Solve the following quadratic equations by factorisation: x - 1 x - 2 + x - 3 x - 4 = 3 1 3 , x ≠ 2 , 4

Given equation is x - 1 x - 2 + x - 3 x - 4 = 3 1 3 ⇒ x - 1 x - 4 + x - 3 x - 2 x - 2 x - 4 = 10 3 ⇒ 3 x 2 - 5 x + 4 + x 2 - 5 x + 6 = 10 x 2 - 6 x + 8 ⇒ 4 x 2 - 30 x + 50 = 0 ⇒ 2 x 2 - 15 x + 25 = 0 Splitting the middle term, we get: ⇒ 2 x 2 - 10 x - 5 x + 25 = 0 ⇒ 2 x ( x - 5 ) - 5 ( x - 5 ) = 0 ⇒ ( 2 x - 5 ) ( x - 5 ) = 0 Thus, 2 x - 5 = 0 ⇒ 2 x = 5 ⇒ x = 5 2 or x - 5 = 0 ⇒ x = 5 Hence, x = 5 2 or x = 5

Solve the following quadratic equation by factorization: 1 2 a + b + 2 x = 1 2 a + 1 b + 1 2 x

Given equation is 1 2 a + b + 2 x = 1 2 a + 1 b + 1 2 x . ⇒ 1 2 a + b + 2 x - 1 2 a = 1 b + 1 2 x ⇒ 2 a - ( 2 a + b + 2 x ) ( 2 a + b + 2 x ) ( 2 a ) = 2 x + b 2 b x ⇒ - b - 2 x 4 a 2 + 2 a b + 4 a x = 2 x + b 2 b x ⇒ - 1 ( 2 x + b ) 4 a 2 + 2 a b + 4 a x = 2 x + b 2 b x ⇒ - 2 b x ( 2 x + b ) = 4 a 2 + 2 a b + 4 a x ( 2 x + b ) ⇒ 4 a 2 + 2 a b + 4 a x ( 2 x + b ) + 2 b x ( 2 x + b ) = 0 ⇒ ( 2 x + b ) 4 a 2 + 2 a b + 4 a x + 2 b x = 0 ⇒ 2 x + b = 0 or 4 a 2 + 2 a b + ( 4 a + 2 b ) x = 0 ⇒ x = - b 2 or x = - 4 a 2 + 2 a b 4 a + 2 b ⇒ x = - b 2 or x = - a ( 4 a + 2 b ) 4 a + 2 b ⇒ x = - b 2 or x = - a Hence, the factors are − a and - b 2 .

Solve the following quadratic equation by factorisation. 3 3 x - 1 2 x + 3 - 2 2 x + 3 3 x - 1 = 5 ; x ≠ 1 3 , - 3 2

Given equation is 3 3 x - 1 2 x + 3 - 2 2 x + 3 3 x - 1 = 5 . Simplifying, 3 ( 3 x - 1 ) 2 - 2 ( 2 x + 3 ) 2 ( 2 x + 3 ) ( 3 x - 1 ) = 5 ⇒ 3 9 x 2 + 1 - 6 x - 2 4 x 2 + 9 + 12 x 6 x 2 - 2 x + 9 x - 3 = 5 ⇒ 27 x 2 + 3 - 18 x - 8 x 2 - 18 - 24 x 6 x 2 + 7 x - 3 = 5 ⇒ 19 x 2 - 42 x - 15 = 5 6 x 2 + 7 x - 3 ⇒ 19 x 2 - 42 x - 15 = 30 x 2 + 35 x - 15 ⇒ 30 x 2 - 19 x 2 + 35 x + 42 x - 15 + 15 = 0 ⇒ 11 x 2 + 77 x = 0 ⇒ x 2 + 7 x = 0 Factorising the quadratic equation x 2 + 7 x = 0 , x ( x + 7 ) = 0 ⇒ x = 0 or x + 7 = 0 ⇒ x = 0 or x = - 7 Hence, the solutions of the quadratic equation are 0 and - 7 .

