Chapter 5: Quadratic Equations

Exercise 5A

Question 1

Find which of the following are the solutions of equation 6x² - x - 2 = 0 ?

$\dfrac{1}{2}$
$\dfrac{-1}{2}$
$\dfrac{2}{3}$

Answer

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

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

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

Question 2

Determine whether x = $-\dfrac{1}{3}$ and x = $\dfrac{2}{3}$ are the solutions of the equation 9x² - 3x - 2 = 0.

Answer

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

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

Question 3

Solve the following equation by factorization:

16x² = 25

Answer

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 = $\dfrac{-5}{4}$ or x = $\dfrac{5}{4}$ .

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

Question 4

Solve the Following equation by factorization:

x² + 2x = 24

Answer

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}.

Question 5

Solve the following equation by factorization:

x² - x = 156

Answer

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}.

Question 6

Solve the following equation by factorization:

x² - 11x = 42

Answer

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}.

Question 7

Solve the following equation by factorization:

x² - 7x + 10 = 0

Answer

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}.

Question 8

Solve the following equation by factorization:

x² + 18x = 40

Answer

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}.

Question 9

Solve the following equation by factorization:

x² + 17 = 18x

Answer

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}.

Question 10

Solve the following equation by factorization:

3x² = 5x

Answer

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 = $\dfrac{5}{3}$ .

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

Question 11

Solve the following equation by factorization:

(x + 3)(x - 3) = 27

Answer

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}.

Question 12

Solve the following equation by factorization:

x² - 30x + 216 = 0

Answer

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}.

Question 13

Solve the following equation by factorization:

12x² + 29x + 14 = 0

Answer

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

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

Question 14

Solve the following equation by factorization:

2x² - 7x = 39

Answer

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 = $\dfrac{13}{2}$ .

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

Question 15

Solve the following equation by factorization:

10x² = 9x + 7

Answer

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

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

Question 16

Solve the following equation by factorization:

15x² - 28 = x

Answer

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

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

Question 17

Solve the following equation by factorization:

8x² + 15 = 26x

Answer

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 = $\dfrac{5}{2}$ or x = $\dfrac{3}{4}$

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

Question 18

Solve the following equation by factorization:

3x² + 8 = 10x

Answer

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 = $\dfrac{4}{3}$ .

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

Question 19

Solve the following equation by factorization:

x(6x - 11) = 35

Answer

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

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

Question 20

Solve the following equation by factorization:

6x(3x - 7) = 7(7 - 3x)

Answer

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

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

Question 21

Solve the following equation by factorization:

2x² - 9x + 10 = 0, when (i) x ∈ N (ii) x ∈ Q.

Answer

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 = $\dfrac{5}{2}$ or x = 2.

(i) Since, x ∈ N

Hence, value of x = {2}.

(ii) Since, x ∈ Q

Hence, x = $\Big\lbrace2, \dfrac{5}{2}\Big\rbrace$ .

Question 22

Solve the following equation by factorization:

4x² - 9x - 100 = 0, when x ∈ Q

Answer

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 = $\dfrac{25}{4}$ or x = -4

Since, x ∈ Q

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

Question 23

Solve the following equation by factorization:

3x² + 11x + 10 = 0, when x ∈ I

Answer

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 = $\dfrac{-5}{3}$ or x = -2.

Since, x ∈ I

x = -2

Hence, x = {-2}.

Question 24

Solve the following equation by factorization:

$x + \dfrac{1}{x} = 3\dfrac{1}{3}$ , x ≠ 0

Answer

Given,

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

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

⇒ x = 3 or 3x = 1

⇒ x = 3 or x = $\dfrac{1}{3}$

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

Question 25

Solve the following equation by factorization:

5x - $\dfrac{35}{x}$ = 18

Answer

Given,

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

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

⇒ 5x = -7 or x = 5

⇒ x = $\dfrac{-7}{5}$ or x = 5.

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

Question 26

Solve the following equation by factorization:

10x - $\dfrac{1}{x}$ = 3

Answer

Given,

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

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

⇒ 5x = -1 or 2x = 1

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

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

Question 27

Solve the following equation by factorization:

3a²x² + 8abx + 4b² = 0, a ≠ 0

Answer

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

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

Question 28

Solve the following equation by factorization:

4x² - 4ax + (a² - b²) = 0, where a, b ∈ R.

Answer

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

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

Question 29

Solve the following equation by factorization:

5x² - 12x - 9 = 0, when (i) x ∈ I (ii) x ∈ Q

Answer

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 = $\dfrac{-3}{5}$ or x = 3.

(i) Since, x ∈ I

Hence, x = {3}.

