac{5}{3}$.

Hence, x = $\Big\lbrace0, \frac{5}{3}\Big\rbrace$.

Given,

⇒ (x + 3)(x - 3) = 27

⇒ (x)² - (3)² = 27

⇒ x² - 9 = 27

⇒ x² = 27 + 9

⇒ x² = 36

⇒ x = $\sqrt{36}$

⇒ x = ± 6

Hence, x = {6, -6}.

Given,

⇒ x² - 30x + 216 = 0

⇒ x² - 18x - 12x + 216 = 0

⇒ x(x - 18) - 12(x - 18) = 0

⇒ (x - 18)(x - 12) = 0

⇒ (x - 18) = 0 or (x - 12) = 0      [Using Zero-product rule]

⇒ x = 18 or x = 12.

Hence, x = {18, 12}.

Given,

⇒ 12x² + 29x + 14 = 0

⇒ 12x² + 21x + 8x + 14 = 0

⇒ 3x(4x + 7) + 2(4x + 7) = 0

⇒ (4x + 7)(3x + 2) = 0

⇒ 4x + 7 = 0 or 3x + 2 = 0      [Using Zero-product rule]

⇒ 4x = -7 or 3x = -2

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

Hence, x = $\Big\lbrace\frac{-7}{4}, \frac{-2}{3}\Big\rbrace$.

Given,

⇒ 2x² - 7x = 39

⇒ 2x² - 7x - 39 = 0

⇒ 2x² + 6x - 13x - 39 = 0

⇒ 2x(x + 3) - 13(x + 3) = 0

⇒ (x + 3)(2x - 13) = 0

⇒ x + 3 = 0 or 2x - 13 = 0      [Using Zero-product rule]

⇒ x = -3 or 2x = 13

⇒ x = -3 or x = $\frac{13}{2}$.

Hence, x = $\Big\lbrace\frac{13}{2}, -3\Big\rbrace$.

Given,

⇒ 10x² = 9x + 7

⇒ 10x² - 9x - 7 = 0

⇒ 10x² + 5x - 14x - 7 = 0

⇒ 5x(2x + 1) - 7(2x + 1) = 0

⇒ (2x + 1)(5x - 7) = 0

⇒ (2x + 1) = 0 or (5x - 7) = 0      [Using Zero-product rule]

⇒ 2x = -1 or 5x = 7

⇒ x = $-\frac{1}{2}$ or x = $\frac{7}{5}$.

Hence, x = $\Big\lbrace-\frac{1}{2}, \frac{7}{5}\Big\rbrace$.

Given,

⇒ 15x² - 28 = x

⇒ 15x² - x - 28 = 0

⇒ 15x² - 21x + 20x - 28 = 0

⇒ 3x(5x - 7) + 4(5x - 7) = 0

⇒ (5x - 7)(3x + 4) = 0

⇒ (5x - 7) = 0 or (3x + 4) = 0      [Using Zero-product rule]

⇒ 5x = 7 or 3x = - 4

⇒ x = $\frac{7}{5}$ or x = $\frac{-4}{3}$

Hence, x = $\Big\lbrace\frac{-4}{3}, \frac{7}{5}\Big\rbrace$.

Given,

⇒ 8x² + 15 = 26x

⇒ 8x² - 26x + 15 = 0

⇒ 8x² - 20x - 6x + 15 = 0

⇒ 4x(2x - 5) - 3(2x - 5) = 0

⇒ (2x - 5)(4x - 3) = 0

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

⇒ 2x = 5 or 4x = 3

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

Hence, x = $\Big\lbrace\frac{5}{2}, \frac{3}{4}\Big\rbrace$.

Given,

⇒ 3x² + 8 = 10x

⇒ 3x² - 10x + 8 = 0

⇒ 3x² - 6x - 4x + 8 = 0

⇒ 3x(x - 2) - 4(x - 2) = 0

⇒ (x - 2)(3x - 4) = 0

⇒ (x - 2) = 0 or (3x - 4) = 0      [Using Zero-product rule]

⇒ x = 2 or 3x = 4

⇒ x = 2 or x = $\frac{4}{3}$.

Hence, x = $\Big\lbrace2, \frac{4}{3}\Big\rbrace$.

Given,

⇒ x(6x - 11) = 35

⇒ 6x² - 11x - 35 = 0

⇒ 6x² + 10x - 21x - 35 = 0

⇒ 2x(3x + 5) - 7(3x + 5) = 0

⇒ (3x + 5)(2x - 7) = 0

⇒ (3x + 5) = 0 or (2x - 7) = 0      [Using Zero-product rule]

⇒ 3x = -5 or 2x = 7

⇒ x = $\frac{-5}{3}$ or x = $\frac{7}{2}$.

Hence, x = $\Big\lbrace\frac{-5}{3}, \frac{7}{2}\Big\rbrace$.

Given,

⇒ 6x(3x - 7) = 7(7 - 3x)

⇒ 18x² - 42x = 49 - 21x

⇒ 18x² - 42x + 21x - 49 = 0

⇒ 6x(3x - 7) + 7(3x - 7) = 0

⇒ (3x - 7)(6x + 7) = 0

⇒ (3x - 7) = 0 or (6x + 7) = 0      [Using Zero-product rule]

⇒ 3x = 7 or 6x = -7

⇒ x = $\frac{7}{3}$ or x = $\frac{-7}{6}$.

Hence, x = $\Big\lbrace\frac{7}{3}, \frac{-7}{6}\Big\rbrace$.

Given,

⇒ 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, x = $\Big\lbrace2, \frac{5}{2}\Big\rbrace$.

Given,

⇒ 4x² - 9x - 100 = 0

⇒ 4x² + 16x - 25x - 100 = 0

⇒ 4x(x + 4) - 25(x + 4) = 0

⇒ (4x - 25)(x + 4) = 0

⇒ 4x - 25 = 0 or x + 4 = 0      [Using Zero-product rule]

⇒ 4x = 25 or x = -4

⇒ x = $\frac{25}{4}$ or x = -4

Since, x ∈ Q

Hence, x = $\Big\lbrace-4, \frac{25}{4}\Big\rbrace$.

Given,

⇒ 3x² + 11x + 10 = 0

⇒ 3x² + 6x + 5x + 10 = 0

⇒ 3x(x + 2) + 5(x + 2) = 0

⇒ (3x + 5)(x + 2) = 0

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

⇒ 3x = -5 or x = -2

⇒ x = $\frac{-5}{3}$ or x = -2.

Since, x ∈ I

x = -2

Hence, x = {-2}.

Given,

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

⇒ x - 3 = 0 or 3x - 1 = 0      [Using Zero-product rule]

⇒ x = 3 or 3x = 1

⇒ x = 3 or x = $\frac{1}{3}$

Hence, x = $\Big\lbrace3, \frac{1}{3}\Big\rbrace$.

Given,

$$\begin{aligned}\Rightarrow 5x - \frac{35}{x} = 18 \\[1em] \Rightarrow \frac{5x^2 - 35}{x} = 18 \\[1em] \Rightarrow 5x^2 - 35 = 18x \\[1em] \Rightarrow 5x^2 - 18x - 35 = 0 \\[1em] \Rightarrow 5x^2 - 25x + 7x - 35 = 0 \\[1em] \Rightarrow 5x(x - 5) + 7(x - 5) = 0 \\[1em] \Rightarrow (5x + 7)(x - 5) = 0.\end{aligned}$$

⇒ 5x + 7 = 0 or (x - 5) = 0      [Using Zero-product rule]

⇒ 5x = -7 or x = 5

⇒ x = $\frac{-7}{5}$ or x = 5.

Hence, x = $\Big\lbrace5, \frac{-7}{5}\Big\rbrace$.

Given,

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

⇒ (5x + 1) = 0 or (2x - 1) = 0      [Using Zero-product rule]

⇒ 5x = -1 or 2x = 1

⇒ x = $-\frac{1}{5}$ or x = $\frac{1}{2}$.

Hence, x = $\Big\lbrace\frac{1}{2}, -\frac{1}{5}\Big\rbrace$.

Given,

⇒ 3a²x² + 8abx + 4b² = 0

⇒ 3a²x² + 6abx + 2abx + 4b² = 0

⇒ 3ax(ax + 2b) + 2b(ax + 2b) = 0

⇒ (3ax + 2b)(ax + 2b) = 0

⇒ (3ax + 2b) = 0 or (ax + 2b) = 0      [Using Zero-product rule]

⇒ 3ax = -2b or ax = -2b

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

Hence, x = $\Big\lbrace-\frac{2b}{a}, -\frac{2b}{3a}\Big\rbrace$.

Given,

⇒ 4x² - 4ax + (a² - b²) = 0

⇒ (4x² - 4ax + a²) - b² = 0

⇒ [(2x)² - 2 × a × 2x + (a)²] - b² = 0

⇒ (2x - a)² - b² = 0

⇒ (2x - a + b)(2x - a - b) = 0

⇒ (2x - a + b) = 0 or (2x - a - b) = 0      [Using Zero-product rule]

⇒ 2x = a - b or 2x = a + b

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

Hence, x = $\Big\lbrace \frac{a + b}{2}, \frac{a - b}{2}\Big\rbrace$.

Given,

⇒ 5x² - 12x - 9 = 0

⇒ 5x² - 15x + 3x - 9 = 0

⇒ 5x(x - 3) + 3(x - 3) = 0

⇒ (5x + 3)(x - 3) = 0

⇒ (5x + 3) = 0 or (x - 3) = 0      [Using Zero-product rule]

⇒ 5x = -3 or x = 3

⇒ x = $\frac{-3}{5}$ or x = 3.

(i) Since, x ∈ I

Hence, x = {3}.

(ii) Since, x ∈ Q

Hence, x = $\Big\lbrace3, \frac{-3}{5}\Big\rbrace$.

Given,

⇒ 2x² - 11x + 15 = 0

⇒ 2x² - 6x - 5x + 15 = 0

⇒ 2x(x - 3) - 5(x - 3) = 0

⇒ (2x - 5)(x - 3) = 0

⇒ (2x - 5) = 0 or (x - 3) = 0      [Using Zero-product rule]

⇒ 2x = 5 or x = 3

⇒ x = $\frac{5}{2}$ or x = 3.

(i) Since, x ∈ N

Hence, x = {3}.

(ii) Since, x ∈ I

Hence, x = {3}.

Given,

$$\begin{aligned}\Rightarrow \sqrt{3}x^2 + 11x + 6\sqrt{3} = 0 \\[1em] \Rightarrow \sqrt{3}x^2 + 9x + 2x + 6\sqrt{3} = 0 \\[1em] \Rightarrow \sqrt{3}x(x + 3\sqrt{3}) + 2(x + 3\sqrt{3}) = 0 \\[1em] \Rightarrow (\sqrt{3}x + 2)(x + 3\sqrt{3}) = 0 \\[1em] \Rightarrow (\sqrt{3}x + 2) = 0 \text{ or } (x + 3\sqrt{3}) = 0 \text{ [Using Zero-product rule] } \\[1em] \Rightarrow \sqrt{3}x = -2 \text{ or } x = -3\sqrt{3} \\[1em] \Rightarrow x = \frac{-2}{\sqrt{3}} \text{ or } x = -3\sqrt{3}.\end{aligned}$$

Hence, $x = \Big\lbrace\frac{-2}{\sqrt{3}}, -3\sqrt{3}\Big\rbrace$.

