Question

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

HELPP MEEEEE ​

Answer 1

Answer:

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

Step-by-step explanation:

Start from the given: $z^{1/2} = x^{1/2} + y^{1/2}$.

In other words: $\sqrt{z} = \sqrt{x} + \sqrt{y}$.

Square both sides of the equation:

$(\sqrt{z})^{2} = (\sqrt{x} + \sqrt{y})^{2}$.

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:

$z = (\sqrt{x})^{2} + 2\, (\sqrt{x}) \cdot (\sqrt{y}) + (\sqrt{y})^{2}$.

$z = x + 2\sqrt{x y} + y$.

Subtract $z$ from both sides of this equation:

$0 = x + 2 \sqrt{x y} + y - z$.

Subtract $2 \sqrt{x y}$ from both sides of this equation:

$-2 \sqrt{x y} = x + y - z$.

Again, square both sides of this equation:

$(-2 \sqrt{x y})^{2} = (x + y - z)^{2}$.

$(-2)^{2}\, (\sqrt{x y})^{2} = (x + y - z)^{2}$.

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