(ii) Since, x ∈ Q

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

Question 30

Solve the following equation by factorization:

2x² - 11x + 15 = 0, when (i) x ∈ N (ii) x ∈ I

Answer

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 = $\dfrac{5}{2}$ or x = 3.

(i) Since, x ∈ N

Hence, x = {3}.

(ii) Since, x ∈ I

Hence, x = {3}.

Question 31

Solve the following equation by factorization:

$\sqrt{3}x^2 + 11x + 6\sqrt{3}$ = 0

Answer

Given,

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

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

Question 32

Solve the following equation by factorization:

$2\sqrt{5}x^2 - 3x - \sqrt{5}$ = 0

Answer

Given,

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

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

Question 33

Solve the following equation by factorization:

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

Answer

Given,

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

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

Question 34

Solve the following equation by factorization:

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

Answer

Given,

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

⇒ (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}.

Question 35

Solve the following equation by factorization:

$\dfrac{3x + 1}{7x + 1} = \dfrac{5x + 1}{7x + 5}$

Answer

Given,

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

⇒ (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 = $\dfrac{-2}{7}$ or x = 1.

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

Question 36

Solve the following equation by factorization:

$\dfrac{5}{(2x + 1)} + \dfrac{6}{(x + 1)} = 3$

Answer

Given,

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

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

Question 37

Solve the following equation by factorization:

$\dfrac{2x}{x - 4} + \dfrac{2x - 5}{x - 3} = \dfrac{25}{3}$

Answer

Given,

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

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

Question 38

Solve the following equation by factorization:

$\dfrac{x + 3}{x - 2} - \dfrac{1 - x}{x} = 4\dfrac{1}{4}$

Answer

Given,

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

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

Question 39

Solve the following equation by factorization:

$\dfrac{1}{x - 2} + \dfrac{2}{x - 1} = \dfrac{6}{x}$

Answer

Given,

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

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

Question 40

Solve the following equation by factorization:

$2\Big(\dfrac{x}{x + 1}\Big)^2 - 5\Big(\dfrac{x}{x + 1}\Big) + 2 = 0$ , x ≠ -1

Answer

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

Substituting y = $\dfrac{x}{x + 1}$ in equation $2\Big(\dfrac{x}{x + 1}\Big)^2 - 5\Big(\dfrac{x}{x + 1}\Big) + 2 = 0$ , 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 = $\dfrac{1}{2}$ or y = 2.

Now we have,

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

$\Rightarrow y = \dfrac{x}{x + 1} \\[1em] \Rightarrow \dfrac{1}{2} = \dfrac{x}{x + 1} \\[1em] \Rightarrow x + 1 = 2x \\[1em] \Rightarrow 2x - x = 1 \\[1em] \Rightarrow x = 1.$

Case 2 : y = 2

⇒ y = $\dfrac{x}{x + 1}$

⇒ 2 = $\dfrac{x}{x + 1}$

⇒ 2(x + 1) = x

⇒ 2x + 2 = x

⇒ 2x - x = -2

⇒ x = -2.

Hence, x = {-2, 1}.

Question 41

Solve the following equation by factorization:

5(3x + 1)² + 6(3x + 1) - 8 = 0

Answer

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 = $\dfrac{4}{5}$ or y = -2.

Now we have,

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

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

Case 2 : y = -2

⇒ y = -2

⇒ 3x + 1 = -2

⇒ 3x = -2 - 1

⇒ 3x = -3

⇒ x = $\dfrac{-3}{3}$

⇒ x = -1.

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

Question 42

Solve the following equation by factorization:

$\sqrt{x + 15}$ = (x + 3)

Answer

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)

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

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}.

Question 43

Solve the following equation by factorization:

$\sqrt{2x + 9}$ = (13 - x)

Answer

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)

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

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}.

Question 44

Solve the following equation by factorization:

$\sqrt{3x^2 - 2}$ = (2x - 1)

Answer

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}.

Question 45

Solve the following equation by factorization:

$\sqrt{3x^2 + x + 5}$ = (x - 3)

Answer

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 = $\dfrac{1}{2}$ or x = -4.

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

Question 46

Find the quadratic equation whose solution set is:

(i) {2, -3}

(ii) $\Big\lbrace-3, \dfrac{2}{5}\Big\rbrace$

(iii) $\Big\lbrace\dfrac{2}{5}, -\dfrac{1}{2}\Big\rbrace$

Answer

(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, $\Big\lbrace-3, \dfrac{2}{5}\Big\rbrace$ is solution set.

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

∴ x = -3 or x = $\dfrac{2}{5}$

⇒ 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 $\Big\lbrace-3, \dfrac{2}{5}\Big\rbrace$ is 5x² + 13x - 6 = 0.