Given,

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

Hence, $x = \Big\lbrace\frac{\sqrt{5}}{2}, \frac{-1}{\sqrt{5}}\Big\rbrace$.

Given,

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

Hence, $x = {1, \sqrt{2}}$.

Given,

$\Rightarrow \frac{x + 1}{x - 1} = \frac{3x - 7}{2x - 5}$

⇒ (x + 1)(2x - 5) = (3x - 7)(x - 1)

⇒ (2x² - 5x + 2x - 5) = (3x² - 3x - 7x + 7)

⇒ (2x² - 3x - 5) = (3x² - 10x + 7)

⇒ (3x² - 10x + 7) - (2x² - 3x - 5) = 0

⇒ 3x² - 10x + 7 - 2x² + 3x + 5 = 0

⇒ x² - 7x + 12 = 0

⇒ x² - 3x - 4x + 12 = 0

⇒ x(x - 3) - 4(x - 3) = 0

⇒ (x - 4)(x - 3) = 0

⇒ (x - 4) = 0 or (x - 3) = 0     [Using Zero-product rule]

⇒ x = 4 or x = 3.

Hence, x = {3, 4}.

Given,

$\Rightarrow \frac{3x + 1}{7x + 1} = \frac{5x + 1}{7x + 5}$

⇒ (3x + 1)(7x + 5) = (5x + 1)(7x + 1)

⇒ (21x² + 15x + 7x + 5) = (35x² + 5x + 7x + 1)

⇒ (21x² + 22x + 5) = (35x² + 12x + 1)

⇒ (35x² + 12x + 1) - (21x² + 22x + 5) = 0

⇒ 35x² + 12x + 1 - 21x² - 22x - 5 = 0

⇒ 14x² - 10x - 4 = 0

⇒ 2(7x² - 5x - 2) = 0

⇒ 7x² - 5x - 2 = 0

⇒ 7x² - 7x + 2x - 2 = 0

⇒ 7x(x - 1) + 2(x - 1) = 0

⇒ (7x + 2)(x - 1) = 0

⇒ (7x + 2) = 0 or (x - 1) = 0     [Using Zero-product rule]

⇒ 7x = -2 or x = 1

⇒ x = $\frac{-2}{7}$ or x = 1.

Hence, $x = \Big\lbrace1, \frac{-2}{7}\Big\rbrace$.

Given,

$$\begin{aligned}\Rightarrow \frac{5}{(2x + 1)} + \frac{6}{(x + 1)} = 3 \\[1em] \Rightarrow \frac{5(x + 1) + 6(2x + 1)}{(2x + 1)(x + 1)} = 3 \\[1em] \Rightarrow \frac{5x + 5 + 12x + 6}{(2x + 1)(x + 1)} = 3 \\[1em] \Rightarrow \frac{17x + 11}{(2x^2 + 2x + x + 1)} = 3 \\[1em] \Rightarrow \frac{17x + 11}{(2x^2 + 3x + 1)} = 3 \\[1em] \Rightarrow 17x + 11 = 3(2x^2 + 3x + 1) \\[1em] \Rightarrow 17x + 11 = 6x^2 + 9x + 3 \\[1em] \Rightarrow 6x^2 + 9x - 17x + 3 - 11 = 0 \\[1em] \Rightarrow 6x^2 - 8x - 8 = 0 \\[1em] \Rightarrow 2(3x^2 - 4x - 4) = 0 \\[1em] \Rightarrow 3x^2 - 4x - 4 = 0 \\[1em] \Rightarrow 3x^2 - 6x + 2x - 4 = 0 \\[1em] \Rightarrow 3x(x - 2) + 2(x - 2) = 0 \\[1em] \Rightarrow (3x + 2)(x - 2) = 0 \\[1em] \Rightarrow (3x + 2) \text{ or } (x - 2) = 0 \text{ [Using Zero-product rule] } \\[1em] \Rightarrow 3x = -2 \text{ or } x = 2 \\[1em] \Rightarrow x = \frac{-2}{3} \text{ or } x = 2.\end{aligned}$$

Hence, $x = \Big\lbrace2, \frac{-2}{3}\Big\rbrace$.

Given,

$$\begin{aligned}\Rightarrow \frac{2x}{x - 4} + \frac{2x - 5}{x - 3} = \frac{25}{3} \\[1em] \Rightarrow \frac{2x(x - 3) + (2x - 5)(x - 4)}{(x - 4)(x - 3)} = \frac{25}{3} \\[1em] \Rightarrow \frac{2x^2 - 6x + 2x^2 - 8x - 5x + 20}{x^2 - 3x - 4x + 12} = \frac{25}{3} \\[1em] \Rightarrow \frac{4x^2 - 19x + 20}{x^2 - 7x + 12} = \frac{25}{3} \\[1em] \Rightarrow 3(4x^2 - 19x + 20) = 25(x^2 - 7x + 12) \\[1em] \Rightarrow 12x^2 - 57x + 60 = 25x^2 - 175x + 300 \\[1em] \Rightarrow 25x^2 - 175x + 300 - (12x^2 - 57x + 60) = 0 \\[1em] \Rightarrow 25x^2 - 175x + 300 - 12x^2 + 57x - 60 = 0 \\[1em] \Rightarrow 13x^2 - 118x + 240 = 0 \\[1em] \Rightarrow 13x^2 - 78x - 40x + 240 = 0 \\[1em] \Rightarrow 13x(x - 6) - 40(x - 6) = 0 \\[1em] \Rightarrow (13x - 40)(x - 6) = 0 \\[1em] \Rightarrow (13x - 40) \text{ or } (x - 6) = 0 \text{ [Using Zero-product rule] } \\[1em] \Rightarrow 13x = 40 \text{ or } x = 6 \\[1em] \Rightarrow x = \frac{40}{13} \text{ or } x = 6.\end{aligned}$$

Hence, $x = \Big\lbrace6, \frac{40}{13}\Big\rbrace$.

Given,

$$\begin{aligned}\Rightarrow \frac{x + 3}{x - 2} - \frac{1 - x}{x} = 4\frac{1}{4} \\[1em] \Rightarrow \frac{x(x + 3) - (1 - x)(x - 2)}{x(x - 2)} = \frac{17}{4} \\[1em] \Rightarrow \frac{x^2 + 3x - (x - 2 - x^2 + 2x)}{x^2 - 2x} = \frac{17}{4} \\[1em] \Rightarrow \frac{x^2 + 3x - (3x - 2 - x^2)}{x^2 - 2x} = \frac{17}{4} \\[1em] \Rightarrow x^2 + 3x - 3x + 2 + x^2 = \frac{17}{4} \times (x^2 - 2x) \\[1em] \Rightarrow 4(2x^2 + 2) = 17 \times (x^2 - 2x) \\[1em] \Rightarrow 8x^2 + 8 = 17x^2 - 34x \\[1em] \Rightarrow 17x^2 - 34x - 8x^2 - 8 = 0 \\[1em] \Rightarrow 9x^2 - 34x - 8 = 0 \\[1em] \Rightarrow 9x^2 - 36x + 2x - 8 = 0 \\[1em] \Rightarrow 9x(x - 4) + 2(x - 4) = 0 \\[1em] \Rightarrow (9x + 2)(x - 4) = 0 \\[1em] \Rightarrow (9x + 2) \text{ or } (x - 4) = 0 \text{ [Using Zero-product rule] } \\[1em] \Rightarrow 9x = -2 \text{ or } x = 4 \\[1em] \Rightarrow x = \frac{-2}{9} \text{ or } x = 4.\end{aligned}$$

Hence, $x = \Big\lbrace4, \frac{-2}{9}\Big\rbrace$.

Given,

$$\begin{aligned}\Rightarrow \frac{1}{x - 2} + \frac{2}{x - 1} = \frac{6}{x} \\[1em] \Rightarrow \frac{(x - 1) + 2(x - 2)}{(x - 2)(x - 1)} = \frac{6}{x} \\[1em] \Rightarrow \frac{x - 1 + 2x - 4}{x^2 - x - 2x + 2} = \frac{6}{x} \\[1em] \Rightarrow \frac{x - 1 + (2x - 4)}{x^2 - 3x + 2} = \frac{6}{x} \\[1em] \Rightarrow \frac{3x - 5}{x^2 - 3x + 2} = \frac{6}{x} \\[1em] \Rightarrow x(3x - 5) = 6(x^2 - 3x + 2) \\[1em] \Rightarrow 3x^2 - 5x = 6x^2 - 18x + 12 \\[1em] \Rightarrow 6x^2 - 3x^2 - 18x + 5x + 12 = 0 \\[1em] \Rightarrow 3x^2 - 13x + 12 = 0 \\[1em] \Rightarrow 3x^2 - 9x - 4x + 12 = 0 \\[1em] \Rightarrow 3x(x - 3) - 4(x - 3) = 0 \\[1em] \Rightarrow (3x - 4)(x - 3) = 0 \\[1em] \Rightarrow (3x - 4) \text{ or } (x - 3) = 0 \text{ [Using Zero-product rule] } \\[1em] \Rightarrow 3x = 4 \text{ or } x = 3 \\[1em] \Rightarrow x = \frac{4}{3} \text{ or } x = 3.\end{aligned}$$

Hence, $x = \Big\lbrace3, \frac{4}{3}\Big\rbrace$.

Let us consider y = $\frac{x}{x + 1}$.

Substituting y = $\frac{x}{x + 1}$ in equation $2\Big(\frac{x}{x + 1}\Big)^2 - 5\Big(\frac{x}{x + 1}\Big) + 2 = 0$, we get :

⇒ 2y² - 5y + 2 = 0

⇒ 2y² - 4y - y + 2 = 0

⇒ 2y(y - 2) - 1(y - 2) = 0

⇒ (2y - 1)(y - 2) = 0

⇒ (2y - 1) = 0 or (y - 2) = 0      [Using Zero-product rule]

⇒ 2y = 1 or y = 2

⇒ y = $\frac{1}{2}$ or y = 2.

Now we have,

Case 1 : y = $\frac{1}{2}$

$$\begin{aligned}\Rightarrow y = \frac{x}{x + 1} \\[1em] \Rightarrow \frac{1}{2} = \frac{x}{x + 1} \\[1em] \Rightarrow x + 1 = 2x \\[1em] \Rightarrow 2x - x = 1 \\[1em] \Rightarrow x = 1.\end{aligned}$$

Case 2 : y = 2

⇒ y = $\frac{x}{x + 1}$

⇒ 2 = $\frac{x}{x + 1}$

⇒ 2(x + 1) = x

⇒ 2x + 2 = x

⇒ 2x - x = -2

⇒ x = -2.

Hence, x = {-2, 1}.

Let us consider y = 3x + 1.

Substituting y = 3x + 1 in equation 5(3x + 1)² + 6(3x + 1) - 8 = 0, we get :

⇒ 5y² + 6y - 8 = 0

⇒ 5y² + 10y - 4y - 8 = 0

⇒ 5y(y + 2) - 4(y + 2) = 0

⇒ (5y - 4)(y + 2) = 0

⇒ (5y - 4) = 0 or (y + 2) = 0      [Using Zero-product rule]

⇒ 5y = 4 or y = -2

⇒ y = $\frac{4}{5}$ or y = -2.

