Solve 4 x + 1 x + 2 = 2 x - 1 2 x + 3

We are given 4 x + 1 x + 2 = 2 x - 1 2 x + 3 We have to cross multiply and bring it to the form a x 2 + b x + c = 0 . Then we can factorize and solve. 4 x + 1 2 x + 3 = ( 2 x - 1 ) ( x + 2 ) 8 x 2 + 12 x + 2 x + 3 = 2 x 2 + 4 x - x - 2 6 x 2 + 11 x + 5 = 0 6 x 2 + 5 x + 6 x + 5 = 0 x 6 x + 5 + 1 ( 6 x + 5 ) = 0 6 x + 5 x + 1 = 0 By zero product rule 6 x + 5 = 0 or x + 1 = 0 x = - 5 6 or x = - 1 x = - 1 , - 5 6 .

Solve 2 x + 1 x - 1 = x - 7 x - 3

We are given 2 x + 1 x - 1 = x - 7 x - 3 We have to cross multiply and bring it to the form a x 2 + b x + c = 0 . Then we can factorize and solve. 2 x + 1 x - 3 = ( x - 7 ) ( x - 1 ) 2 x 2 - 6 x + x - 3 = x 2 - x - 7 x + 7 2 x 2 - 5 x - 3 = x 2 - 8 x + 7 x 2 + 3 x - 10 = 0 x 2 + 5 x - 2 x - 10 = 0 x x + 5 - 2 ( x + 5 ) = 0 x + 5 x - 2 = 0 By zero product rule x + 5 = 0 or x - 2 = 0 x = - 5 or x = 2 x = - 5,2 .

Solve 3 x + 1 + 2 x - 1 = 1 3 4

We will factorize 3 x + 1 + 2 x - 1 = 7 4 Taking L.C.M on L.H.S, we get 3 x - 3 + 2 x + 2 x 2 - 1 = 7 4 7 x 2 - 7 = 4 5 x - 1 7 x 2 - 7 = 20 x - 4 7 x 2 - 20 x - 3 = 0 7 x 2 - 21 x + x - 3 = 0 7 x x - 3 + 1 ( x - 3 ) = 0 7 x + 1 x - 3 = 0 By zero product rule 7 x + 1 = 0 or x - 3 = 0 x = - 1 7 , 3 .

Solve 4 x + 4 + 1 1 - x = 1 14

We are given 4 x + 4 + 1 1 - x = 1 14 4 - 4 x + x + 4 ( 1 - x ) ( x + 4 ) = 1 14 - 3 x + 8 ( 1 - x ) ( x + 4 ) = 1 14 - 42 x + 112 = ( 1 - x ) ( x + 4 ) 4 + x - x 2 - 4 x = - 42 x + 112 - x 2 - 3 x + 4 + 42 x - 112 = 0 x 2 - 39 x + 108 = 0 x 2 - 36 x - 3 x + 108 = 0 x x - 36 - 3 ( x - 36 ) = 0 x - 3 x - 36 = 0 x = 3 , 36

Solve 5 x + 1 - 2 x - 1 = 1 3

We are given 5 x + 1 - 2 x - 1 = 1 3 5 x - 5 - 2 x - 2 ( x - 1 ) ( x + 1 ) = 1 3 3 x - 7 ( x 2 - 1 ) = 1 3 x 2 - 1 - 9 x + 21 = 0 x 2 - 9 x + 20 = 0 x 2 - 5 x - 4 x + 20 = 0 x ( x - 5 ) - 4 ( x - 5 ) = 0 x - 5 x - 4 = 0 x = 5 , 4

Solve x x + 1 + 5 6 = x + 1 x

We are given x x + 1 + 5 6 = x + 1 x To solve the addition of fractions take L.C.M of denominators and divide the denominator with L.C.M value and then multiply the resultant with the numerator. 5 x + 5 + 6 x 6 ( x + 1 ) = x + 1 x 11 x + 5 6 x + 6 = x + 1 x We have to cross multiply and bring it to the form a x 2 + b x + c = 0 . Then we can factorize and solve. x 11 x + 5 = 6 x + 6 x + 1 11 x 2 + 5 x = 6 x 2 + 6 x + 6 x + 6 11 x 2 + 5 x - 6 x 2 - 6 - 12 x = 0 5 x 2 - 7 x - 6 = 0 Now, 5 x 2 - 7 x - 6 = 0 5 x 2 - 10 x + 3 x - 6 = 0 5 x x - 2 + 3 ( x - 2 ) = 0 5 x + 3 x - 2 = 0 By zero product rule 5 x + 3 = 0 or x - 2 = 0 5 x = - 3 or x = 2 x = - 3 5 , 2 .

Solve x x - 2 + x - 2 x = 2 4 5

We are given x x - 2 + x - 2 x = 2 4 5 x x - 2 + x - 2 x = 14 5 x 2 + x 2 + 4 - 4 x x ( x - 2 ) = 14 5 5 ( 2 x 2 - 4 x + 4 ) = 14 ( x 2 - 2 x ) 14 x 2 - 28 x - 10 x 2 + 20 x - 20 = 0 4 x 2 - 8 x - 20 = 0 x 2 + 2 x - 5 = 0 5 x 2 - 10 x + 3 x - 6 = 0 5 x x - 2 + 3 ( x - 2 ) = 0 5 x + 3 x - 2 = 0 By zero product rule 5 x + 3 = 0 or x - 2 = 0 x = - 3 5 , 2

Solve x x - 2 + x - 2 x = 2 4 5

x x - 2 + x - 2 x = 14 5 x 2 + x 2 + 4 - 4 x x ( x - 2 ) = 14 5 14 x 2 - 28 x - 10 x 2 + 20 x - 20 = 0 4 x 2 - 8 x - 20 = 0 x 2 + 2 x - 5 = 0 5 x 2 - 10 x + 3 x - 6 = 0 5 x x - 2 + 3 ( x - 2 ) = 0 5 x + 3 x - 2 = 0 x = - 3 5 , 2

E M B I B E

ICSE Skills In Mathematics>Algebra>Chapter 20 - Quadratic Equations>EXERCISE 20(A)

V K Raman and S N Sharma Solutions for Exercise 1: EXERCISE 20(A)

Author:V K Raman & S N Sharma

V K Raman Mathematics Solutions for Exercise - V K Raman and S N Sharma Solutions for Exercise 1: EXERCISE 20(A)

Attempt the practice questions from Exercise 1: EXERCISE 20(A) with hints and solutions to strengthen your understanding. ICSE Skills In Mathematics solutions are prepared by Experienced Embibe Experts.