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

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

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

⇒ 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 $\Big\lbrace\dfrac{2}{5}, -\dfrac{1}{2}\Big\rbrace$ is 10x² + x - 2 = 0.

Question 47

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

Thus, find the other root of the equation.

Answer

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 = $\dfrac{-24}{6}$

⇒ 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 = $\dfrac{-6}{-2}$

⇒ x = -1 or x = 3.

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

Exercise 5B

Question 1

Solve the following equation using quadratic formula:

x² - 4x + 1 = 0

Answer

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

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

By formula,

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

Substituting values we get :

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

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

Question 2

Solve the following equation using quadratic formula:

9x² + 7x - 2 = 0

Answer

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

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

By formula,

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

Substituting values we get :

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

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

Question 3

Solve the following equation using quadratic formula:

$\dfrac{3}{4}$ x² - x - 1 = 0

Answer

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

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

By formula,

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

Substituting values we get :

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

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

Question 4

Solve the following equation using quadratic formula:

4 - 11x = 3x²

Answer

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

Substituting values we get :

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

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

Question 5

Solve the following equation using quadratic formula:

25x² + 30x + 7 = 0

Answer

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

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

By formula,

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

Substituting values we get :

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

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

Question 6

Solve the following equation using quadratic formula:

5x² - 19x + 17 = 0

Answer

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

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

By formula,

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

Substituting values we get :

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

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

Question 7

Solve the following equation using quadratic formula:

3x² - 8x + 2 = 0

Answer

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

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

By formula,

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

Substituting values we get :

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

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

Question 8

Solve the following equation using quadratic formula:

$\sqrt{3}x^2 + 10x - 8\sqrt{3}$ = 0

Answer

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

Substituting values we get :

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

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

Question 9

Solve the following equation using quadratic formula:

2x² + $\sqrt{7}x$ - 7 = 0

Answer

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

Substituting values we get :

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

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

Question 10

Solve the following equation using quadratic formula:

6x² - 31x = 105

Answer

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

Substituting values we get :

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

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

Question 11

Solve the following equation using quadratic formula:

$\dfrac{x + 3}{2x + 3} = \dfrac{x + 1}{3x + 2}$

Answer

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

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

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

By formula,

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

Substituting values we get :

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

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

Question 12

Solve the following equation using quadratic formula:

$\dfrac{x - 1}{x - 2} + \dfrac{x - 3}{x - 4} = 3\dfrac{1}{3}$

Answer

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

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

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

By formula,

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

Substituting values we get :

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

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

Question 13

Solve the following equation using quadratic formula:

x² - 10x + 6 = 0

Answer

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

Substituting values we get :

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

Hence, x = {9.36, 0.64}.

Question 14

Solve the following equation using quadratic formula:

2x² - 6x + 3 = 0

Answer

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

Substituting values we get :

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

Hence, x = {2.37, 0.64}.

Question 15

Solve the following equation using quadratic formula:

3x² - 32x + 12 = 0

Answer

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

Substituting values we get :

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

Hence, x = {10.28, 0.39}.

Question 16

Solve the following equation using quadratic formula:

x² + 7x = 7

Answer

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

Substituting values we get :

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

Hence, x = {0.89, -7.89}.

Question 17

Solve the following equation using quadratic formula:

3x² - x - 7 = 0

Answer

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

Substituting values we get :

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

Hence, x = {1.70, -1.37}.

Question 18

Solve the following equation using quadratic formula:

4x² - 7x + 2 = 0

Answer

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

Substituting values we get :

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

Hence, x = {1.39, 0.36}.

Question 19

Solve the following equation using quadratic formula:

x² - 7x + 3 = 0

Answer

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

Substituting values we get :

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

Hence, x = {6.54, 0.46}.

Question 20

Solve for x the quadratic equation x² - 4x - 8 = 0. Give your answer correct to three significant figures.

Answer

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

Substituting values we get :

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

Correcting the value of x to three significant figures.

Hence, x = {5.46, -1.46}.

Question 21

Solve the following quadratic equation : x² + 4x - 8 = 0. Give your answer correct to one decimal place.

Answer

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

Substituting values we get :

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

Correcting the value of x to one decimal place.

Hence, x = {1.5, -5.5}.

Exercise 5C

Question 1

Discuss the nature of the roots of the following equation without actually solving it:

x² - 8x + 7 = 0

Answer

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.

Question 2

Discuss the nature of the roots of the following equation without actually solving it:

6x² + 7x - 10 = 0

Answer

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.

Question 3

Discuss the nature of the roots of the following equation without actually solving it:

25x² + 30x + 7 = 0

Answer

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.

Question 4

Discuss the nature of the roots of the following equation without actually solving it:

15x² - 28 = x

Answer

⇒ 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.

