How to remove square roots from denominator in $\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}$?

Question

I have this math problem: Remove all square roots in the denominator of

$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}} = ?$

The obvious solution is to multiply the fraction with $\frac{\sqrt{x}-\sqrt{y}}{\sqrt{x}-\sqrt{y}}$.

$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}} = \frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}} \cdot \frac{\sqrt{x}-\sqrt{y}}{\sqrt{x}-\sqrt{y}}$

$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}} = \frac{(\sqrt{x}+\sqrt{y})(\sqrt{x}-\sqrt{y})}{(\sqrt{x}-\sqrt{y})(\sqrt{x}-\sqrt{y})}$

$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}} = \frac{\sqrt{x}\sqrt{x} - \sqrt{x}\sqrt{y} + \sqrt{y}\sqrt{x}-\sqrt{y}\sqrt{y}}{(\sqrt{x}-\sqrt{y})(\sqrt{x}-\sqrt{y})}$

$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}} = \frac{x - \sqrt{x}\sqrt{y} + \sqrt{y}\sqrt{x}-y}{(\sqrt{x}-\sqrt{y})(\sqrt{x}-\sqrt{y})}$

$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}} = \frac{x - y}{(\sqrt{x}-\sqrt{y})(\sqrt{x}-\sqrt{y})}$

$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}} = \frac{x - y}{\sqrt{x}\sqrt{x}-\sqrt{x}\sqrt{y} - \sqrt{y}\sqrt{x}+\sqrt{y}\sqrt{y}}$

$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}} = \frac{x - y}{x-\sqrt{x}\sqrt{y} - \sqrt{y}\sqrt{x}+y}$

$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}} = \frac{x - y}{x-2\sqrt{x}\sqrt{y}+y}$

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

Then I tried to multiply the fraction by $\frac{(\sqrt{x} - \sqrt{y})^2}{(\sqrt{x} - \sqrt{y})^2}$.

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

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

$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}} = \frac{(x - y)(\sqrt{x} - \sqrt{y})^2}{x^2-x2\sqrt{x}\sqrt{y}+xy -2\sqrt{x}\sqrt{y}x + (2\sqrt{x}\sqrt{y})(2\sqrt{x}\sqrt{y}) - 2\sqrt{x}\sqrt{y}y + yx - y2\sqrt{x}\sqrt{y}+y^2}$

$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}} = \frac{(x - y)(\sqrt{x} - \sqrt{y})^2}{x^2-x2\sqrt{x}\sqrt{y} -2\sqrt{x}\sqrt{y}x + (2\sqrt{x}\sqrt{y})(2\sqrt{x}\sqrt{y}) - 2\sqrt{x}\sqrt{y}y - y2\sqrt{x}\sqrt{y}+y^2}$

$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}} = \frac{(x - y)(\sqrt{x} - \sqrt{y})^2}{x^2-2x\sqrt{x}\sqrt{y} -2x\sqrt{x}\sqrt{y} + (2\sqrt{x}\sqrt{y})(2\sqrt{x}\sqrt{y}) - 2\sqrt{x}\sqrt{y}y - y2\sqrt{x}\sqrt{y}+y^2}$

$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}} = \frac{(x - y)(\sqrt{x} - \sqrt{y})^2}{x^2-4x\sqrt{x}\sqrt{y} + (2\sqrt{x}\sqrt{y})(2\sqrt{x}\sqrt{y}) - 2\sqrt{x}\sqrt{y}y - y2\sqrt{x}\sqrt{y}+y^2}$

$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}} = \frac{(x - y)(\sqrt{x} - \sqrt{y})^2}{x^2-4x\sqrt{x}\sqrt{y} + (2\sqrt{x}\sqrt{y})(2\sqrt{x}\sqrt{y}) - 2y\sqrt{x}\sqrt{y} - 2y\sqrt{x}\sqrt{y}+y^2}$

$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}} = \frac{(x - y)(\sqrt{x} - \sqrt{y})^2}{x^2-4x\sqrt{x}\sqrt{y} + (2\sqrt{x}\sqrt{y})(2\sqrt{x}\sqrt{y}) - 4y\sqrt{x}\sqrt{y}+y^2}$

$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}} = \frac{(x - y)(\sqrt{x} - \sqrt{y})^2}{x^2-4x\sqrt{x}\sqrt{y} + 2\cdot2\sqrt{x}\sqrt{x}\sqrt{y}\sqrt{y} - 4y\sqrt{x}\sqrt{y}+y^2}$

$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}} = \frac{(x - y)(\sqrt{x} - \sqrt{y})^2}{x^2-4x\sqrt{x}\sqrt{y} + 4xy - 4y\sqrt{x}\sqrt{y}+y^2}$

This seems like a dead end. Where exactly did I make a mistake?

Answer 1

Multiply denominator and numerator with $\sqrt x+\sqrt y$.

Answer 2

Instead multiply by $\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{x}+\sqrt{y}}$.