Solve the given equation using factorisation method. x - 3 x + 3 + x + 3 x - 3 = 2 1 2

Given, x - 3 x + 2 + x + 3 x - 3 = 2 1 2 ⇒ ( x - 3 ) 2 + ( x + 3 ) 2 ( x + 3 ) ( x - 3 ) = 5 2 ⇒ x 2 - 6 x + 9 + x 2 + 6 x + 9 x 2 - 9 = 5 2 ⇒ 2 ( 2 x 2 + 18 ) = 5 ( x 2 - 9 ) ⇒ 4 x 2 + 36 = 5 x 2 - 45 ⇒ x 2 - 81 = 0 ⇒ x 2 - 9 2 = 0 ⇒ ( x + 9 ) ( x - 9 ) = 0 (Zero Product Rule) ⇒ x + 9 = 0 OR x - 9 = 0 ⇒ x = 9 OR x = - 9 Therefore, x = 9 , - 9 .

Solve the given equation using factorisation method. 4 x + 2 - 1 x + 3 = 4 2 x + 1

4 x + 2 - 1 x + 3 = 4 2 x + 1 4 ( x + 3 ) - ( x + 2 ) ( x + 2 ) ( x + 3 ) = 4 2 x + 1 4 x + 12 - x - 2 x 2 + 2 x + 3 x + 6 = 4 2 x + 1 3 x + 10 x 2 + 5 x + 6 = 4 2 x + 1 ( 3 x + 10 ) ( 2 x + 1 ) = 4 ( x 2 + 5 x + 6 ) 6 x 2 + 3 x + 20 x + 10 = 4 x 2 + 20 x + 24 2 x 2 + 3 x - 14 = 0 2 x 2 + 7 x - 4 x - 14 = 0 x ( 2 x + 7 ) - 2 ( 2 x + 7 ) = 0 ( 2 x + 7 ) ( x - 2 ) = 0 2 x + 7 = 0 OR ( x - 2 ) = 0 x = - 7 2 OR x = 2 Therefore, x = - 7 2 , 2

Solve the given equation using factorisation method. 5 x - 2 - 3 x + 6 = 4 x

5 x - 2 - 3 x + 6 = 4 x 5 ( x + 6 ) - 3 ( x - 2 ) ( x - 2 ) ( x + 6 ) = 4 x 5 x + 30 - 3 x + 6 x 2 + 6 x - 2 x - 12 = 4 x 2 x + 36 x 2 + 4 x - 12 = 4 x 4 x 2 + 16 x - 48 = 2 x 2 + 36 x 2 x 2 - 20 x - 48 = 0 x 2 - 10 x - 24 = 0 x 2 - 12 x + 2 x - 24 = 0 x ( x - 12 ) + 2 ( x - 12 ) = 0 ( x - 12 ) ( x + 2 ) = 0 x - 12 = 0 OR x + 2 = 0 (Zero Product Rule) x = 12 OR x = - 2 Therefore, x = 12 , - 2

Solve the given equation using factorisation method. 1 + 1 x + 1 1 - 1 x - 1 = 7 8

1 + 1 x + 1 1 - 1 x - 1 = 7 8 x + 1 + 1 x + 1 x - 1 - 1 x - 1 = 7 8 x + 2 x + 1 x - 2 x - 1 = 7 8 x 2 - 4 x 2 - 1 = 7 8 8 x 2 - 32 = 7 x 2 - 7 x 2 = 25 x 2 - 25 = 0 ( x - 5 ) ( x + 5 ) = 0 x - 5 = 0 OR x + 5 = 0 (Zero Product Rule) x = 5 OR x = - 5

Solve : x 3 + 3 6 - x = 2 ( 6 + x ) 15 ; ( x ≠ 6 )

Given, x 3 + 3 6 - x = 2 ( 6 + x ) 15 x ( 6 - x ) + 3 × 3 3 ( 6 - x ) = 12 + 2 x 15 6 x - x 2 + 9 6 - x = 12 + 2 x 5 30 x - 5 x 2 + 45 = 72 + 12 x - 12 x - 2 x 2 3 x 2 - 30 x + 27 = 0 x 2 - 10 x + 9 = 0 Now, by splitting middle term method, x 2 - 9 x - x + 9 = 0 x ( x - 9 ) - 1 ( x - 9 ) = 0 ( x - 9 ) ( x - 1 ) = 0 x - 9 = 0 or x - 1 = 0 (Zero Product Rule) Hence, x = 9 or x = 1

Solve the equation 9 x 2 + 3 x 4 + 2 = 0 if possible, for real values x .

Given, 9 x 2 + 3 x 4 + 2 = 0 36 x 2 + 3 x + 8 4 = 0 36 x 2 + 3 x + 8 = 0 Now a = 36 , b = 3 , c = 8 D = b 2 - 4 a c = 9 - 4 × 36 × 8 D = - 1143 D = Negative number So, there will be no real roots for this quadratic equation.

E M B I B E

Concise Mathematics>Algebra>Chapter 5 - Quadratic Equations>EXERCISE

R K Bansal Solutions for Exercise 1: EXERCISE

Author:R K Bansal

R K Bansal Mathematics Solutions for Exercise - R K Bansal Solutions for Exercise 1: EXERCISE

Attempt the practice questions from Exercise 1: EXERCISE with hints and solutions to strengthen your understanding. Concise Mathematics solutions are prepared by Experienced Embibe Experts.