Question

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

HELPP MEEEEE ​

Answer 1

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.)