Question 5

Discuss the nature of the roots of the following equation without actually solving it:

16x² = 24x + 1

Answer

⇒ 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.

Question 6

Discuss the nature of the roots of the following equation without actually solving it:

2x² - $2\sqrt{6}x$ + 3 = 0

Answer

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.

Question 7

Discuss the nature of the roots of the following equation without actually solving it:

2x² + 2x + 3 = 0

Answer

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.

Question 8

Discuss the nature of the roots of the following equation without actually solving it:

2x² - 5x - 4 = 0

Answer

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.

Question 9

Discuss the nature of the roots of the following equation without actually solving it:

5x² - 13x - 6 = 0

Answer

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.

Question 10

Discuss the nature of the roots of the following equation without actually solving it:

9x² - 6x + 1 = 0

Answer

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.

Question 11

Discuss the nature of the roots of the following equation without actually solving it:

3x² - 2x + 5 = 0

Answer

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.

Question 12

Discuss the nature of the roots of the following equation without actually solving it:

x² + $2\sqrt{3}x$ - 1 = 0

Answer

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.

Question 13

Find the values of k for which the following equation has equal roots:

9x² + kx + 1 = 0

Answer

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}.

Question 14

Find the values of k for which the following equation has equal roots:

x² - 2kx + 7k - 12 = 0

Answer

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 = $\dfrac{12}{4}$

⇒ k = 4 or k = 3.

Hence, k = {4, 3}.

Question 15

Find the values of k for which the following equation has equal roots:

(3k + 1)x² + 2(k + 1)x + k = 0

Answer

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 = $\dfrac{4}{-8}$

⇒ k = 1 or k = $-\dfrac{1}{2}$

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

Question 16

Find the values of k for which the following equation has equal roots:

x² - 2(5 + 2k)x + 3(7 + 10k) = 0

Answer

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 = $\dfrac{1}{2}$ or k = $\dfrac{16}{8}$

⇒ k = $\dfrac{1}{2}$ or k = 2

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

Question 17

Find the values of k for which the following equation has equal roots:

(k + 1)x² + 2(k + 3)x + (k + 8) = 0

Answer

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 = $\dfrac{-4}{-12}$

⇒ k = $\dfrac{1}{3}$ .

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

Question 18

Find the values of k for which the following equation has equal roots:

kx² + kx + 1 = -4x² - x

Answer

⇒ 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}.

Question 19

Find the values of k for which the following equation has equal roots:

3kx² = 4(kx - 1)

Answer

⇒ 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}.

Question 20

Find the values of k for which the following equation has equal roots:

x² + 4kx + (k² - k + 2) = 0

Answer

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 = $\dfrac{8}{12}$

⇒ k = -1 or k = $\dfrac{2}{3}$ .

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

Question 21

Show that the equation x² + ax - 1 = 0 has real and distinct roots for all real values of a.

Answer

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.

Question 22

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

Answer

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.

Question 23

If the roots of the equation (c² - ab)x² - 2(a² - bc)x + (b² - ac) = 0 are real and equal, show that either a = 0 or a³ + b³ + c³ = 3abc.

Answer

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.

Question 24

If a, b, c ∈ R, show that the roots of the equation (a - b)x² + (b + c - a)x - c = 0 are rational.

Answer

(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.

Question 25

If a, b, c are rational, prove that the roots of the equation (b - c)x² + (c - a)x + (a - b) = 0 are also rational.

Answer

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.

Multiple Choice Questions

Question 1

The degree of a quadratic equation is :

1
2
3
none of these

Answer

The degree of a quadratic equation is the highest power of the variable in it.

A quadratic equation has the general form:

⇒ ax² + bx + c = 0

Here, the highest power of x is 2.

Hence, option 2 is the correct option.

Question 2

The roots of the equation px² + qx + r = 0, where p ≠ 0, are given by:

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

Answer

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

Substituting values we get :

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

Hence, option 3 is the correct option.

Question 3

The discriminant of the quadratic equation ax² + bx + c = 0, a ≠ 0 is given by:

b² - 2ac
b² - ac
b² - 4ac
none of these

Answer

We know that,

Discriminant (D) = b² - 4ac.

Hence, option 3 is the correct option.

Question 4

For real roots of a quadratic equation, the discriminant must be:

greater than or equal to zero
greater than zero
less than or equal to zero
less than zero

Answer

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.