Now we have,

Case 1 : y = $\frac{4}{5}$

$$\begin{aligned}\Rightarrow y = \frac{4}{5} \\[1em] \Rightarrow 3x + 1 = \frac{4}{5} \\[1em] \Rightarrow 5(3x + 1) = 4 \\[1em] \Rightarrow 15x + 5 = 4 \\[1em] \Rightarrow 15x = 4 - 5 \\[1em] \Rightarrow 15x = -1 \\[1em] \Rightarrow x = \frac{-1}{15}.\end{aligned}$$

Case 2 : y = -2

⇒ y = -2

⇒ 3x + 1 = -2

⇒ 3x = -2 - 1

⇒ 3x = -3

⇒ x = $\frac{-3}{3}$

⇒ x = -1.

Hence, x = $\Big\lbrace-1, \frac{-1}{15}\Big\rbrace$.

Given,

⇒ $\sqrt{x + 15}$ = (x + 3)

Squaring both sides, we get :

⇒ (x + 15) = (x + 3)²

⇒ x + 15 = (x)² + (3)² + 2 × x × 3

⇒ x + 15 = x² + 9 + 6x

⇒ x² + 9 + 6x - x - 15 = 0

⇒ x² + 5x - 6 = 0

⇒ x² + 6x - x - 6 = 0

⇒ x(x + 6) - 1(x + 6) = 0

⇒ (x + 6)(x - 1) = 0

⇒ (x + 6) = 0 or (x - 1) = 0      [Using Zero-product rule]

⇒ x = -6 or x = 1.

Substituting x = -6 in the L.H.S. of this equation $\sqrt{x + 15}$ = (x + 3)

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

Substituting x = -6 in the R.H.S. of this equation $\sqrt{x + 15}$ = (x + 3)

⇒ x + 3

⇒ -6 + 3

⇒ -3.

L.H.S ≠ R.H.S .

∴ x = -6 is not valid.

Hence, x = {1}.

Given,

⇒ $\sqrt{2x + 9}$ = (13 - x)

Squaring both sides we get :

⇒ (2x + 9) = (13 - x)²

⇒ 2x + 9 = (13²) + (x)² - 2 × 13 × x

⇒ 2x + 9 = 169 + x² - 26x

⇒ x² - 26x + 169 - 2x - 9 = 0

⇒ x² - 28x + 160 = 0

⇒ x² - 20x - 8x + 160 = 0

⇒ x(x - 20) - 8(x - 20) = 0

⇒ (x - 20)(x - 8) = 0

⇒ (x - 20) = 0 or (x - 8) = 0      [Using Zero-product rule]

⇒ x = 20 or x = 8

⇒ x = 8.

Substituting x = 20 in the L.H.S. of this equation $\sqrt{2x + 9}$ = (13 - x)

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

Substituting x = 20 in the R.H.S. of this equation $\sqrt{2x + 9}$ = (13 - x)

⇒ 13 - x

⇒ 13 - 20

⇒ -7.

L.H.S ≠ R.H.S .

∴ x = 20 is not valid.

Hence, x = {8}.

Given,

⇒ $\sqrt{3x^2 - 2}$ = (2x - 1)

Squaring both sides we get :

⇒ (3x² - 2) = (2x - 1)²

⇒ 3x² - 2 = (2x)² + (1)² - 2 × 2x × 1

⇒ 3x² - 2 = 4x² + 1 - 4x

⇒ 4x² + 1 - 4x - 3x² + 2 = 0

⇒ x² - 4x + 3 = 0

⇒ x² - x - 3x + 3 = 0

⇒ x(x - 1) - 3(x - 1) = 0

⇒ (x - 1)(x - 3) = 0

⇒ (x - 1) = 0 or (x - 3) = 0      [Using Zero-product rule]

⇒ x = 1 or x = 3.

Hence, x = {1, 3}.

Given,

⇒ $\sqrt{3x^2 + x + 5}$ = (x - 3)

Squaring both sides we get :

⇒ (3x² + x + 5) = (x - 3)²

⇒ 3x² + x + 5 = (x²) + (3²) - 2 × x × 3

⇒ 3x² + x + 5 = x² + 9 - 6x

⇒ 3x² + x + 5 - x² - 9 + 6x = 0

⇒ 2x² + 7x - 4 = 0

⇒ 2x² - x + 8x - 4 = 0

⇒ x(2x - 1) + 4(2x - 1) = 0

⇒ (2x - 1)(x + 4) = 0

⇒ (2x - 1) = 0 or (x + 4) = 0      [Using Zero-product rule]

⇒ 2x = 1 or x = -4

⇒ x = $\frac{1}{2}$ or x = -4.

Hence, x = $\Big\lbrace-4, \frac{1}{2}\Big\rbrace$.

(i) 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} is x² + x - 6 = 0.

(ii) Since, $\Big\lbrace-3, \frac{2}{5}\Big\rbrace$ is solution set.

It means -3 and $\frac{2}{5}$ are roots of the equation,

∴ x = -3 or x = $\frac{2}{5}$

⇒ x = -3 or 5x = 2

⇒ x + 3 = 0 or 5x - 2 = 0

⇒ (x + 3)(5x - 2) = 0

⇒ (5x² - 2x + 15x - 6) = 0

⇒ 5x² + 13x - 6 = 0.

Hence, quadratic equation with solution set $\Big\lbrace-3, \frac{2}{5}\Big\rbrace$ is 5x² + 13x - 6 = 0.

(iii) Since, $\Big\lbrace\frac{2}{5}, -\frac{1}{2}\Big\rbrace$ is solution set.

It means $\frac{2}{5}$ and $-\frac{1}{2}$ are roots of the equation,

∴ x = $\frac{2}{5}$ or x = $-\frac{1}{2}$

⇒ 5x = 2 or 2x = -1

⇒ 5x - 2 = 0 or 2x + 1 = 0

⇒ (5x - 2)(2x + 1) = 0

⇒ (10x² + 5x - 4x - 2) = 0

⇒ 10x² + x - 2 = 0.

Hence, quadratic equation with solution set $\Big\lbrace\frac{2}{5}, -\frac{1}{2}\Big\rbrace$ is 10x² + x - 2 = 0.

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

⇒ (k + 2)(3)² - 3k + 6 = 0

⇒ (k + 2)(9) - 3k + 6 = 0

⇒ 9k + 18 - 3k + 6 = 0

⇒ 6k + 24 = 0

⇒ 6k = -24

⇒ k = $\frac{-24}{6}$

⇒ k = -4.

Substitute the value of k = -4 in (k + 2)x² - kx + 6 = 0 we get,

⇒ (-4 + 2)(x)² - (-4)x + 6 = 0

⇒ (-2)(x)² - (-4)x + 6 = 0

⇒ -2x² + 4x + 6 = 0

⇒ -2x² - 2x + 6x + 6 = 0

⇒ -2x(x + 1) + 6(x + 1) = 0

⇒ (x + 1)(-2x + 6) = 0

⇒ (x + 1) = 0 or (-2x + 6) = 0      [Using Zero-product rule]

⇒ x = -1 or -2x = -6

⇒ x = -1 or x = $\frac{-6}{-2}$

⇒ x = -1 or x = 3.

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

Comparing equation x² - 4x + 1 = 0 with ax² + bx + c = 0, we get :

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

By formula,

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

Substituting values we get :

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

Hence, $x = {2 + \sqrt{3}, 2 - \sqrt{3}}$.

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

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

By formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow x = \frac{-(7) \pm \sqrt{(7)^2 - 4 \times 9 \times (-2)}}{2(9)} \\[1em] = \frac{-7 \pm \sqrt{49 + 72}}{18} \\[1em] \Rightarrow x = \frac{-7 \pm \sqrt{121}}{18} \\[1em] = \frac{-7 \pm 11}{18} \\[1em] = \frac{-7 + 11}{18} \text{ or } \frac{-7 - 11}{18}\\[1em] = \frac{4}{18} \text{ or } \frac{-18}{18}\\[1em] = \frac{2}{9} \text{ or } -1.\end{aligned}$$

Hence, $x = \Big\lbrace\frac{2}{9}, -1\Big\rbrace$.

Comparing equation $\frac{3}{4}$x² - x - 1 = 0 with ax² + bx + c = 0, we get :

a = $\frac{3}{4}$, b = -1 and c = -1.

By formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4\Big(\frac{3}{4}\Big)(-1)}}{2\Big(\frac{3}{4}\Big)} \\[1em] = \frac{1 \pm \sqrt{1 + 3}}{\Big(\frac{3}{2}\Big)} \\[1em] = \frac{1 \pm \sqrt{4}}{\Big(\frac{3}{2}\Big)} \\[1em] = \frac{1 \pm 2}{\Big(\frac{3}{2}\Big)} \\[1em] = \frac{2(1 \pm 2)}{3} \\[1em] = \frac{2 \pm 4}{3} \\[1em] = \frac{2 + 4}{3} \text{ or } \frac{2 - 4}{3} \\[1em] = \frac{6}{3} \text{ or } \frac{-2}{3} \\[1em] = 2 \text{ or } \frac{-2}{3}.\end{aligned}$$

Hence, $x = \Big\lbrace2, \frac{-2}{3}\Big\rbrace$.

⇒ 3x² + 11x - 4 = 0

Comparing equation 3x² + 11x - 4 = 0 with ax² + bx + c = 0, we get :

a = 3, b = 11 and c = -4.

By formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow x = \frac{-(11) \pm \sqrt{(11)^2 - 4 \times (3) \times (-4)}}{2(3)} \\[1em] = \frac{-11 \pm \sqrt{121 + 48}}{6} \\[1em] = \frac{-11 \pm \sqrt{169}}{6} \\[1em] = \frac{-11 \pm 13}{6} \\[1em] = \frac{-11 + 13}{6} \text{ or } \frac{-11 - 13}{6} \\[1em] = \frac{2}{6} \text{ or } \frac{-24}{6} \\[1em] = \frac{1}{3} \text{ or } -4.\end{aligned}$$

Hence, $x = \Big\lbrace\frac{1}{3}, -4\Big\rbrace$.

Comparing equation 25x² + 30x + 7 = 0 with ax² + bx + c = 0, we get :

a = 25, b = 30 and c = 7.

By formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow x = \frac{-(30) \pm \sqrt{(30)^2 - 4 \times 25 \times 7}}{2(25)} \\[1em] = \frac{-30 \pm \sqrt{900 - 700}}{50} \\[1em] = \frac{-30 \pm \sqrt{200}}{50} \\[1em] = \frac{-30 \pm \sqrt{2 \times 100}}{50} \\[1em] = \frac{-30 \pm 10\sqrt{2}}{50} \\[1em] = \frac{10(-3 \pm \sqrt{2})}{50} \\[1em] = \frac{(-3 \pm \sqrt{2})}{5} \\[1em] = \frac{(-3 + \sqrt{2})}{5} \text{ or } \frac{(-3 - \sqrt{2})}{5}.\end{aligned}$$

Hence, $x = \Big\lbrace\frac{-3 + \sqrt{2}}{5}, \frac{-3 - \sqrt{2}}{5}\Big\rbrace$.

