Factorize the following by completing the squares: x 2 - 2 5 x y + y 2 25 - 1

Given, x 2 - 2 5 x y + y 2 25 - 1 Here, coefficient of x 2 is 1 . x 2 - 2 x × y 5 + y 2 25 - 1 ⇒ x - y 5 2 - 1 Now, using identity, we get: ⇒ ( x - y 5 - 1 ) ( x - y 5 + 1 ) Hence, we get the factors as x - y 5 - 1 x - y 5 + 1 .

Factorize the following by completing the squares: 1 3 x 2 - 2 x - 9

Given, 1 3 x 2 - 2 x - 9 Taking the coefficient of x 2 out, we get: ⇒ 1 3 x 2 - 6 x - 27 Now, add and subtract - 6 2 2 = - 3 2 = 9 . ⇒ 1 3 x 2 - 6 x + 9 - 9 - 27 ⇒ 1 3 x - 3 2 - 36 ⇒ 1 3 x - 3 2 - 6 2 Therefore, using identity, we get: ⇒ 1 3 x - 3 + 6 x - 3 - 6 ⇒ 1 3 x + 3 x - 9 Hence, we get the factors as 1 3 x + 3 x - 9 .

Factorize the following by completing the squares: 1 4 a 2 + a - 3

Given, 1 4 a 2 + a - 3 Taking out the coeffitient of a 2 , we get: ⇒ 1 4 a 2 + 4 a - 12 Adding subtracting 4 2 2 = 2 2 = 4 , we get: ⇒ 1 4 a 2 + 4 a + 4 - 4 - 12 ⇒ 1 4 a + 2 2 - 4 - 12 ⇒ 1 4 a + 2 2 - 16 Now, using identity, we get: ⇒ 1 4 a + 2 2 - 4 2 ⇒ 1 4 a + 2 + 4 a + 2 - 4 ⇒ 1 4 a + 6 a - 2 Hence, we get the factors as 1 4 a + 6 a - 2 .

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START UP MATHEMATICS>Chapter 7 - Factorisation of Algebraic Expressions>EXERCISE 7.6

Monicaa Abhijit Solutions for Chapter: Factorisation of Algebraic Expressions, Exercise 6: EXERCISE 7.6

Author:Monicaa Abhijit

Monicaa Abhijit Mathematics Solutions for Exercise - Monicaa Abhijit Solutions for Chapter: Factorisation of Algebraic Expressions, Exercise 6: EXERCISE 7.6

Attempt the practice questions on Chapter 7: Factorisation of Algebraic Expressions, Exercise 6: EXERCISE 7.6 with hints and solutions to strengthen your understanding. START UP MATHEMATICS solutions are prepared by Experienced Embibe Experts.