Question 5

If the roots of the quadratic equation, ax² + bx + c = 0, a ≠ 0 are real and equal, then each root is equal to:

$\dfrac{-a}{2b}$
$\dfrac{-b}{2a}$
$\dfrac{-2a}{b}$
$\dfrac{-c}{2a}$

Answer

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 = $\dfrac{-b + \sqrt{D}}{2a}$

r₂ = $\dfrac{-b - \sqrt{D}}{2a}$

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

r₁ = $\dfrac{-b + \sqrt{0}}{2a}$

r₁ = $\dfrac{-b}{2a}$

r₂ = $\dfrac{-b - \sqrt{0}}{2a}$

r₂ = $\dfrac{-b}{2a}$

r₁ = r₂ = $\dfrac{-b}{2a}$

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

Hence, option 2 is the correct option.

Question 6

If the discriminant of the quadratic equation, ax² + bx + c = 0, a ≠ 0 is greater than zero and a perfect square and a, b, c are rational, then the roots are:

rational and equal
irrational and unequal
irrational and equal
rational and unequal

Answer

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.

Question 7

If the discriminant of a quadratic equation, ax² + bx + c = 0, is greater than zero and a perfect square and b is irrational, then the roots are:

irrational and unequal
irrational and equal
rational and unequal
rational and equal

Answer

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.

Question 8

Which of the following is a quadratic equation?

x² - $2\sqrt{x}$ + 7 = 0
2x² - 5x = (x - 1)²
$x - \dfrac{1}{x} = 2x^{2}$
x² + $\dfrac{1}{x^{2}}$ = 2

Answer

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.

Question 9

Which of the following is a quadratic equation?

x² + 1 = (2 - x)² + 3
2x² + 3 = (5 + x)(2x - 3)
x³ - x² = (x - 1)³
none of these

Answer

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.

Question 10

Which of the following is not a quadratic equation?

3x - x² = x² + 5
(x + 2)² = 2(x² - 5)
$(\sqrt{2}x + 3)^{2} = 2x^{2} + 6$
(x - 1)² = 3x² + x - 2

Answer

Solving,

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

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.

Question 11

The roots of the quadratic equation 2x² - x - 6 = 0 are:

-2, $\dfrac{3}{2}$
2, $\dfrac{-3}{2}$
-2, $\dfrac{-3}{2}$
2, $\dfrac{3}{2}$

Answer

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 = $-\dfrac{3}{2}$ or x = 2

Hence, option 2 is the correct option.

Question 12

Which of the following quadratic equations has 2 and 3 as its roots?

x² - 5x + 6 = 0
x² + 5x + 6 = 0
x² - 5x - 6 = 0
x² + 5x - 6 = 0

Answer

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.

Question 13

Which of the following is a root of the quadratic equation, 3x² + 13x + 14 = 0 ?

$-\dfrac{1}{3}$
$-\dfrac{3}{2}$
$-\dfrac{5}{3}$
$-\dfrac{7}{3}$

Answer

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 = $-\dfrac{7}{3}$ or x = -2

Hence, option 4 is the correct option.

Question 14

If x = $\dfrac{-1}{2}$ is a solution of the quadratic equation 3x² + 2kx - 3 = 0, then the value of k is:

$-\dfrac{3}{4}$
$-\dfrac{5}{4}$
$-\dfrac{9}{4}$
$-\dfrac{4}{5}$

Answer

Given,

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

Equation : 3x² + 2kx - 3 = 0

Substituting value of x in equation:

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

Hence, option 3 is the correct option.

Question 15

If ax² + bx + c = 0 has equal roots, then c = ?

$\dfrac{b}{2a}$
$\dfrac{b^{2}}{4a}$
$-\dfrac{b}{2a}$
$-\dfrac{b^{2}}{4a}$

Answer

Given,

ax² + bx + c = 0 has equal roots

Since equations has equal roots,

Discriminant (D) = 0

⇒ b² - 4ac = 0

⇒ b² = 4ac

⇒ c = $\dfrac{b^2}{4a}$ .

Hence, option 2 is the correct option.

Question 16

If the equation, x² - ax + 1 = 0 has two distinct and real roots, then:

|a| ≥ 2
|a| ≤ 2
|a| > 2
|a| < 2

Answer

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.

Question 17

The positive value of k for which the equations x² + kx + 64 = 0 and x² - 8x + k = 0 will both have real roots, is

Answer

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 ≤ $\dfrac{64}{4}$

⇒ k ≤ 16 .........(4)

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

⇒ k = 16.

Hence, option 1 is the correct option.

Question 18

If 3 is a root of the quadratic equation x² - px + 3 = 0, then p is equal to:

Answer

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 = $\dfrac{12}{3}$

⇒ p = 4.

Hence, option 1 is the correct option.

Question 19

If the roots of the quadratic equation 2x² + 8x + k = 0 are equal, then the value of k is:

2
8
4
none of these

Answer

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

a = 2, b = 8 and c = k.

