Question

IF X+Y=12 AND XY=32 ,FIND THE VALUE OF X^2+Y^2​

Answer 1

Answer:

Given that: x+y=12 and xy = 32

Using the identity, (x+y)

2

= x

2

+2xy+y

2

⇒ (12)

2

=x

2

+64+y

2

⇒ 144 = x

2

+y

2

+64

⇒ 144−64 = x

2

+y

2

⇒ 80 = x

2

+y

2

The answer is x

2

+y

2

= 80