Comparing equation 5x² - 19x + 17 = 0 with ax² + bx + c = 0, we get :

a = 5, b = -19 and c = 17.

By formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow x = \frac{-(-19) \pm \sqrt{(-19)^2 - 4 \times 5 \times 17}}{2 \times 5} \\[1em] = \frac{19 \pm \sqrt{361 - 340}}{10} \\[1em] = \frac{19 \pm \sqrt{21}}{10} \\[1em] = \frac{19 + \sqrt{21}}{10} \text{ or } \frac{19 - \sqrt{21}}{10}.\end{aligned}$$

Hence, $x = \Big\lbrace\frac{19 + \sqrt{21}}{10}, \frac{19 - \sqrt{21}}{10}\Big\rbrace$.

Comparing equation 3x² - 8x + 2 = 0 with ax² + bx + c = 0, we get :

a = 3, b = -8 and c = 2.

By formula,

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

Substituting values we get :

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

Hence, $x = \Big\lbrace\frac{4 + \sqrt{10}}{3}, \frac{4 - \sqrt{10}}{3}\Big\rbrace$.

Comparing equation $\sqrt{3}x^2 + 10x - 8\sqrt{3}$ = 0 with ax² + bx + c = 0, we get :

a = $\sqrt{3}$, b = 10 and c = $-8\sqrt{3}$.

By formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow x = \frac{-(10) \pm \sqrt{(10)^2 - 4 \times \sqrt{3} \times (-8\sqrt{3})}}{2\times(\sqrt{3})} \\[1em] = \frac{-10 \pm \sqrt{100 + 96}}{2\sqrt{3}} \\[1em] = \frac{-10 \pm \sqrt{196}}{2\sqrt{3}} \\[1em] = \frac{-10 \pm 14}{2\sqrt{3}} \\[1em] = \frac{2(-5 \pm 7)}{2\sqrt{3}} \\[1em] = \frac{-5 \pm 7}{\sqrt{3}} \\[1em] = \frac{-5 + 7}{\sqrt{3}} \text{ or } \frac{-5 - 7}{\sqrt{3}} \\[1em] = \frac{2}{\sqrt{3}} \text{ or } \frac{-12}{\sqrt{3}} \\[1em] = \frac{2}{\sqrt{3}} \text{ or } \frac{-4 \times 3}{\sqrt{3}} \\[1em] = \frac{2}{\sqrt{3}} \text{ or } -4\sqrt{3}.\end{aligned}$$

Hence, $x = \Big\lbrace\frac{2}{\sqrt{3}}, -4\sqrt{3}\Big\rbrace$.

Comparing equation 2x² + $\sqrt{7}x$ - 7 = 0 with ax² + bx + c = 0, we get :

a = 2, b = $\sqrt{7}$ and c = -7.

By formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow x = \frac{-(\sqrt{7}) \pm \sqrt{(\sqrt{7})^2 - 4 \times 2 \times (-7)}}{2 \times 2} \\[1em] = \frac{-\sqrt{7} \pm \sqrt{7 + 56}}{4} \\[1em] = \frac{-\sqrt{7} \pm \sqrt{63}}{4} \\[1em] = \frac{-\sqrt{7} \pm \sqrt{7 \times 9}}{4} \\[1em] = \frac{-\sqrt{7} \pm 3\sqrt{7}}{4} \\[1em] = \frac{-\sqrt{7} + 3\sqrt{7}}{4} \text{ or } \frac{-\sqrt{7} - 3\sqrt{7}}{4} \\[1em] = \frac{2\sqrt{7}}{4} \text{ or } \frac{-4\sqrt{7}}{4} \\[1em] = \frac{\sqrt{7}}{2} \text{ or } -\sqrt{7}.\end{aligned}$$

Hence, $x = \Big\lbrace-\sqrt{7}, \frac{\sqrt{7}}{2}\Big\rbrace$.

⇒ 6x² - 31x - 105 = 0

Comparing equation 6x² - 31x - 105 = 0 with ax² + bx + c = 0, we get :

a = 6, b = -31 and c = -105.

By formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow x = \frac{-(-31) \pm \sqrt{(-31)^2 - 4 \times 6 \times(-105)}}{2\times(6)} \\[1em] = \frac{31 \pm \sqrt{961 + 2520}}{12} \\[1em] = \frac{31 \pm \sqrt{3481}}{12} \\[1em] = \frac{31 \pm 59}{12} \\[1em] = \frac{31 + 59}{12} \text{ or } \frac{31 - 59}{12} \\[1em] = \frac{90}{12} \text{ or } \frac{-28}{12} \\[1em] = \frac{15}{2} \text{ or } \frac{-7}{3}.\end{aligned}$$

Hence, $x = \Big\lbrace\frac{15}{2}, \frac{-7}{3}\Big\rbrace$.

$$\begin{aligned}\Rightarrow \frac{x + 3}{2x + 3} = \frac{x + 1}{3x + 2} \\[1em] \Rightarrow (x + 3)(3x + 2) = (x + 1)(2x + 3) \\[1em] \Rightarrow (3x^2 + 2x + 9x + 6) = (2x^2 + 3x + 2x + 3) \\[1em] \Rightarrow 3x^2 + 11x + 6 = 2x^2 + 5x + 3 \\[1em] \Rightarrow 3x^2 - 2x^2 + 11x - 5x + 6 - 3 = 0 \\[1em] \Rightarrow x^2 + 6x + 3 = 0.\end{aligned}$$

Comparing equation x² + 6x + 3 = 0 with ax² + bx + c = 0, we get :

a = 1, b = 6 and c = 3.

By formula,

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

Substituting values we get :

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

Hence, $x = \Big\lbrace(-3 + \sqrt{6}), (-3 - \sqrt{6})\Big\rbrace$.

$$\begin{aligned}\Rightarrow \frac{x - 1}{x - 2} + \frac{x - 3}{x - 4} = 3\frac{1}{3} \\[1em] \Rightarrow \frac{(x - 1)(x - 4) + (x - 3)(x - 2)}{(x - 2)(x - 4)} = \frac{10}{3} \\[1em] \Rightarrow \frac{x^2 - 4x - x + 4 + (x^2 - 2x - 3x + 6)}{x^2 - 4x - 2x + 8} = \frac{10}{3} \\[1em] \Rightarrow \frac{x^2 - 5x + 4 + (x^2 - 5x + 6)}{x^2 - 6x + 8} = \frac{10}{3} \\[1em] \Rightarrow \frac{2x^2 - 10x + 10 }{x^2 - 6x + 8} = \frac{10}{3} \\[1em] \Rightarrow 3(2x^2 - 10x + 10) = 10(x^2 - 6x + 8) \\[1em] \Rightarrow 6x^2 - 30x + 30 = 10x^2 - 60x + 80 \\[1em] \Rightarrow 10x^2 - 60x + 80 - (6x^2 - 30x + 30 ) = 0 \\[1em] \Rightarrow 10x^2 - 60x + 80 - 6x^2 + 30x - 30 = 0 \\[1em] \Rightarrow 4x^2 - 30x + 50 = 0 \\[1em] \Rightarrow 2(2x^2 - 15x + 25) = 0 \\[1em] \Rightarrow 2x^2 - 15x + 25 = 0.\end{aligned}$$

Comparing equation 2x² - 15x + 25 = 0 with ax² + bx + c = 0, we get :

a = 2, b = -15 and c = 25.

By formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow x = \frac{-(-15) \pm \sqrt{(-15)^2 - 4(2)(25)}}{2(2)} \\[1em] = \frac{15 \pm \sqrt{225 - 200}}{4} \\[1em] = \frac{15 \pm \sqrt{25}}{4} \\[1em] = \frac{15 \pm 5}{4} \\[1em] = \frac{15 + 5}{4} \text{ or } \frac{15 - 5}{4} \\[1em] = \frac{20}{4} \text{ or } \frac{10}{4} \\[1em] = 5 \text{ or } \frac{5}{2}.\end{aligned}$$

Hence, x = $\Big\lbrace5, \frac{5}{2}\Big\rbrace$.

Given,

⇒ x² - 10x + 6 = 0

Comparing equation x² - 10x + 6 = 0 with ax² + bx + c = 0, we get :

a = 1, b = -10 and c = 6.

By formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow x = \frac{-(-10) \pm \sqrt{(-10)^2 - 4 \times 1 \times (6)}}{2 \times (1)} \\[1em] = \frac{10 \pm \sqrt{100 - 24}}{2} \\[1em] = \frac{10 \pm \sqrt{76}}{2} \\[1em] = \frac{10 \pm \sqrt{19 \times 4}}{2} \\[1em] = \frac{10 \pm 2\sqrt{19}}{2} \\[1em] = \frac{10 + 2\sqrt{19}}{2} \text{ or } \frac{10 - 2\sqrt{19}}{2} \\[1em] = \frac{2(5 + \sqrt{19})}{2} \text{ or } \frac{2(5 - \sqrt{19})}{2} \\[1em] = 5 + \sqrt{19} \text{ or } 5 - \sqrt{19} \\[1em] = 5 + 4.36 \text{ or } 5 - 4.36 \\[1em] = 9.36 \text{ or } 0.64\end{aligned}$$

Hence, x = {9.36, 0.64}.

Given,

⇒ 2x² - 6x + 3 = 0

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

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

By formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4 \times 2 \times (3)}}{2 \times (2)} \\[1em] = \frac{6 \pm \sqrt{36 - 24}}{4} \\[1em] = \frac{6 \pm \sqrt{12}}{4} \\[1em] = \frac{6 \pm \sqrt{3 \times 4}}{4} \\[1em] = \frac{6 \pm 2\sqrt{3}}{4} \\[1em] = \frac{6 + 2\sqrt{3}}{4} \text{ or } \frac{6 - 2\sqrt{3}}{4} \\[1em] = \frac{2(3 + \sqrt{3})}{4} \text{ or } \frac{2(3 - \sqrt{3})}{4} \\[1em] = \frac{3 + \sqrt{3}}{2} \text{ or } \frac{3 - \sqrt{3}}{2} \\[1em] = \frac{3 + 1.73}{2} \text{ or } \frac{3 - 1.73}{2} \\[1em] = \frac{4.73}{2} \text{ or } \frac{1.27}{2} \\[1em] = 2.365 \text{ or } 0.635 \\[1em] \approx 2.37 \text{ or } 0.64\end{aligned}$$

Hence, x = {2.37, 0.64}.

Given,

⇒ 3x² - 32x + 12 = 0

Comparing equation 3x² - 32x + 12 = 0 with ax² + bx + c = 0, we get :

a = 3, b = -32 and c = 12.

By formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow x = \frac{-(-32) \pm \sqrt{(-32)^2 - 4 \times 3 \times (12)}}{2 \times (3)} \\[1em] = \frac{32 \pm \sqrt{1024 - 144}}{6} \\[1em] = \frac{32 \pm \sqrt{880}}{6} \\[1em] = \frac{32 \pm \sqrt{4 \times 220}}{6} \\[1em] = \frac{32 \pm 2\sqrt{220}}{6} \\[1em] = \frac{32 + 2\sqrt{220}}{6} \text{ or } \frac{32 - 2\sqrt{220}}{6} \\[1em] = \frac{2(16 + \sqrt{220})}{6} \text{ or } \frac{2(16 - \sqrt{220})}{6} \\[1em] = \frac{16 + \sqrt{220}}{3} \text{ or } \frac{16 - \sqrt{220}}{3} \\[1em] = \frac{16 + 14.83}{3} \text{ or } \frac{16 - 14.83}{3} \\[1em] = \frac{30.83}{3} \text{ or } \frac{1.17}{3} \\[1em] = 10.28 \text{ or } 0.39\end{aligned}$$

Hence, x = {10.28, 0.39}.

Given,

⇒ x² + 7x - 7 = 0

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

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

By formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow x = \frac{-(7) \pm \sqrt{(7)^2 - 4 \times (1) \times (-7)}}{2 \times (1)} \\[1em] = \frac{-7 \pm \sqrt{49 + 28}}{2} \\[1em] = \frac{-7 \pm \sqrt{77}}{2} \\[1em] = \frac{-7 + \sqrt{77}}{2} \text{ or } \frac{-7 - \sqrt{77}}{2} \\[1em] = \frac{-7 + 8.77}{2} \text{ or } \frac{-7 - 8.77}{2} \\[1em] = \frac{1.77}{2} \text{ or } \frac{-15.77}{2} \\[1em] = 0.89 \text{ or } -7.89.\end{aligned}$$

Hence, x = {0.89, -7.89}.

Given,

⇒ 3x² - x - 7 = 0

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

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

By formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \times (3) \times (-7)}}{2 \times (3)} \\[1em] = \frac{1 \pm \sqrt{1 + 84}}{6} \\[1em] = \frac{1 \pm \sqrt{85}}{6} \\[1em] = \frac{1 + \sqrt{85}}{6} \text{ or } \frac{1 - \sqrt{85}}{6} \\[1em] = \frac{1 + 9.2195}{6} \text{ or } \frac{1 - 9.2195}{6} \\[1em] = \frac{10.2195}{6} \text{ or } \frac{-8.2195}{6} \\[1em] = 1.703 \approx 1.70 \text{ or } -1.369 \approx -1.37.\end{aligned}$$

Hence, x = {1.70, -1.37}.

Given,

⇒ 4x² - 7x + 2 = 0

Comparing equation 4x² - 7x + 2 = 0 with ax² + bx + c = 0, we get :

a = 4, b = -7 and c = 2.

By formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4 \times (4) \times (2)}}{2 \times (4)} \\[1em] = \frac{7 \pm \sqrt{49 - 32}}{8} \\[1em] = \frac{7 \pm \sqrt{17}}{8} \\[1em] = \frac{7 + \sqrt{17}}{8} \text{ or } \frac{7 - \sqrt{17}}{8} \\[1em] = \frac{7 + 4.12}{8} \text{ or } \frac{7 - 4.12}{8} \\[1em] = \frac{11.12}{8} \text{ or } \frac{2.88}{8} \\[1em] = 1.39 \text{ or } 0.36\end{aligned}$$

Hence, x = {1.39, 0.36}.

Given,

⇒ x² - 7x + 3 = 0

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

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

By formula,

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

Substituting values we get :

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

Hence, x = {6.54, 0.46}.

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

a = 1, b = -5 and c = -10.

By formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow x = \frac{-(-5) \pm \sqrt{(-5)^2 + 4 \times (1) \times (-10)}}{2 \times (1)} \\[1em] \Rightarrow x = \frac{5 \pm \sqrt{25 + 40}}{2} \\[1em] \Rightarrow x = \frac{5 \pm \sqrt{65}}{2} \\[1em] \Rightarrow x = \frac{5 \pm 8.06}{2} \\[1em] \Rightarrow x = \frac{5 + 8.06}{2} \text{ or } \frac{5 - 8.06}{2} \\[1em] \Rightarrow x = \frac{13.06}{2} \text{ or } \frac{-3.06}{2} \\[1em] \Rightarrow x = 6.53 \text{ or } -1.53.\end{aligned}$$

Hence, x = 6.53 and -1.53

Given,

⇒ x² - 4x - 8 = 0

Comparing equation x² - 4x - 8 = 0 with ax² + bx + c = 0, we get :

a = 1, b = -4 and c = -8.

By formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \times (1) \times (-8)}}{2 \times (1)} \\[1em] = \frac{4 \pm \sqrt{16 + 32}}{2} \\[1em] = \frac{4 \pm \sqrt{48}}{2} \\[1em] = \frac{4 + \sqrt{48}}{2} \text{ or } \frac{4 - \sqrt{48}}{2} \\[1em] = \frac{4 + 6.928}{2} \text{ or } \frac{4 - 6.928}{2} \\[1em] = \frac{10.928}{2} \text{ or } \frac{-2.928}{2} \\[1em] = 5.464 \text{ or } -1.464\end{aligned}$$

Correcting the value of x to three significant figures.

Hence, x = {5.46, -1.46}.

Given,

⇒ x² + 4x - 8 = 0

Comparing equation x² + 4x - 8 = 0 with ax² + bx + c = 0, we get :

a = 1, b = 4 and c = -8.

By formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow x = \frac{-(4) \pm \sqrt{(4)^2 - 4 \times (1) \times (-8)}}{2 \times (1)} \\[1em] = \frac{-4 \pm \sqrt{16 + 32}}{2} \\[1em] = \frac{-4 \pm \sqrt{48}}{2} \\[1em] = \frac{-4 + \sqrt{48}}{2} \text{ or } \frac{-4 - \sqrt{48}}{2} \\[1em] = \frac{-4 + 6.9}{2} \text{ or } \frac{-4 - 6.9}{2} \\[1em] = \frac{2.9}{2} \text{ or } \frac{-10.9}{2} \\[1em] = 1.45 \text{ or } -5.45\end{aligned}$$

Correcting the value of x to one decimal place.

Hence, x = {1.5, -5.5}.

Comparing x² - 8x + 7 = 0 with ax² + bx + c = 0 we get,

a = 1, b = -8 and c = 7.

We know that,

Discriminant (D) = b² - 4ac = (-8)² - 4 × (1) × (7)

= 64 - 28 = 36; which is positive, and a perfect square.

Hence, the roots are rational and unequal.

Comparing 6x² + 7x - 10 = 0 with ax² + bx + c = 0 we get,

a = 6, b = 7 and c = -10.

We know that,

Discriminant (D) = b² - 4ac = (7)² - 4 × (6) × (-10)

= 49 -(-240)

= 49 + 240 = 289; which is positive, and a perfect square.

Hence, the roots are rational and unequal.

Comparing 25x² + 30x + 7 = 0 with ax² + bx + c = 0 we get,

a = 25, b = 30 and c = 7.

We know that,

Discriminant (D) = b² - 4ac = (30)² - 4 × 25 × 7

= 900 - 700 = 200; which is positive, but not a perfect square.

Hence, the roots are irrational and unequal.

⇒ 15x² - 28 = x

⇒ 15x² - x - 28 = 0

Comparing 15x² - x - 28 = 0 with ax² + bx + c = 0 we get,

a = 15, b = -1 and c = -28.

We know that,

Discriminant (D) = b² - 4ac = (1)² - 4 × 15 × -28

= 1 - (-1680) = 1 + 1680 = 1681; which is positive and is a perfect square.

Hence, the roots are rational and unequal.

⇒ 16x² = 24x + 1

⇒ 16x² - 24x - 1 = 0

Comparing 16x² - 24x - 1 = 0 with ax² + bx + c = 0 we get,

a = 16, b = -24 and c = -1.

We know that,

Discriminant (D) = b² - 4ac

= (-24)² - 4 × 16 × -1

= 576 - (-64) = 576 + 64 = 640; which is positive and is not a perfect square.

Hence, the roots are irrational and unequal.

Comparing 2x² - $2\sqrt{6}x$ + 3 = 0 with ax² + bx + c = 0 we get,

a = 2, b = $-2\sqrt{6}$ and c = 3.

We know that,

Discriminant (D) = b² - 4ac = $(-2\sqrt{6})^2$ - 4 × 2 × 3

= (4 × 6) - 24 = 24 - 24 = 0;

Since, b is irrational and discriminant equals to zero.

Hence, the roots are irrational and equal.

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

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

We know that,

Discriminant (D) = b² - 4ac = (2)² - 4 × 2 × 3

= 4 - 24 = -20; which is negative.

Hence, the roots are imaginary and unequal.

Comparing 2x² - 5x - 4 = 0 with ax² + bx + c = 0 we get,

a = 2, b = -5 and c = -4.

We know that,

Discriminant (D) = b² - 4ac = (-5)² - 4 × 2 × -4

= 25 - (-32) = 25 + 32 = 57 ; which is positive and is not a perfect square.

Hence, the roots are irrational and unequal.

Comparing 5x² - 13x - 6 = 0 with ax² + bx + c = 0 we get,

a = 5, b = -13 and c = -6.

We know that,

Discriminant (D) = b² - 4ac = (-13)² - 4 × 5 × -6

= 169 - (-120) = 169 + 120 = 289 ; which is positive and is a perfect square.

Hence, the roots are rational and unequal.

Comparing 9x² - 6x + 1 = 0 with ax² + bx + c = 0 we get,

a = 9, b = -6 and c = 1.

We know that,

Discriminant (D) = b² - 4ac = (-6)² - 4 × 9 × 1

= 36 - 36 = 0.

Hence, the roots are rational and equal.

Comparing 3x² - 2x + 5 = 0 with ax² + bx + c = 0 we get,

a = 3, b = -2 and c = 5.

We know that,

Discriminant (D)= b² - 4ac = (-2)² - 4 × 3 × 5

= 4 - 60 = -56; which is negative.

Hence, the roots are imaginary and unequal.

Comparing x² + $2\sqrt{3}x$ - 1 = 0 with ax² + bx + c = 0 we get,

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

We know that,

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

= (2 × 3) + 4 = 10 ; which is positive and is not a perfect square.

Since, b is irrational and discriminant is greater than zero.

Hence, the roots are irrational and unequal.

Comparing 9x² + kx + 1 = 0 with ax² + bx + c = 0 we get,

a = 9, b = k and c = 1.

Since equations has equal roots,

∴ D = 0

⇒ (k)² - 4 × 9 × 1 = 0

⇒ k² - 36 = 0

⇒ k² = 36

⇒ k = $\sqrt{36}$

⇒ k = ± 6

Hence, k = {6, -6}.

Comparing x² - 2kx + 7k - 12 = 0 with ax² + bx + c = 0 we get,

a = 1, b = -2k and c = (7k - 12).

Since equations has equal roots,

∴ D = 0

⇒ (-2k)² - 4 × 1 × (7k - 12) = 0

⇒ 4k² - (28k - 48) = 0

⇒ 4k² - 28k + 48 = 0

⇒ 4k² - 16k - 12k + 48 = 0

⇒ 4k(k - 4) - 12(k - 4) = 0

⇒ (k - 4)(4k - 12) = 0

⇒ (k - 4) = 0 or (4k - 12) = 0      [Using Zero-product rule]

⇒ k = 4 or 4k = 12

⇒ k = 4 or k = $\frac{12}{4}$

⇒ k = 4 or k = 3.

