Question

"Question 5 Show that (i) (3x + 7)^2 − 84x = (3x − 7)^2 (ii) (9p − 5q)^2 + 180pq = (9p + 5q)^2 (iii) (iv) (4pq + 3q)^2 − (4pq − 3q)^2 = 48pq^2 (v) (a − b) (a + b) + (b − c) (b + c) + (c − a) (c + a) = 0

Class 8 Algebraic Expressions and Identities Page 151"

Answer 1

An identity is true only for certain values of its variables. An equation is not an identity.

The following are the identities

(a + b)² = a² + 2ab + b² 
(a – b)² = a² – 2ab + b² 
(a – b)(a + b) = a² – b²

 Another useful identity is

 (x + a) (x + b) = x² + (a + b) x + ab

If the given expression is the difference of two squares we use the formula

a² –b² = (a+b)(a-b)

 

• The above four identities are useful in carrying out squares and products of algebraic expressions. They also allow easy alternative methods to calculate products of numbers and so on.

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Solution:

1) (3x + 7)² – 84x = (3x – 7)²

LHS = (3x)²+2×3x×7+(7)²-84x

9x² + 42x + 49 - 84x

9x² + 42x  - 84x+49

= 9x² - 42x + 49

RHS = (3x)²-2×3x×7+(7)²

 9x² - 42x + 49

LHS = RHS

2)  (9p – 5q)²+ 180pq = (9p + 5q)²

LHS =(9p – 5q)²+ 180pq

(9p)²-2×9p×5q+(5q)²

 = 81p² - 90pq + 25q² + 180pq
= 81p² - 90pq  + 180pq + 25q²

= 81p² + 90pq  + 25q²

RHS = (9p + 5q)²

(9p)²+2×9p×5q+(5q)²

81p² + 90pq + 25q²

LHS = RHS

 

4)  LHS (4pq + 3q)²– (4pq – 3q)²

(4pq)²+2×4pq×3q+(3q)² - ((4pq)²- 2×4pq×3q+(3q)²)

= 16p²q²+ 24pq²+ 9q² - 16p²q² + 24pq² - 9q²

= 48pq²

RHS= 48pq²

LHS = RHS

 

5) LHS= (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a) = 0

a² - b² + b² - c² + c² - a²
= 0

RHS= 0

LHS= RHS

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Hope this will help you.....