SChand CLASS 10 Chapter 5 Quadratic Equations Exercise 5A

Exercise 5A

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Q1 | Ex-5A | Class 8 | S.Chand | Mathematics | Chapter 5 | Quadratic Equations | myhelper

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Question 1

(i) (x – 3)(x + 7) = 0

(ii) (x – 3)(x + 7) = 0

Sol :

⇒ (x – 3)(x + 7) = 0

⇒ x – 3 = 0 or x + 7 = 0

⇒ x = 3 or x = -7

⇒ (3x + 4)(2x – 11) = 0

⇒ 3x + 4 = 0 or 2x – 11 = 0

⇒ $x =\frac{ -4}{3}$ or $x = \frac{11}{2}$

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Question 2

$x^2 = 4x$

Sol :

⇒ $x^2 = 4x$

⇒ $(x^2 – 4x) = 0$

⇒ x(x – 4) = 0

⇒ x = 0 or x – 4 = 0

⇒ x = 0 or x = 4

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Question 3

$\left(\frac{x}{3} – 1\right)\left(\frac{x}{2}+ 7\right) = 0$

Sol :

⇒ $\left(\frac{x}{3} – 1\right)\left(\frac{x}{2} + 7\right) = 0$

⇒ $\frac{x}{3} – 1 = 0$ or $\frac{x}{2}+ 7 = 0$

⇒ $\frac{x}{3} = 1$ or $x/2 = -7$

⇒ x = 3 or x = -14

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Question 4

$\frac{(x^2 – 5x) }{2 }= 0$

Sol :

⇒ $\frac{(^x2 – 5x) }{ 2} = 0$

⇒$(x^2 – 5x) = 0$

⇒ x(x – 5) = 0

⇒ x = 0 or x – 5 = 0

⇒ x = 0 or 5

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Question 5

$x^2 – 3x – 10 = 0$

Sol :

⇒ $x^2 – 3x – 10 = 0$

⇒ $x^2 – 5x + 2x – 10 = 0$

⇒ x(x – 5) + 2(x – 5) = 0

⇒ (x – 5)(x + 2) = 0

⇒ x = 5 or -2

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Question 6

$x^2 + x – 12 = 0$

Sol :

⇒ $x^2 + x – 12 = 0$

⇒ $x^2 + 4x – 3x – 12 = 0$

⇒ x(x + 4) – 3(x + 4) = 0

⇒ (x + 4)(x – 3) = 0

⇒ x = -4 or 3

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Question 7

$2(x^2 + 1) = 5x$

Sol :

⇒ $2(x^2 + 1) = 5x$

⇒ $2x^2 – 5x + 2 = 0$

⇒ $2x^2 – 4x – x + 2 = 0$

⇒ 2x(x – 2) – 1(x – 2) = 0

⇒ (x – 2)(2x – 1) = 0

⇒ x = 2 or $\frac{1}{2}$

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Question 8

x(2x + 5) = 3

Sol :

⇒ x(2x + 5) = 3

⇒ $2x^2 + 5x – 3 = 0$

⇒ $2x^2 + 6x – x – 3 = 0$

⇒ 2x(x + 3) – 1(x + 3) = 0

⇒ (x + 3)(2x – 1) = 0

⇒ x + 3 = 0 or 2x – 1 = 0

⇒ x = -3 or $\frac{1}{2}$

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Q9 | Ex-5A | Class 8 | S.Chand | Mathematics | Chapter 5 | Quadratic Equations | myhelper

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Question 9

$4x^2 – 3x – 1 = 0$

Sol :

⇒ $4x^2 – 3x – 1 = 0$

⇒ $4x^2 – 4x + x – 1 = 0$

⇒ 4x(x – 1) + 1(x – 1) = 0

⇒ (x – 1)(4x + 1) = 0

⇒ x – 1 = 0 or 4x – 1 = 0

⇒ x = 1 or $\frac{1}{4}$

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Question 10

$6x^2 – 13x + 5 = 0$

Sol :

⇒ $6x^2 – 13x + 5 = 0$

⇒ $6x^2 – 10x – 3x + 5 = 0$

⇒ 2x(3x – 5) – 1(3x – 5) = 0

⇒ (3x – 5)(2x – 1) = 0

⇒ 3x – 5 = 0 or 2x – 1 = 0

⇒ $x = \frac{5}{3}$ or $\frac{1}{2}$

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Question 11

$3x^2 – 5x – 12 = 0$

Sol :

⇒ $3x^2 – 5x – 12 = 0$

⇒ $3x^2 – 9x + 4x – 12 = 0$

⇒ 3x(x – 3) + 4(x – 3) = 0

⇒ (x – 3)(3x + 4) = 0

⇒ x – 3 = 0 or 3x + 4 = 0

⇒ x = 3 or $\frac{-4}{3}$

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Question 12

$2x^2 – 11x + 5 = 0$

Sol :

⇒ $2x^2 – 11x + 5 = 0$

⇒ $2x^2 – 10x – x + 5 = 0$

⇒ 2x(x – 5) – 1(x – 5) = 0

⇒ (x – 5)(2x – 1) = 0

⇒ x – 5 = 0 or 2x – 1 = 0

⇒ x = 5 or $\frac{1}{2}$

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Question 13

$\frac{x}{2} +\frac{6}{x}= 4$

Sol :

⇒ $\frac{x}{2} + \frac{6}{x} = 4$

⇒ $\frac{(x^2 + 12)}{2x} = 4$

⇒ $x^2 – 8x + 12 = 0$

⇒ $x^2 – 6x – 2x + 12 = 0$

⇒ x(x – 6) – 2(x – 6) = 0

⇒ (x – 6)(x – 2) = 0

⇒ x – 6 = 0 or x – 2 = 0

⇒ x = 6 or 2.

