Solve equation for x by factorisation 1 x - 3 + 2 x - 2 = 8 x , x ≠ 0 , 2 , 3

The given equation is: 1 x - 3 + 2 x - 2 = 8 x Taking LCM in the denominator of LHS, we get ⇒ x - 2 + 2 x - 3 x - 3 x - 2 = 8 x ⇒ x - 2 + 2 x - 6 x 2 - 2 x - 3 x + 6 = 8 x ⇒ 3 x - 8 x 2 - 5 x + 6 = 8 x Transposing x 2 - 5 x + 6 from LHS to RHS and x from RHS to LHS, we get ⇒ 8 x 2 - 40 x + 48 = 3 x 2 - 8 x ⇒ 5 x 2 - 32 x + 48 = 0 Factorising above quadratic equation by splitting the middle term, we get ⇒ 5 x 2 - 20 x - 12 x + 48 = 0 ⇒ 5 x x - 4 - 12 x - 4 = 0 ⇒ 5 x - 12 x - 4 = 0 ⇒ 5 x - 12 = 0 or x - 4 = 0 ⇒ 5 x = 12 or x = 4 ⇒ x = 12 5 or x = 4 Hence, x = 4 and x = 12 5 are the roots of the given equation.

Solve equation for x by factorisation x - 2 x - 3 + x - 4 x - 5 = 3 1 3 , x ≠ 3 , 5

The given equation is: x - 2 x - 3 + x - 4 x - 5 = 10 3 Taking LCM in the denominator, we get ⇒ x - 2 x - 5 + x - 3 x - 4 x - 3 x - 5 = 10 3 ⇒ x 2 - 5 x - 2 x + 10 + x 2 - 4 x - 3 x + 12 x 2 - 5 x - 3 x + 15 = 10 3 ⇒ 2 x 2 - 14 x + 22 x 2 - 8 x + 15 = 10 3 Taking 2 as a common factor in the numerator, we get ⇒ x 2 - 7 x + 11 x 2 - 8 x + 15 = 5 3 ⇒ 3 x 2 - 21 x + 33 = 5 x 2 - 40 x + 75 ⇒ 2 x 2 - 19 x + 42 = 0 ⇒ 2 x 2 - 12 x - 7 x + 42 = 0 ⇒ 2 x x - 6 - 7 x - 6 = 0 ⇒ x - 6 2 x - 7 = 0 ⇒ x - 6 = 0 or 2 x - 7 = 0 ⇒ x = 6 or x = 7 2 Hence, x = 6 and x = 7 2 are the roots of the given equation.

Solve equation for x by factorisation 3 x + 1 + 4 x - 1 = 29 4 x - 1 , x ≠ - 1 , 1 , 1 4

The given equation is: 3 x + 1 + 4 x - 1 = 29 4 x - 1 ⇒ 3 x - 1 + 4 ( x + 1 ) x - 1 ( x + 1 ) = 29 4 x - 1 ⇒ 3 x - 3 + 4 x + 4 x 2 - 1 = 29 4 x - 1 ⇒ 7 x + 1 x 2 - 1 = 29 4 x - 1 ⇒ 7 x + 1 4 x - 1 = 29 x 2 - 29 ⇒ 28 x 2 - 7 x + 4 x - 1 = 29 x 2 - 29 ⇒ x 2 + 3 x - 28 = 0 Now split the middle term ⇒ x 2 + 7 x - 4 x - 28 = 0 ⇒ x x + 7 - 4 x + 7 = 0 ⇒ x - 4 x + 7 = 0 ⇒ x - 4 = 0 or x + 7 = 0 ⇒ x = 4 or x = - 7 Hence, x = 4 and x = - 7 are the roots of the given equation.