We know that,

Since equations has equal roots,

⇒ Discriminant = 0

⇒ b² - 4ac = 0

⇒ (8)² - 4(2)(k) = 0

⇒ 64 - 8k = 0

⇒ 8k = 64

⇒ k = $\dfrac{64}{8}$

⇒ k = 8.

Hence, option 2 is the correct option.

Question 20

The solution set for the quadratic equation 2x² - x + $\dfrac{1}{8}$ = 0 is:

$\Big\lbrace\dfrac{1}{4}, \dfrac{1}{4}\Big\rbrace$
$\Big\lbrace-\dfrac{1}{4}, \dfrac{1}{4}\Big\rbrace$
$\Big\lbrace-\dfrac{1}{2}, \dfrac{1}{4}\Big\rbrace$
{4, 4}

Answer

Given,

⇒ 2x² - x + $\dfrac{1}{8}$ = 0

Multiply the equation with 16, we get:

⇒ 16(2x² - x + $\dfrac{1}{8}$ = 0)

⇒ 32x² - 16x + $\dfrac{16}{8}$ = 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 = $\dfrac{1}{4}$ or x = $\dfrac{1}{4}$

Hence, option 1 is the correct option.

Question 21

The solution set for the quadratic equation 2x² = 288, is:

{12, 12}
{-12, -12}
{-12, 18}
{-12, 12}

Answer

Given,

2x² = 288

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

The solution set = {-12, 12}

Hence, option 4 is the correct option.

Question 22

The solution set for the quadratic equation 3x² - 18x = 0 is:

{6, 6}
{0, -6}
{0, 6}
none of these

Answer

Given,

⇒ 3x² - 18x = 0

⇒ 3x(x - 6) = 0

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

⇒ x = 0 or x = 6.

Hence, option 3 is the correct option.

Question 23

The values of k for which the quadratic equation 9x² - 3kx + k = 0, has equal roots, are:

0, 1
0, 2
2, 4
0, 4

Answer

Comparing 9x² - 3kx + k = 0 with ax² + bx + c = 0 we get,

a = 9, b = -3k and c = k.

We know that,

Since equations has equal roots,

⇒ D = 0

⇒ b² - 4ac = 0

⇒ (-3k)² - 4(9)(k) = 0

⇒ 9k² - 36k = 0

⇒ 9k(k - 4) = 0

⇒ 9k = 0 or (k - 4) = 0 [Using Zero-product rule]

⇒ k = 0 or k = 4.

Hence, option 4 is the correct option.

Question 24

If the roots of the quadratic equation, px(x - 2) + 6 = 0 are equal, then the value of p is:

0
4
6
none of these

Answer

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.

Question 25

If the quadratic equation, px² - $2\sqrt{5}$ px + 15 = 0 has two equal roots, then the value of p is:

0
3
6
both 0 and 3

Answer

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.

Question 26

If 1 is a root of the quadratic equation, ky² + ky + 3 = 0, then the value of k is:

$-\dfrac{2}{3}$
$-\dfrac{1}{3}$
$-\dfrac{1}{2}$
$-\dfrac{3}{2}$

Answer

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 = $-\dfrac{3}{2}$ .

Hence, option 4 is the correct option.

Question 27

If the equation x² + 5kx + 16 = 0 has no real roots, then :

k > $\dfrac{8}{5}$
k < $-\dfrac{8}{5}$
$-\dfrac{8}{5} \lt k \lt \dfrac{8}{5}$
none of these

Answer

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

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

Hence, option 3 is the correct option.

Question 28

The discriminant of the equation, 3x² - 2x + $\dfrac{1}{3}$ = 0 is :

Answer

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

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

We know that,

⇒ D = b² - 4ac

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

⇒ D = 4 - 4

⇒ D = 0.

Hence, option 1 is the correct option.

Question 29

What is the nature of the roots of the equation, 2x² - 6x + 3 = 0?

rational and unequal
irrational and unequal
real and equal
imaginary and unequal

Answer

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.

Question 30

The nature of the roots of the equation, 3x² - $4\sqrt{3}x$ + 4 = 0 is:

real and equal
irrational and unequal
rational and unequal
imaginary and unequal

Answer

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.

Question 31

The value of the discriminant of the equation 2x² - 3x + 5 = 0, is:

31
$\sqrt{-31}$
$\sqrt{31}$
-31

Answer

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.

Question 32

If the equation, ax² + 2x + a = 0 has two real and equal roots, then:

a = 0, 1
a = 1, 1
a = 0, -1
a = -1, 1

Answer

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 = $\dfrac{4}{4}$

⇒ a² = 1

⇒ a = $\sqrt{1}$

⇒ a = ± 1

⇒ a = -1, 1.

Hence, option 4 is the correct option.