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Question 14

$10x – \frac{1}{x} = 3$

Sol :

⇒ $10x – \frac{1}{x} = 3$

⇒ $10x^2 – 1 = 3x$

⇒ $10x^2 – 3x – 1 = 0$

⇒ $10x^2 – 5x + 2x – 1 = 0$

⇒ 5x(2x – 1) + 1(2x – 1) = 0

⇒ (2x – 1)(5x + 1) = 0

⇒ 2x – 1 = 0 or 5x + 1 = 0

⇒ $x = \frac{1}{2}$ or $\frac{-1}{5}$

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Question 15

$9x +\frac{1}{x} = 6$

Sol :

⇒ $9x + \frac{1}{x}= 6$

⇒ $9x^2 + 1 = 6x$

⇒ $9x^2 – 6x + 1 = 0$

⇒ $(3x)^2 – 2(3x)(1) + 1 = 0$

⇒ $(3x – 1)^2 = 0$

⇒ (3x – 1)(3x – 1) = 0

⇒ (3x – 1) = 0 or (3x – 1) = 0

⇒ $x =\frac{1}{3}$ or $\frac{1}{3}$

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Question 16

$\frac{x}{5} + \frac{28}{(x + 2)}= 5$

Sol :

⇒ $\frac{[x(x + 2) + 140]}{ 5(x + 2) }= 5$

⇒ $x^2 + 2x + 140 = 25(x + 2)$

⇒ $x^2 + 2x + 140 = 25x + 50$

⇒ $x^2 + 2x – 25x + 140 – 50 = 0$

⇒ $x^2 – 23x + 90 = 0$

⇒ $x^2 – 18x – 5x + 90 = 0$

⇒ x(x – 18) – 5(x – 18) = 0

⇒ (x – 18)(x – 5) = 0

⇒ (x – 18) = 0 or (x – 5) = 0

⇒ x = 18 or x = 5

⇒ x = 18 or 5

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Question 17

$\frac{x}{(x – 1)}+ \frac{(x – 1)}{x }= 2\frac{1}{2}$

Sol :

⇒ $\frac{x}{(x – 1)} + \frac{(x – 1)}{x}= 2 \frac{1}{2}$

⇒ $\frac{[x^2 + (x – 1)^2] }{ x(x – 1) }=\frac{5}{2}$

⇒ $2[x^2 + x^2 – 2x + 1] = 5x(x – 1)$

⇒ $4x^2 – 4x + 2 = 5x2 – 5x$

⇒ $x^2 – x – 2 = 0$

⇒ $x^2 – 2x + x – 2 = 0$

⇒ x(x – 2) + 1(x – 2) = 0

⇒ (x + 1)(x – 2) = 0

⇒ (x + 1) = 0 or (x – 2) = 0

⇒ x = -1 or 2

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Q18 | Ex-5A | Class 8 | S.Chand | Mathematics | Chapter 5 | Quadratic Equations | myhelper

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Question 18

$\frac{x}{a}– \frac{(a + b)}{x} = \frac{b(a + b)}{ax}$

Sol :

⇒ $\frac{x}{a} – \frac{(a + b)}{x}= \frac{b(a + b)}{ax}$

⇒ $[x^2 – \frac{a(a + b)]}{ax} = \frac{b(a + b)}{ax}$

⇒ $x^2 – a(a + b) = b(a + b)$

⇒ $x^2 – a(a + b) – b(a + b) = 0$

⇒ $x^2 – (a + b)^2 = 0$

⇒ [x + (a + b)][x – (a + b)] = 0

⇒ [x + (a + b)] = 0 or [x – (a + b)] = 0

⇒ x = (a + b) or -(a + b)

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Question 19

$\frac{1}{(a + b + x)} – \frac{1}{x}= \frac{1}{a} + \frac{1}{b}$

Sol : 

⇒ $\frac{1}{(a + b + x) }– \frac{1}{x}=\frac{1}{a} + \frac{1}{b}$

⇒ $\frac{(x – a – b – x)}{x(a + b + x) }= \frac{(a + b)}{ab}$

⇒ $\frac{-(a + b)}{x(a + b + x) }=\frac{(a + b)}{ab}$

⇒ $\frac{-1}{x(a + b + x)} = \frac{1}{ab}$

⇒ -x(a + b + x) = ab

⇒ $-x^2 – ax – bx – ab = 0$

⇒ $x^2 + ax + bx + ab = 0$

⇒ x(x + a) + b(x + a) = 0

⇒ (x + a)(x + b) = 0

⇒ (x + a) = 0 or (x + b) = 0

⇒ x = -a or -b

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Question 20

$\frac{(x + 3)}{(x – 2)} – \frac{(1 – x)}{x} = \frac{17}{4}$, x ≠ 0, 2.