Hence, k = {4, 3}.

Comparing (3k + 1)x² + 2(k + 1)x + k = 0 with ax² + bx + c = 0 we get,

a = (3k + 1), b = 2(k + 1) and c = k.

Since equations has equal roots,

∴ D = 0

⇒ [2(k + 1)]² - 4 × (3k + 1) × (k) = 0

⇒ 4(k + 1)² - (12k + 4) × (k) = 0

⇒ 4[(k)² + (1)² + 2 × k × 1] - (12k² + 4k) = 0

⇒ 4(k² + 1 + 2k) - 12k² - 4k = 0

⇒ 4k² + 4 + 8k - 12k² - 4k = 0

⇒ -8k² + 4k + 4 = 0

⇒ -8k² + 8k - 4k + 4 = 0

⇒ -8k(k - 1) - 4(k - 1) = 0

⇒ (k - 1)(-8k - 4) = 0

⇒ (k - 1) = 0 or (-8k - 4) = 0      [Using Zero-product rule]

⇒ k = 1 or -8k = 4

⇒ k = 1 or k = $\frac{4}{-8}$

⇒ k = 1 or k = $-\frac{1}{2}$

Hence, k = $\Big\lbrace-\frac{1}{2}, 1\Big\rbrace$.

Comparing x² - 2(5 + 2k)x + 3(7 + 10k) = 0 with ax² + bx + c = 0 we get,

a = 1, b = -2(5 + 2k) and c = 3(7 + 10k).

Since equations has equal roots,

∴ D = 0

⇒ [-2(5 + 2k)]² - 4 × 1 × 3(7 + 10k) = 0

⇒ 4(5 + 2k)² - 12(7 + 10k) = 0

⇒ 4[(5)² + (2k)² + 2 × 5 × 2k] - (84 + 120k) = 0

⇒ 4(25 + 4k² + 20k) - 84 - 120k = 0

⇒ 100 + 16k² + 80k - 84 - 120k = 0

⇒ 16k² - 40k + 16 = 0

⇒ 16k² - 8k - 32k + 16 = 0

⇒ 8k(2k - 1) - 16(2k - 1) = 0

⇒ (2k - 1)(8k - 16)= 0

⇒ (2k - 1) = 0 or (8k - 16)= 0      [Using Zero-product rule]

⇒ 2k = 1 or 8k = 16

⇒ k = $\frac{1}{2}$ or k = $\frac{16}{8}$

⇒ k = $\frac{1}{2}$ or k = 2

Hence, k = $\Big\lbrace2 , \frac{1}{2}\Big\rbrace$.

Comparing (k + 1)x² + 2(k + 3)x + (k + 8) = 0 with ax² + bx + c = 0 we get,

a = (k + 1), b = 2(k + 3) and c = (k + 8).

Since equations has equal roots,

∴ D = 0

⇒ [2(k + 3)]² - 4 × (k + 1) × (k + 8) = 0

⇒ 4(k + 3)² - (4k + 4) × (k + 8) = 0

⇒ 4[(k)² + (3)² + 2 × k × 3] - (4k² + 32k + 4k + 32) = 0

⇒ 4(k² + 9 + 6k) - (4k² + 36k + 32) = 0

⇒ 4k² + 36 + 24k - 4k² - 36k - 32 = 0

⇒ -12k + 4 = 0

⇒ -12k = -4

⇒ k = $\frac{-4}{-12}$

⇒ k = $\frac{1}{3}$.

Hence, k = $\Big\lbrace\frac{1}{3}\Big\rbrace$.

⇒ kx² + kx + 1 = -4x² - x

⇒ kx² + kx + 1 + 4x² + x = 0

⇒ kx² + 4x² + kx + x + 1 = 0

⇒ x² (k + 4) + x(k + 1) + 1 = 0

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

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

Since equations has equal roots,

∴ D = 0

⇒ (k + 1)² - 4.(k + 4).1 = 0

⇒ [(k)² + (1)² + 2 × k × 1] - (4k + 16) = 0

⇒ k² + 1 + 2k - 4k - 16 = 0

⇒ k² - 2k - 15 = 0

⇒ k² - 5k + 3k - 15 = 0

⇒ k(k - 5) + 3(k - 5) = 0

⇒ (k - 5)(k + 3) = 0

⇒ (k - 5) = 0 or (k + 3) = 0      [Using Zero-product rule]

⇒ k = 5 or k = -3

Hence, k = {5 , -3}.

⇒ 3kx² = 4(kx - 1)

⇒ 3kx² = 4kx - 4

⇒ 3kx² - 4kx + 4 = 0

Comparing 3kx² - 4kx + 4 = 0 with ax² + bx + c = 0 we get,

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

Since equations has equal roots,

∴ D = 0

⇒ (-4k)² - 4 × (3k) × 4 = 0

⇒ 16k² - 48k = 0

⇒ 16k(k - 3) = 0

⇒ 16k = 0 or (k - 3) = 0      [Using Zero-product rule]

⇒ k = 0 or k = 3

Hence, k = {0, 3}.

Comparing x² + 4kx + (k² - k + 2) = 0 with ax² + bx + c = 0 we get,

a = 1, b = 4k and c = (k² - k + 2).

Since equations has equal roots,

∴ D = 0

⇒ (4k)² - 4.1.(k² - k + 2) = 0

⇒ 16k² - (4k² - 4k + 8) = 0

⇒ 16k² - 4k² + 4k - 8 = 0

⇒ 12k² + 4k - 8 = 0

⇒ 12k² + 12k - 8k - 8 = 0

⇒ 12k(k + 1) - 8(k + 1) = 0

⇒ (k + 1)(12k - 8) = 0

⇒ (k + 1) = 0 or (12k - 8) = 0      [Using Zero-product rule]

⇒ k = -1 or 12k = 8

⇒ k = -1 or k = $\frac{8}{12}$

⇒ k = -1 or k = $\frac{2}{3}$.

Hence, k = $\Big\lbrace-1,\frac{2}{3}\Big\rbrace$.

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

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

We know that,

Discriminant (D) = b² - 4ac = (a)² - 4.(1).(-1) = a² + 4

Since a² ≥ 0 for all real values of a, we have D = a² + 4 > 0

The equation has real and distinct roots for all real values of a.

Hence, equation x² + ax - 1 = 0 has real and distinct roots for all real values of a.

Comparing 3x² + 7x + 8 = 0 with ax² + bx + c = 0 we get,

a = 3, b = 7 and c = 8.

We know that,

Discriminant (D) = b² - 4ac = (7)² - 4.(3).(8) = 49 - 96 = - 47; which is negative.

The equation has imaginary and unequal roots for all real values of x.

Hence, the equation 3x² + 7x + 8 = 0 is not true for any real value of x.

Comparing (c² - ab)x² - 2(a² - bc)x + (b² - ac) = 0 with ax² + bx + c = 0 we get,

a = (c² - ab), b = -2(a² - bc) and c = (b² - ac).

Since equations has equal roots,

∴ D = 0

⇒ [-2(a² - bc)]² - 4 × (c² - ab) ×(b² - ac) = 0

⇒ 4(a² - bc)² - 4 × (c²b² - ac³ - ab³ + a²bc ) = 0

⇒ 4[(a²)² + (bc)² - 2 × a² × bc] - (4c²b² - 4ac³ - 4ab³ + 4a²bc) = 0

⇒ 4(a⁴ + b²c² - 2a²bc) - 4c²b² + 4ac³ + 4ab³ - 4a²bc = 0

⇒ 4a⁴ + 4b²c² - 8a²bc - 4c²b² + 4ac³ + 4ab³ - 4a²bc = 0

⇒ 4a⁴ + 4b²c² - 4c²b² - 8a²bc - 4a²bc + 4ac³ + 4ab³ = 0

⇒ 4a⁴ - 12a²bc + 4ac³ + 4ab³ = 0

⇒ 4a(a³ - 3abc + c³ + b³) = 0

⇒ 4a = 0 or (a³ + c³ + b³ - 3abc) = 0      [Using Zero-product rule]

⇒ a = 0 or a³ + c³ + b³ = 3abc

Hence, proved a = 0 or a³ + b³ + c³ = 3abc.

(a - b)x² + (b + c - a)x - c = 0, a ≠ b

Comparing (a - b)x² + (b + c - a)x - c = 0 with ax² + bx + c = 0 we get,

a = (a - b), b = (b + c - a) and c = -c.

We know that,

Discriminant (D) = b² - 4ac

= (b + c - a)² - 4 × (a - b) × (-c)

= [(b + c) - a]² - 4 × (-ac + bc)

= [(b + c)² + (a)² - 2 × (b + c) × (a)] - (-4ac + 4bc)

= [(b)² + (c)² + 2 × b × c + a² - 2 × (ab + ac)] + 4ac - 4bc

= (b² + c² + 2bc + a² - 2ab - 2ac) + 4ac - 4bc

= a² + b² + c² + 2bc - 2ab - 2ac + 4ac - 4bc

= a² + b² + c² + 2bc - 4bc - 2ab - 2ac + 4ac

= a² + b² + c² - 2bc - 2ab + 2ac

= a² + c² + 2ac + b² - 2ab - 2bc

= a² + c² + 2ac + b² - 2.(a + c).b

= (a + c - b)²

Thus D = (a + c - b)², which is a perfect square.

The equation has rational roots. The roots are unequal if b ≠ a + c and equal if b = a + c (as then discriminant equals to zero).

Hence, (a - b)x² + (b + c - a)x - c = 0 has rational roots. The roots are unequal if b ≠ a + c and equal if b = a + c.

Comparing (b - c)x² + (c - a)x + (a - b) = 0 with ax² + bx + c = 0 we get,

a = (b - c), b = (c - a) and c = (a - b).

We know that,

Discriminant (D) = b² - 4ac

= (c - a)² - 4 × (b - c) × (a - b)

= [(c)² + (a)² - 2 × c × a] - 4 × (ba - b² - ac + bc)

= (c² + a² - 2ac) -(4ba - 4b² - 4ac + 4bc)

= c² + a² - 2ac - 4ba + 4b² + 4ac - 4bc

= c² + a² - 2ac + 4ac - 4ba + 4b² - 4bc

= c² + a² + 2ac - 4ba + 4b² - 4bc

= c² + a² + 2ac + (2b)² - 2.(c + a).2b

= (a + c - 2b)²

Thus, D = (a + c - 2b)², which is a perfect square.

The equation has rational roots.

Hence, proved that (b - c)x² + (c - a)x + (a - b) = 0 has rational roots.

Given,

⇒ px² + qx + r = 0

Comparing equation px² + qx + r = 0 with ax² + bx + c = 0, we get :

a = p, b = q and c = r.

By formula,

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

Substituting values we get :

x = $\frac{-q \pm \sqrt{q^2 - 4pr}}{2p}$.

Hence, option 3 is the correct option.

We know that,

Discriminant (D) = b² - 4ac.

Hence, option 3 is the correct option.

If the Discriminant is greater than 0, then the roots are real and distinct.

If the Discriminant is equal to 0, then the roots are real and equal.

Thus, for real roots of a quadratic equation, the discriminant must be greater than or equal to zero.

Hence, option 1 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.