Question 33

The value of the discriminant of the equation, $\sqrt{3}x^2 + 10x + 7\sqrt{3}$ = 0 is:

Answer

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.

Question 34

The value of the discriminant of the equation, x² - ( $\sqrt{2}$ + 1)x + $\sqrt{2}$ = 0 is:

3 + $2\sqrt{2}$
1 - $2\sqrt{2}$
3 - $2\sqrt{2}$
2 - $\sqrt{2}$

Answer

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,

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

Hence, option 3 is the correct option.

Question 35

The roots of the equation $x + \dfrac{1}{x}$ = 3 are:

$\dfrac{2 \pm \sqrt{5}}{2}$
$\dfrac{3 \pm \sqrt{5}}{2}$
$\dfrac{1 \pm \sqrt{3}}{2}$
none of these

Answer

Given,

$x + \dfrac{1}{x}$ = 3

Multiplying whole equation by x, we get:

$\Rightarrow \dfrac{x^2 + 1}{x} = 3$

⇒ x² + 1 = 3x

⇒ x² - 3x + 1 = 0

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

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

By formula,

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

Substituting values we get :

$\Rightarrow x = \dfrac{-(-3) \pm \sqrt{(-3)^2 - 4 \times 1 \times 1}}{2 \times 1} \\[1em] = \dfrac{3 \pm \sqrt{9 - 4}}{2} \\[1em] = \dfrac{3 \pm \sqrt{5}}{2}.$

Hence, option 2 is the correct option.

Question 36

The solution set for the quadratic equation 2x² + kx - k² = 0 is:

{k, k}
{-k, k}
$\Big\lbrace-k, \dfrac{k}{2}\Big\rbrace$
$\Big\lbrace\dfrac{-k}{2}, k\Big\rbrace$

Answer

⇒ 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 = $\dfrac{k}{2}$ or x = -k

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

Hence, option 3 is the correct option.

Question 37

The solution set for the equation, 25x(x + 1) = -4, is:

$\Big\lbrace\dfrac{1}{5}, \dfrac{4}{5}\Big\rbrace$
$\Big\lbrace-\dfrac{1}{5}, \dfrac{4}{5}\Big\rbrace$
$\Big\lbrace-\dfrac{4}{5}, -\dfrac{1}{5}\Big\rbrace$
$\Big\lbrace-\dfrac{4}{5}, \dfrac{1}{5}\Big\rbrace$

Answer

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

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

Hence, option 3 is the correct option.

Question 38

The value of the discriminant of the equation, 3x² = - 11x - 10, is:

Answer

Given,

⇒ 3x² = - 11x - 10

⇒ 3x² + 11x + 10 = 0

Comparing 3x² + 11x + 10 = 0 with ax² + bx + c = 0 we get,

a = 3, b = 11 and c = 10.

We know that,

⇒ D = b² - 4ac

⇒ D = (11)² - 4 × 3 × 10

⇒ D = 121 - 120

⇒ D = 1.

Hence, option 1 is the correct option.

Question 39

If -5 is a root of the quadratic equation 2x² + px - 15 = 0 and the quadratic equation p(x² + x) + k = 0 has equal roots, then the value of k is:

$\dfrac{7}{4}$
$\dfrac{5}{4}$
$\dfrac{3}{4}$
$\dfrac{1}{4}$

Answer

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 = $\dfrac{35}{5}$

⇒ 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 = $\dfrac{49}{28}$

⇒ k = $\dfrac{7}{4}$ .

Hence, option 1 is the correct option.

Case Study Based Questions

Question 1

Case Study I

Shridharacharya was an Indian mathematician, Sanskrit Pandit and philosopher from Bengal. He is known for his treatises – Trisatika and Patiganita. He was the first to give an algorithm for solving quadratic equations. His other major works were on algebra, particularly fractions and he gave an exposition on zero. He separated Algebra from Arithmetic. The quadratic formula which is used to find the roots of a quadratic equation is known as Shridharacharya’s rule.

1. A quadratic equation of the form ax² + bx + c = 0, a ≠ 0, has two roots given by:

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

2. A quadratic equation of the form ax² + bx + c = 0, a ≠ 0, has rational roots, if the value of (b² - 4ac) is:

less than 0
greater than 0
equal to 0
equal to 0 or a perfect square

3. The maximum number of roots that a quadratic equation can have is:

4. A quadratic equation ax² + bx + c = 0, a ≠ 0, having real coefficients, cannot have real roots if:

b² - 4ac < 0
b² - 4ac = 0
b² - 4ac > 0
b² - 4ac ≥ 0

5. The quadratic equation, x² + 26x + 169 = 0, has:

non real roots
rational and unequal roots
equal roots
irrational roots

Answer

1. By formula,

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

Hence, option (d) is the correct option.