Sol  :

⇒ $\frac{(x + 3)}{(x – 2) }– \frac{(1 – x)}{x} = \frac{17}{4}$

⇒ $\frac{[x(x + 3) – (x – 2)(1 – x)] }{x(x – 2) }= \frac{17}{4}$

⇒ $\frac{[x^2 + 3x – x + x^2 + 2 – 2x] }{(x^2 – 2x)} = \frac{17}{4}$

⇒ $4[2x^2 + 2] = 17(x^2 – 2x)$

⇒ $8x^2 + 8 = 17x^2 – 34x$

⇒ $17x^2 – 8x^2 – 34x – 8 = 0$

⇒ $9x^2 – 34x – 8 = 0$

⇒ $9x^2 – 36x + 2x – 8 = 0$

⇒ 9x(x – 4) + 2(x – 4) = 0

⇒ (x – 4)(9x + 2) = 0

⇒ (x – 4) = 0 or (9x + 2) = 0

⇒ x = 4 or $\frac{-2}{9}$

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Question 21

$\frac{2x}{(x – 4)} + \frac{(2x – 5)}{(x – 3) }= \frac{25}{3}$

Sol :

⇒ $\frac{2x}{(x – 4)}+ \frac{(2x – 5)}{(x – 3) }= \frac{25}{3}$

⇒ $\frac{[2x(x – 3) + (2x – 5)(x – 4)]}{ (x – 4)(x – 3)} = \frac{25}{3}$

⇒ $\frac{[2x^2 – 6x + 2x^2 – 8x – 5x + 20]}{ (x^2 – 7x + 12) }= \frac{25}{3}$

⇒ $3[4x^2 – 19x + 20] = 25(x^2 – 7x + 12)$

⇒ $12x^2 – 57x + 60 = 25x^2 – 175x + 300$

⇒ $25x^2 – 12x^2 – 175x + 57x + 300 – 60 = 0$

⇒ $13x^2 – 118x + 240 = 0$

⇒ $13x^2 – 78x – 40x + 240 = 0$

⇒ 13x(x – 6) – 40(x – 6) = 0

⇒ (x – 6)(13x – 40) = 0

⇒ (x – 6) = 0 or (13x – 40) = 0

⇒ x = 6 or $\frac{40}{13}$

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Question 22

$\frac{(x + 1)}{(x – 1)} – \frac{(x – 1)}{(x + 1)} =\frac{5}{6}$, x ≠ 1, -1.

Sol :

⇒ $\frac{(x + 1)}{(x – 1)} – \frac{(x – 1)}{(x + 1)} = \frac{5}{6}$

⇒ $\frac{[(x + 1)^2 – (x – 1)^2] }{(x – 1)(x + 1) }= \frac{5}{6}$

⇒ $6[x^2 + 2x + 1 – x^2 +2x – 1] = 5(x^2 – 1)$

⇒ $6[4x] = 5(x^2 – 1)$

⇒ $5x^2 – 24x – 5 = 0$

⇒ $5x^2 – 25x + x – 5 = 0$

⇒ 5x(x – 5) + 1(x – 5) = 0

⇒ (5x + 1)(x – 5) = 0

⇒ (5x + 1) = 0 or (x – 5) = 0

⇒ $x =\frac{-1}{5}$ or 5

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Question 23

$√2x^2 – 3x – 2√2 = 0$

Sol :

⇒ $√2x^2 – 3x – 2√2 = 0$

⇒ $√2x^2 – 4x + x – 2√2 = 0$

⇒ √2x(x – 2√2) + 1(x – 2√2) = 0

⇒ (√2x + 1)(x – 2√2) = 0

⇒ (√2x + 1) = 0 or (x – 2√2) = 0

⇒ $x =\frac{-1}{√2}$ or \frac{2√2}$

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Question 24

$a(x^2 + 1) – x(a^2 + 1) = 0$

Sol :

⇒ $a(x^2 + 1) – x(a^2 + 1) = 0$

⇒ $ax^2 + a – xa^2 – x = 0$

⇒ $ax^2 – xa^2 – x + a = 0$

⇒ ax(x – a) – 1(x – a) = 0

⇒ (ax – 1)(x – a) = 0

⇒ (ax – 1) = 0 or (x – a) = 0

⇒ $x =\frac{1}{a}$ or a

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Question 25

Find the solution set of ${x : x^2 – 2x – 35 = 0}$

Sol :

⇒ $x^2 – 2x – 35 = 0$

⇒ $x^2 – 7x + 5x – 35 = 0$

⇒ x(x – 7) + 5(x – 7) = 0

⇒ (x + 5)(x – 7) = 0

⇒ (x + 5) = 0 or (x – 7) = 0

⇒ x = -5 or 7

⇒ x ∈ {-5, 7}