Solve the following quadratic equation by factorisation. 3 7 x + 1 5 x - 3 - 4 5 x - 3 7 x + 1 = 11 ; x ≠ 3 5 , - 1 7

Given equation is 3 7 x + 1 5 x - 3 - 4 5 x - 3 7 x + 1 = 11 . Simplifying, 3 ( 7 x + 1 ) 2 - 4 ( 5 x - 3 ) 2 ( 5 x - 3 ) ( 7 x + 1 ) = 11 ⇒ 3 49 x 2 + 1 + 14 x - 4 25 x 2 + 9 - 30 x 35 x 2 + 5 x - 21 x - 3 = 11 ⇒ 147 x 2 + 3 + 42 x - 100 x 2 - 36 + 120 x 35 x 2 - 16 x - 3 = 11 ⇒ 47 x 2 + 162 x - 33 = 11 35 x 2 - 16 x - 3 ⇒ 47 x 2 + 162 x - 33 = 385 x 2 - 176 x - 33 ⇒ 385 x 2 - 47 x 2 - 176 x - 162 x - 33 + 33 = 0 ⇒ 338 x 2 - 338 x = 0 ⇒ x 2 - x = 0 Factorising the quadratic equation x 2 - x = 0 , ⇒ x ( x - 1 ) = 0 ⇒ x = 0 or x - 1 = 0 ⇒ x = 0 or x = 1 Hence, the solutions of the quadratic equation are 0 and 1 .

E M B I B E

MATHEMATICS CLASS X>Chapter 4 - Quadratic Equations>EXERCISE 4.3

R. D. Sharma Solutions for Chapter: Quadratic Equations, Exercise 3: EXERCISE 4.3

Author:R. D. Sharma

R. D. Sharma Mathematics Solutions for Exercise - R. D. Sharma Solutions for Chapter: Quadratic Equations, Exercise 3: EXERCISE 4.3

Attempt the free practice questions on Chapter 4: Quadratic Equations, Exercise 3: EXERCISE 4.3 with hints and solutions to strengthen your understanding. MATHEMATICS CLASS X solutions are prepared by Experienced Embibe Experts.

Questions from R. D. Sharma Solutions for Chapter: Quadratic Equations, Exercise 3: EXERCISE 4.3 with Hints & Solutions

MEDIUM

10th CBSE

IMPORTANT

MATHEMATICS CLASS X>Chapter 4 - Quadratic Equations>EXERCISE 4.3>Q 55

Solve the following quadratic equations by factorization:

$a b x^{2} + (b^{2} - a c) x - b c = 0$

MEDIUM

10th CBSE

IMPORTANT

MATHEMATICS CLASS X>Chapter 4 - Quadratic Equations>EXERCISE 4.3>Q 56

Solve the following quadratic equations by factorization:

$a^{2} b^{2} x^{2} + b^{2} x - a^{2} x - 1 = 0$

MEDIUM

10th CBSE

IMPORTANT

MATHEMATICS CLASS X>Chapter 4 - Quadratic Equations>EXERCISE 4.3>Q 57

Solve the following quadratic equations by factorisation:

$\frac{x - 1}{x - 2} + \frac{x - 3}{x - 4} = 3 \frac{1}{3}, x \neq 2, 4$

HARD

10th CBSE

IMPORTANT

MATHEMATICS CLASS X>Chapter 4 - Quadratic Equations>EXERCISE 4.3>Q 58

Solve the following quadratic equation by factorization:

$\frac{1}{2 a + b + 2 x} = \frac{1}{2 a} + \frac{1}{b} + \frac{1}{2 x}$

MEDIUM

10th CBSE

IMPORTANT

MATHEMATICS CLASS X>Chapter 4 - Quadratic Equations>EXERCISE 4.3>Q 59

Solve the following quadratic equation by factorisation.
$3 (\frac{3 x - 1}{2 x + 3}) - 2 (\frac{2 x + 3}{3 x - 1}) = 5; x \neq \frac{1}{3}, \frac{- 3}{2}$