Solve equation for x by factorisation x + 1 x - 1 + x - 2 x + 2 = 4 - 2 x + 3 x - 2 , x ≠ - 1 , - 2 , 2

The given equation is: x + 1 x - 1 + x - 2 x + 2 = 4 - 2 x + 3 x - 2 Taking LCM on LHS and RHS ⇒ x + 1 x + 2 + x - 2 x - 1 x - 1 x + 2 = 4 x - 2 - 2 x + 3 x - 2 ⇒ x 2 + 2 x + x + 2 + x 2 - x - 2 x + 2 x 2 + x - 2 = 4 x - 8 - 2 x - 3 x - 2 ⇒ 2 x 2 + 4 x 2 + x - 2 = 2 x - 11 x - 2 On cross multiplication ⇒ 2 x 2 + 4 x - 2 = 2 x - 11 x 2 + x - 2 ⇒ 2 x 3 - 4 x 2 + 4 x - 8 = 2 x 3 - 2 x 2 - 4 x - 11 x 2 - 11 x + 22 ⇒ 5 x 2 + 19 x - 30 = 0 Now, split the middle ⇒ 5 x 2 + 25 x - 6 x - 30 = 0 ⇒ 5 x ( x + 5 ) - 6 ( x + 5 ) = 0 ⇒ ( 5 x - 6 ) ( x + 5 ) = 0 So, 5 x - 6 = 0 ⇒ x = 6 5 x + 5 = 0 ⇒ x = - 5 Thus, the value of x is 6 5 and - 5

Solve for x : 1 a + b + x = 1 a + 1 b + 1 x , a ≠ 0 , b ≠ 0 , c ≠ 0

We have, 1 a + b + x = 1 a + 1 b + 1 x ⇒ 1 a + b + x - 1 x = 1 a + 1 b ⇒ x - a - b - x x x + a + b = a + b a b ⇒ - a + b x 2 + a x + b x = a + b a b ⇒ - 1 x 2 + a x + b x = 1 a b ⇒ x 2 + a x + b x = - a b ⇒ x 2 + a x + b x + a b = 0 ⇒ x x + a + b x + a = 0 ⇒ x + a x + b = 0 ⇒ x + a = 0 or x + b = 0 ⇒ x = - a or x = - b Therefore, the required values of x are - a , - b .

Solve for x : x x + 1 + x + 1 x = 2 1 12 , x ≠ 0 , - 1

Given, x x + 1 + x + 1 x = 2 1 12 Take LCM and convert the equation to a x 2 + b x + c = 0 ⇒ x × x + x + 1 x + 1 x + 1 x = 25 12 ⇒ x 2 + x 2 + 2 x + 1 x 2 + x = 25 12 ⇒ 2 x 2 + 2 x + 1 x 2 + x = 25 12 ⇒ 24 x 2 + 24 x + 12 = 25 x 2 + 25 x ⇒ x 2 + x - 12 = 0 Split the middle term ⇒ x 2 - 3 x + 4 x - 12 = 0 ⇒ x ( x - 3 ) + 4 ( x - 3 ) = 0 ⇒ ( x - 3 ) ( x + 4 ) = 0 Therefore, value of x is 3 and 4 .

Solve for x : 2 x + 9 + x = 13

We have, 2 x + 9 + x = 13 ⇒ 2 x + 9 = 13 - x On squaring both sides 2 x + 9 2 = 13 - x 2 ⇒ 2 x + 9 = 13 2 + x 2 - 2 × 13 × x ⇒ 2 x + 9 = x 2 - 26 x + 169 ⇒ x 2 - 26 x + 169 - 2 x - 9 = 0 Now split the middle term ⇒ x 2 - 28 x + 160 = 0 ⇒ x 2 - 20 x - 8 x + 160 = 0 ⇒ x x - 20 - 8 x - 20 = 0 ⇒ x - 20 x - 8 = 0 ⇒ x = 20 o r x = 8 Thus, the value of x is 20 and 8

E M B I B E

CBSE Syllabus Standard Mathematics for Class X>Chapter 4 - Quadratic Equations>Exercise 4.2

M L Aggarwal Solutions for Chapter: Quadratic Equations, Exercise 2: Exercise 4.2

Author:M L Aggarwal

M L Aggarwal Mathematics Solutions for Exercise - M L Aggarwal Solutions for Chapter: Quadratic Equations, Exercise 2: Exercise 4.2

Attempt the practice questions on Chapter 4: Quadratic Equations, Exercise 2: Exercise 4.2 with hints and solutions to strengthen your understanding. CBSE Syllabus Standard Mathematics for Class X solutions are prepared by Experienced Embibe Experts.