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

Mathematics

1 answer:

Rudik [331]2 years ago

3 0

Answer:

Step-by-step 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

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