Let us consider the quadratic equation ax² + bx + c = 0, a ≠ 0 and a, b, c are real numbers.

Let r₁ and r₂ be the roots of this equation.

r₁ = $\frac{-b + \sqrt{D}}{2a}$

r₂ = $\frac{-b - \sqrt{D}}{2a}$

if the roots are real and equal then D = 0,

r₁ = $\frac{-b + \sqrt{0}}{2a}$

r₁ = $\frac{-b}{2a}$

r₂ = $\frac{-b - \sqrt{0}}{2a}$

r₂ = $\frac{-b}{2a}$

r₁ = r₂ = $\frac{-b}{2a}$

Hence, each root can be given by $\frac{-b}{2a}$.

Hence, option 2 is the correct option.

We know that,

If the Discriminant is greater than 0 and a perfect square and a, b, c are rational, then the roots are rational and unequal.

Hence, option 4 is the correct option.

We know that,

If the Discriminant is greater than 0 and a perfect square and b is irrational, then the roots are irrational and unequal.

Hence, option 1 is the correct option.

Solving,

⇒ 2x² - 5x = (x - 1)²

⇒ 2x² - 5x = [(x)² + (1)² - 2 × x × 1]

⇒ 2x² - 5x = x² + 1 - 2x

⇒ 2x² - 5x - x² - 1 + 2x = 0

⇒ 2x² - x² - 5x + 2x - 1 = 0

⇒ x² - 3x - 1 = 0

The highest power of equation is 2 and it's equivalent to ax² + bx + c = 0 . Hence the equation is a quadratic equation.

Hence, option 2 is the correct option.

Solving,

⇒ x³ - x² = (x - 1)³

⇒ x³ - x² = [(x)³ - (1)³ - 3x × 1 × (x - 1)]

⇒ x³ - x² = x³ - 1 - 3x(x - 1)

⇒ x³ - x² = x³ - 1 - 3x² + 3x

⇒ x³ - x² - x³ + 1 + 3x² - 3x = 0

⇒ x³ - x³ - x² + 3x² - 3x + 1 = 0

⇒ 2x² - 3x + 1 = 0

The highest power of equation is 2 and it's equivalent to ax² + bx + c = 0 . Hence the equation is a quadratic equation.

Hence, option 3 is the correct option.

Solving,

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

The highest power of equation is 1 and it's not equivalent to ax² + bx + c = 0 . Thus it is a linear equation.

Hence, option 3 is the correct option.

Given,

⇒ 2x² - x - 6 = 0

⇒ 2x² - 4x + 3x - 6 = 0

⇒ 2x(x - 2) + 3(x - 2) = 0

⇒ (2x + 3)(x - 2) = 0

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

⇒ 2x = -3 or x = 2

⇒ x = $-\frac{3}{2}$ or x = 2

Hence, option 2 is the correct option.

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 - 5x + 6 = 0.

Hence, option 1 is the correct option.

Given,

⇒ 3x² + 13x + 14 = 0

⇒ 3x² + 6x + 7x + 14 = 0

⇒ 3x(x + 2) + 7(x + 2) = 0

⇒ (3x + 7)(x + 2) = 0

⇒ 3x + 7 = 0 or x + 2 = 0      [Using Zero-product rule]

⇒ 3x = -7 or x = -2

⇒ x = $-\frac{7}{3}$ or x = -2

Hence, option 4 is the correct option.

Given,

x = $\frac{-1}{2}$

Equation : 3x² + 2kx - 3 = 0

Substituting value of x in equation:

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

Hence, option 3 is the correct option.

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

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

We know that,

Discriminant (D) = b² - 4ac

= (-a)² - 4 × 1 × 1

= a² - 4

Since equations has distinct real roots,

⇒ D > 0

⇒ a² - 4 > 0

⇒ a² > 4

⇒ |a| > $\sqrt{4}$

⇒ |a| > 2.

Hence, option 3 is the correct option.

Given,

x² + kx + 64 = 0     ....(1)

x² - 8x + k = 0     ....(2)

Comparing x² + kx + 64 = 0 with ax² + bx + c = 0 we get,

a = 1, b = k and c = 64.

We know that,

Discriminant (D) = b² - 4ac

= (k)² - 4 × (1) × (64)

= k² - 256

Since equations has real roots,

⇒ D ≥ 0

⇒ k² - 256 ≥ 0

⇒ k² ≥ 256

⇒ |k| ≥ $\sqrt{256}$

⇒ |k| ≥ 16

⇒ 16 ≤ k ≤ -16

Taking only positive value,

⇒ k ≥ 16 .........(3)

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

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

We know that,

Discriminant (D) = b² - 4ac

= (-8)² - 4 × (1) × (k)

= (64) - 4k

Since equations has real roots,

⇒ D ≥ 0

⇒ 64 - 4k ≥ 0

⇒ 4k ≤ 64

⇒ k ≤ $\frac{64}{4}$

⇒ k ≤ 16 .........(4)

From (3) and (4), we get :

⇒ k = 16.

Hence, option 1 is the correct option.

Given,

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

Substituting value of x = 3 in equation, we get :

⇒ (3)² - p(3) + 3 = 0

⇒ 9 - 3p + 3 = 0

⇒ 12 - 3p = 0

⇒ 12 = 3p

⇒ p = $\frac{12}{3}$

⇒ p = 4.

Hence, option 1 is the correct option.

Given,

⇒ 2x² - x + $\frac{1}{8}$ = 0

Multiply the equation with 16, we get:

⇒ 16(2x² - x + $\frac{1}{8}$ = 0)

⇒ 32x² - 16x + $\frac{16}{8}$ = 0

⇒ 32x² - 16x + 2 = 0

⇒ 2(16x² - 8x + 1) = 0

⇒ 16x² - 8x + 1 = 0

⇒ 16x² - 4x - 4x + 1 = 0

⇒ 4x(4x - 1) - 1(4x - 1) = 0

⇒ (4x - 1)(4x - 1) = 0

⇒ (4x - 1) = 0 or (4x - 1) = 0      [Using Zero-product rule]

⇒ 4x = 1 or 4x = 1

⇒ x = $\frac{1}{4}$ or x = $\frac{1}{4}$

Hence, option 1 is the correct option.

Given,

2x² = 288

$$\begin{aligned}\Rightarrow 2x^2 = 288 \\[1em] \Rightarrow x^2 = \frac{288}{2} \\[1em] \Rightarrow x^2 = 144 \\[1em] \Rightarrow x = \sqrt{144} \\[1em] \Rightarrow x = \pm 12 \\[1em] \Rightarrow x = 12 \text{ or } x = -12.\end{aligned}$$

The solution set = {-12, 12}

Hence, option 4 is the correct option.

Given,

Equation : 2x² - kx + k = 0

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

For equal roots,

⇒ Discriminant (D) = 0

⇒ b² - 4ac = 0

⇒ (-k)² - 4 × 2 × k = 0

⇒ k² - 8k = 0

⇒ k(k - 8) = 0

⇒ k = 0 or k - 8 = 0

⇒ k = 0 or k = 8.

Hence, Option 4 is the correct option.

x = 233.356

On rounding off to two significant figures

x = 230.

Hence, Option 4 is the correct option.

Given,

⇒ px(x - 2) + 6 = 0

⇒ px² - 2px + 6 = 0

Comparing px² - 2px + 6 = 0 with ax² + bx + c = 0 we get,

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

We know that,

Since equations has equal roots,

⇒ D = 0

⇒ b² - 4ac = 0

⇒ (-2p)² - 4(p)(6) = 0

⇒ 4p² - 24p = 0

⇒ 4p(p - 6) = 0

⇒ 4p = 0 or p - 6 = 0      [Using Zero-product rule]

⇒ p = 0 or p = 6.

Since p = 0 would make the equation no longer quadratic, we take:

p = 6

Hence, option 3 is the correct option.

Comparing px² - $2\sqrt{5}$ px + 15 = 0 with ax² + bx + c = 0 we get,

a = p, b = $-2\sqrt{5}$ p and c = 15.

We know that,

Since equations has equal roots,

⇒ D = 0

⇒ b² - 4ac = 0

⇒ ($-2\sqrt{5}$ p)² - 4(p)(15) = 0

⇒ 20p² - 60p = 0

⇒ 20p(p - 3) = 0

⇒ 20p = 0 or (p - 3) = 0      [Using Zero-product rule]

⇒ p = 0 or p = 3.

Since p = 0 would make the equation no longer quadratic, we take:

p = 3

Hence, option 2 is the correct option.

Given,

1 is a root of the quadratic equation, ky² + ky + 3 = 0.

Substituting value of y = 1 in equation :

⇒ k(1)² + k(1) + 3 = 0

⇒ k + k + 3 = 0

⇒ 2k + 3 = 0

⇒ 2k = -3

⇒ k = $-\frac{3}{2}$.

Hence, option 4 is the correct option.

Comparing x² + 5kx + 16 = 0 with ax² + bx + c = 0 we get,

a = 1, b = 5k and c = 16.

We know that,

Since equations has no real roots,

⇒ D < 0

⇒ b² - 4ac < 0

⇒ (5k)² - 4(1)(16) < 0

⇒ 25k² - 64 < 0

⇒ 25k² < 64

$$\begin{aligned}\Rightarrow k^2 \lt \frac{64}{25} \\[1em] \Rightarrow |k| \lt \sqrt{\frac{64}{25}} \\[1em] \Rightarrow |k| \lt \frac{8}{5} \\[1em] \Rightarrow -\frac{8}{5} \lt k \lt \frac{8}{5}.\end{aligned}$$

Hence, option 3 is the correct option.

⇒ 3x² = 6x

⇒ 3x² - 6x = 0

⇒ 3x(x - 2) = 0      [Using Zero-product rule]

⇒ 3x = 0 or (x - 2) = 0

⇒ x = 0 or x = 2.

Hence, option 3 is the correct option.

⇒ 2x² + kx - k² = 0

⇒ 2x² + 2kx - kx - k² = 0

⇒ 2x(x + k) - k(x + k) = 0

⇒ (2x - k)(x + k) = 0

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

⇒ 2x = k or x = -k

⇒ x = $\frac{k}{2}$ or x = -k

The solution set: $\Big\lbrace-k, \frac{k}{2}\Big\rbrace$.

Hence, option 3 is the correct option.

⇒ 25x(x + 1) = -4

⇒ 25x² + 25x + 4 = 0

⇒ 25x² + 5x + 20x + 4 = 0

⇒ 5x(5x + 1) + 4(5x + 1) = 0

⇒ (5x + 4)(5x + 1) = 0

⇒ (5x + 4) = 0 or (5x + 1) = 0      [Using Zero-product rule]

⇒ 5x = -4 or 5x = -1

⇒ x = $\frac{-4}{5}$ or x = $\frac{-1}{5}$.

The solution set: $\Big\lbrace-\frac{4}{5}, -\frac{1}{5}\Big\rbrace$.

Hence, option 3 is the correct option.

Comparing 3x² - 2x + $\frac{1}{3}$ = 0 with ax² + bx + c = 0 we get,

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

We know that,

⇒ D = b² - 4ac

⇒ D = (-2)² - 4(3) $\Big(\frac{1}{3}\Big)$

⇒ D = 4 - 4

⇒ D = 0.

Hence, option 1 is the correct option.

