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

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User ShinyuX
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1 Answer

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Answer:

Make use of the fact that
(a + b)^(2) = a^(2) + 2\, a\, b + b^(2).

Explanation:

Start from the given:
z^(1/2) = x^(1/2) + y^(1/2).

In other words:
√(z) = √(x) + √(y).

Square both sides of the equation:


(√(z))^(2) = (√(x) + √(y))^(2).

Equality would still hold because:


\begin{aligned}& a = b \\ \implies \; & a\cdot a = a\cdot b \\ \implies \; & a \cdot a = b \cdot b \\ \implies \; & a^(2) = b^(2)\end{aligned}.

Simplify:


z = (√(x))^(2) + 2\, (√(x)) \cdot (√(y)) + (√(y))^(2).


z = x + 2√(x y) + y.

Subtract
z from both sides of this equation:


0 = x + 2 √(x y) + y - z.

Subtract
2 √(x y) from both sides of this equation:


-2 √(x y) = x + y - z.

Again, square both sides of this equation:


(-2 √(x y))^(2) = (x + y - z)^(2).


(-2)^(2)\, (√(x y))^(2) = (x + y - z)^(2).


(x + y - z)^(2) = 4\, x \, y (by the symmetric property of equalities.)

User Mfjones
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