Question

5.b) If z^(1/2)=x^(1/2)+y^(1/2) , show that (x+y-z)^2=4xy

Answer 1

Explanation:

• (x+y-z)²= 4xy
• (x+y-z)²- 4xy = 0
• (x+y-z)²-(2√x√y)² = 0
• (x+y-z-2√x√y)(x+y-z+2√x√y) =0
• [(√x-√y)²-z]*[(√x+√y)²-z]=0
• (√x-√y)²-z = 0 or (√x+√y)²-z = 0

We have : z^(1/2)= x^(1/2)+y^(1/2) ⇒ √z = √x + √y ⇒ z = (√x + √y)²

• so (√x+√y)²-z = 0

so (x+y-z)²= 4xy