2. Given,

ax² + bx + c = 0, a ≠ 0

We know that,

For the roots to be rational, the discriminant must be a perfect square or zero.

Hence, option (d) is the correct option.

3. A polynomial of degree n has at most n roots.

Given, equation :

ax² + bx + c = 0

Here, the highest power of x is 2. Therefore, maximum number of roots for quadratic equation is 2.

Hence, option (b) is the correct option.

4. Given,

A quadratic equation cannot have real roots if discriminant is less than zero.

Thus, b² − 4ac < 0.

Hence, option (a) is the correct option.

5. Given,

x² + 26x + 169 = 0

Comparing x² + 26x + 169 = 0 with ax² + bx + c = 0 we get,

a = 1, b = 26 and c = 169.

We know that,

⇒ D = b² - 4ac

⇒ D = (26)² - 4(1)(169)

⇒ D = 676 - 676

⇒ D = 0, the roots are real and equal.

Hence, option (c) is the correct option.

Question 2

Case Study II

Raman Lal runs a stationery shop in Pune. The analysis of his sales, expenditures and profits showed that for x number of notebooks sold, the weekly profit (in ₹) was P(x) = - 2x² + 88x - 680. Raman Lal found that:

He has a loss if he does not sell any notebook in a week.
There is no profit no loss for a certain value x₀ of x.
The profit goes on increasing with an increase in x i.e. the number of notebooks sold. But he gets a maximum profit at a sale of 22 notebooks in a week.

Now answer the following questions :

1. What will be Raman Lal’s profit if he sold 20 notebooks in a week?

₹ 144
₹ 280
₹ 340
₹ 560

2. What is the maximum profit that Raman Lal can earn in a week?

₹ 144
₹ 288
₹ 340
₹ 680

3. What is Raman Lal’s loss if he does not sell any notebooks in a particular week?

₹ 0
₹ 340
₹ 680
₹ 960

4. Write a quadratic equation for the condition when Raman Lal does not have any profit or loss during a week.

2x² - 44x + 340 = 0
x² + 44x - 340 = 0
x² - 88x + 340 = 0
x² - 44x + 340 = 0

5. What is the minimum number of notebooks x₀ that Raman Lal should sell in a week so that he does not incur any loss?

Answer

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.

Assertion Reason Questions

Question 1

Assertion (A): The discriminant of the quadratic equation $x^2 + 2\sqrt{2}x + 1 = 0$ is less than zero.

Reason (R): The discriminant of the quadratic equation ax² + bx + c = 0 is $\sqrt{b^{2} - 4ac}$ .

A is true, R is false
A is false, R is true
Both A and R are true
Both A and R are false

Answer

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.

So, Assertion (A) is false.

The discriminant of the quadratic equation ax² + bx + c = 0 is b² - 4ac.

So, Reason (R) is false.

Thus, Both A and R are false.

Hence, option 4 is the correct option.

Question 2

Assertion (A): The quadratic equation 3kx² - 4kx + 4 = 0 has equal roots, if k = 3.

Reason (R): For equal roots of a quadratic equation, we must have D = 0.

A is true, R is false
A is false, R is true
Both A and R are true
Both A and R are false

Answer

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.

Hence, option 3 is the correct option.

Question 3

Assertion (A): The roots of the quadratic equation 3x² + 7x + 8 = 0 are imaginary.

Reason (R): The discriminant of a quadratic equation is always positive.

A is true, R is false
A is false, R is true
Both A and R are true
Both A and R are false

Answer

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 1 is the correct option.

Question 4

Assertion (A): The roots of the quadratic equation 8x² + 2x - 3 = 0 are - $\dfrac{1}{2}$ and $\dfrac{3}{4}$ .

Reason (R): The roots of the quadratic equation ax² + bx + c = 0 are given by $x = \dfrac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$ .

A is true, R is false
A is false, R is true
Both A and R are true
Both A and R are false

Answer

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

Substituting values we get :

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

Thus, A is false, R is true.

Hence, option 2 is the correct option.

Chapter 5

Quadratic Equations

Class - 10 RS Aggarwal Mathematics Solutions

Exercise 5A

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13

Question 14

Question 15

Question 16

Question 17

Question 18

Question 19

Question 20

Question 21

Question 22

Question 23

Question 24

Question 25

Question 26

Question 27

Question 28

Question 29

Question 30

Question 31

Question 32

Question 33

Question 34

Question 35

Question 36

Question 37

Question 38

Question 39

Question 40

Question 41

Question 42

Question 43

Question 44

Question 45

Question 46

Question 47

Exercise 5B

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13

Question 14

Question 15

Question 16

Question 17

Question 18

Question 19

Question 20

Question 21

Exercise 5C

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6