Question

if x+2y=8 and xy=2 ,then find the value of x^2+4y^2

Answer 1

Given,
x+2y=8
xy=2
Now,
x+2y=8
=>(x+2y)^2=(8)^2
=>x^2+2.x.2y+(2y)^2=64
=>x^2+4xy+4y^2=64
=>x^2+4y^2=64-4xy
=>x^2+4y^2=64-4×2
=>x^2+4y^2=64-8
=>x^2+4y^2=56

Answer 2

Given: x+2y =8

xy =2

To find: x² + 4y²

Solution: According to question

x+2y = 8

By squaring both sides,

(x+2y)² = 8²

x² + 4y² +4xy = 64

x² + 4y² = 64 - 4xy

x² + 4y² = 64 - 4(2) (xy = 2 given)

x² + 4y² = 64 - 8

x² + 4y² = 56

Hence the value of x² + 4y² = 56