Comparing $\sqrt{3}x^2 + 10x + 7\sqrt{3}$ = 0 with ax² + bx + c = 0 we get,

a = $\sqrt{3}$, b = 10 and c = $7\sqrt{3}$.

We know that,

⇒ D = b² - 4ac

⇒ D = (10)² - $4 \times (\sqrt{3}) \times (7\sqrt{3})$

⇒ D = 100 - 84

⇒ D = 16.

Hence, option 2 is the correct option.

Comparing x² - ($\sqrt{2}$ + 1)x + $\sqrt{2}$ = 0 with ax² + bx + c = 0 we get,

a = 1, b = -($\sqrt{2}$ + 1) and c = $\sqrt{2}$.

We know that,

$$\begin{aligned}\Rightarrow D = b^2 - 4ac \\[1em] = [-(\sqrt{2} + 1)]^2 - 4 \times 1 \times (\sqrt{2}) \\[1em] = (\sqrt2)^2 + 2 \times \sqrt2 \times 1 + 1^2 - 4(\sqrt{2}) \\[1em] = 2 + 2\sqrt2 + 1 - 4\sqrt{2} \\[1em] = 3 - 2\sqrt2.\end{aligned}$$

Hence, option 3 is the correct option.

Comparing 2x² - 3x + 5 = 0 with ax² + bx + c = 0 we get,

a = 2, b = -3 and c = 5.

We know that,

⇒ D = b² - 4ac

⇒ D = (-3)² - 4(2)(5)

⇒ D = 9 - 40

⇒ D = -31.

Hence, option 4 is the correct option.

Comparing ax² + 2x + a = 0 with ax² + bx + c = 0 we get,

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

We know that,

Since equations has real and equal roots,

⇒ D = 0

⇒ b² - 4ac = 0

⇒ (2)² - 4(a)(a) = 0

⇒ 4 - 4a² = 0

⇒ 4 = 4a²

⇒ a² = $\frac{4}{4}$

⇒ a² = 1

⇒ a = $\sqrt{1}$

⇒ a = ± 1

⇒ a = -1, 1.

Hence, option 4 is the correct option.

Comparing the given quadratic equation with ax2 + bx + c = 0, we get :

a = 3, b = $\sqrt7$ and c = 2

Discriminant = b2 - 4ac

= $(\sqrt7)^2$ - 4 x 3 x 2

= 7 - 24

= -17.

Since, D < 0, there are no real roots.

Hence, option 4 is the correct option.

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

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

We know that,

⇒ D = b² - 4ac

⇒ D = (-6)² - 4(2)(3)

⇒ D = 36 - 24

⇒ D = 12

⇒ D > 0 roots are real and unequal.

⇒ D = 12 is not a perfect square, roots are irrational.

Thus, roots are irrational and unequal.

Hence, option 2 is the correct option.

Comparing 3x² - $4\sqrt{3}x$ + 4 = 0 with ax² + bx + c = 0 we get,

a = 3, b = -$4\sqrt{3}$ and c = 4.

We know that,

⇒ D = b² - 4ac

⇒ D = (-$4\sqrt{3}$)² - 4(3)(4)

⇒ D = 16(3) - 48

⇒ D = 48 - 48

⇒ D = 0.

Thus, roots are real and equal.

Hence, option 1 is the correct option.

Given,

-5 is a root of the quadratic equation 2x² + px - 15 = 0.

Substituting value of x = -5 in 2x² + px - 15 = 0, we get:

⇒ 2(-5)² + p(-5) - 15 = 0

⇒ 2 × 25 - 5p - 15 = 0

⇒ 50 - 5p - 15 = 0

⇒ 35 - 5p = 0

⇒ 5p = 35

⇒ p = $\frac{35}{5}$

⇒ p = 7.

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

⇒ 7(x² + x) + k = 0

⇒ 7x² + 7x + k = 0

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

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

Since equation has equal roots,

⇒ Discriminant = 0

⇒ b² - 4ac = 0

⇒ (7)² - 4(7)(k) = 0

⇒ 49 - 28k = 0

⇒ 28k = 49

⇒ k = $\frac{49}{28}$

⇒ k = $\frac{7}{4}$.

Hence, option 1 is the correct option.

Given,

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

Comparing $x^2 + 2\sqrt{2}x + 1 = 0$ with ax² + bx + c = 0 we get,

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

We know that,

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

= 8 - 4 = 4; which is positive.

Since, D = 4 > 0, the discriminant is greater than zero.

So, Assertion (A) is true.

D > 0 Real and distinct roots

D = 0 Real and equal roots

D < 0 Imaginary roots

If Discriminant is grater than zero,

Real and distinct roots

So, Reason (R) is true.

Thus, Both A and R are true, but R is not the correct explanation of A.

Hence, option 2 is the correct option.

Given,

⇒ 3kx² - 4kx + 4 = 0

when k = 3

⇒ 3 × (3) × x² - 4 × (3) × x + 4 = 0

⇒ 9x² - 12x + 4 = 0

Comparing 9x² - 12x + 4 = 0 with ax² + bx + c = 0 we get,

a = 9, b = -12 and c = 4.

We know that,

Discriminant (D) = b² - 4ac

= (-12)² - 4 × (9) × (4)

= 144 - 144 = 0.

Therefore, the equation has rational and equal roots.

So, Assertion (A) is true.

The Discriminant is given by b² - 4ac, if the discriminant of any quadratic equation is zero. Then it is said have equal and real roots.

So, Reason (R) is true.

Thus, both A and R are true and R is the correct explanation of A.

Hence, option 1 is the correct option.

Given,

⇒ 3x² + 7x + 8 = 0

Comparing 3x² + 7x + 8 = 0 with ax² + bx + c = 0 we get,

a = 3, b = 7 and c = 8.

We know that,

Discriminant (D) = b² - 4ac

= (7)² - 4 × (3) × (8)

= 49 - 96 = -47; which is negative.

Therefore, the equation has imaginary and unequal roots.

So, Assertion (A) is true.

The Discriminant of quadratic equation can be positive, negative or equal to zero.

So, Reason (R) is false.

A is true, R is false.

Hence, option 3 is the correct option.

Given,

⇒ 8x² + 2x - 3 = 0

Comparing equation 8x² + 2x - 3 = 0 with ax² + bx + c = 0, we get :

a = 8, b = 2 and c = -3.

By formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow x = \frac{-(2) \pm \sqrt{(2)^2 - 4 \times (8) \times (-3)}}{2 \times (8)} \\[1em] = \frac{-2 \pm \sqrt{4 + 96}}{16} \\[1em] = \frac{-2 \pm \sqrt{100}}{16} \\[1em] = \frac{-2 \pm 10}{16} \\[1em] = \frac{-2 + 10}{16} \text{ or } \frac{-2 - 10}{16} \\[1em] = \frac{8}{16} \text{ or } \frac{-12}{16} \\[1em] = \frac{1}{2} \text{ or } \frac{-3}{4}.\end{aligned}$$

Thus, A is false, R is true.

Hence, option 4 is the correct option.

Given,

$\frac{5}{x} + 4\sqrt3 = \frac{2\sqrt3}{x^2}$

Substituting $\frac{1}{x} = t$ in the given equation, we get :

$$\begin{aligned}\Rightarrow 5t + 4\sqrt3 = 2\sqrt3t^2 \\[1em] \Rightarrow 5t + 4\sqrt3 - 2\sqrt3t^2 = 0 \\[1em] \Rightarrow 2\sqrt3t^2 - 5t - 4\sqrt3 = 0 \\[1em] \Rightarrow 2\sqrt3t^2 - 8t + 3t - 4\sqrt3 = 0 \\[1em] \Rightarrow 2t(\sqrt3t - 4) + \sqrt3(\sqrt3t - 4) = 0 \\[1em] \Rightarrow (2t + \sqrt3) = 0 \text{ or } (\sqrt3t - 4) = 0 \\[1em] \Rightarrow 2t = -\sqrt3 \text{ or } \sqrt3t = 4 \\[1em] \Rightarrow t = \frac{-\sqrt3}{2} \text{ or } t = \frac{4}{\sqrt3} \\[1em] \Rightarrow \frac{1}{x} = \frac{-\sqrt3}{2} \text{ or } \frac{1}{x} = \frac{4}{\sqrt3} \\[1em] \Rightarrow x = \frac{2}{-\sqrt3} \text{ or } x = \frac{\sqrt3}{4}\end{aligned}$$

Hence, $x = \frac{2}{-\sqrt3} \text{ or } x = \frac{\sqrt3}{4}$.

(a) Given, equation : 5x² − 9x + 4 = 0

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

a = 5, b = -9 and c = 4.

Discriminant (D) = b² - 4ac = (-9)² - 4 × 5 × 4 = 81 - 80 = 1.

Since, D > 0 and a perfect square.

Hence, equation 5x² − 9x + 4 = 0 has real roots.

(b) Solving,

⇒ 5x² − 9x + 4 = 0

⇒ 5x² - 5x - 4x + 4 = 0

⇒ 5x(x - 1) - 4(x - 1) = 0

⇒ (5x - 4)(x - 1) = 0

⇒ 5x - 4 = 0 or x - 1 = 0

⇒ 5x = 4 or x = 1

⇒ x = $\frac{4}{5}$ or x = 1.

Hence, roots of the equation are 1 and $\frac{4}{5}$.

1. Given,

⇒ P(x) = -2x² + 88x - 680

Books sold by Raman lal is 20.Therefore x = 20

⇒ P(20) = −2(20)² + 88(20) − 680

⇒ P(20) = −2(400) + 1760 − 680

⇒ P(20) = −800 + 1080

⇒ P(20) = 280.

Hence, option (b) is the correct option.

2. Given,

⇒ P(x) = - 2x² + 88x - 680

Maximum profit occurs when 22 books are sold, thus x = 22.

⇒ P(22) = −2(22)² + 88(22) − 680

= −2(484) + 1936 − 680

= −968 + 1256

= ₹ 288.

Hence, option (b) is the correct option.

3. Given,

⇒ Number of notebooks sold in a particular week = 0

Thus, x = 0

The weekly profit P(x) = -2x² + 88x - 680.

Raman Lal’s loss if no notebooks sold in a week,

⇒ P(0) = -2(0)² + 88(0) - 680

= −680

= loss of ₹ 680.

Hence, option (c) is the correct option.

4. Given,

⇒ P(x) = -2x² + 88x - 680

If there is no profit/loss, then profit = ₹ 0.

⇒ P(x) = 0

⇒ −2x² + 88x − 680 = 0

⇒ −2(x² - 44x + 340) = 0

⇒ x² − 44x + 340 = 0.

Hence, option (d) is the correct option.

5. When Raman lal has no profit/loss we get the equation x² − 44x + 340 = 0.

⇒ x² − 44x + 340 = 0

⇒ x² − 10x - 34x + 340 = 0

⇒ x(x − 10) - 34(x - 10) = 0

⇒ (x − 10)(x - 34) = 0

⇒ (x − 10) = 0 or (x - 34) = 0      [Using Zero-product rule]

⇒ x = 10 or x = 34.

The minimum value of x with no loss is x = 10. Raman lal has to sell 10 books so that he does not incur any loss.

Hence, option (b) is the correct option.