Question

If (x-y) = 4, and xy =32, find the value of (i) (x+y); (ii) x^2+y^2; (iii) x^2-y^2

Answer 1

Answer:

Step-by-step explanation:

Given that,

(x-y) = 4 and xy = 32

We know the Identity : a² + b² = ( a - b )² + 2ab

2) x²+y² = ( x - y )² + 2xy

= (4)² + 2 (32)

= 16 + 64

= 80

1) ( x + y )² = x ² + y² + 2xy

= 80 ÷ (2 × 32 ) ( ∵x²+y² = 80 )

= 80 + 64

(x +y )² = 144

( x + y ) = √144 =12 ( ∵√ 144 = √12 × 12 = 12 )

∴ ( x + y ) = 12

3) x² - y² = ( x - y ) ( x + y ) (∵ a² - b² = ( a - b ) ( a + b ) )

= 4 × 12

= 48