Question

if
xy = 56 and x^2+y^2=128,find the value of x, y.​

Answer 1

Answer:

(x-y)^2 = x^2-2xy+y^2

(x-y)^2 = 128-2x56

= 128-112

= 16

x-y=v16

x-y=4. (i)

(x+y)^2-(x-y) 2 = 4xy

(x + y)^2-4^2 =224

(x+y)^2-16 =224

(x+y)^2 = 24ottttttttxx