The properties of exponents, which we've talked about earlier, tell us among other things that

$\begin{pmatrix} xy \end{pmatrix}^{a}=x^{a}y^{a}$

$\begin{pmatrix} \frac{x}{y} \end{pmatrix}^{a}=\frac{x^{a}}{y^{a}}$

We also know that

$\sqrt[a]{x}=x^{\frac{1}{a}}$

$or$

$\sqrt{x}=x^{\frac{1}{2}}$

If we combine these two things then we get the product property of radicals and the quotient property of radicals. These two properties tell us that the square root of a product equals the product of the square roots of the factors.

$\sqrt{xy}=\sqrt{x}\cdot \sqrt{y}$

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

$where\:\: x\geq 0,y\geq 0$

The answer can't be negative and x and y can't be negative since we then wouldn't get a real answer. In the same way we know that

$\sqrt{x^{2}}=x\: \: where\: \: x\geq 0$

These properties can be used to simplify radical expressions. A radical expression is said to be in its simplest form if there are

no perfect square factors other than 1 in the radicand

$\sqrt{16x}=\sqrt{16}\cdot \sqrt{x}=\sqrt{4^{2}}\cdot \sqrt{x}=4\sqrt{x}$

no fractions in the radicand and

$\sqrt{\frac{25}{16}x^{2}}=\frac{\sqrt{25}}{\sqrt{16}}\cdot \sqrt{x^{2}}=\frac{5}{4}x$

no radicals appear in the denominator of a fraction.

$\sqrt{\frac{15}{16}}=\frac{\sqrt{15}}{\sqrt{16}}=\frac{\sqrt{15}}{4}$

If the denominator is not a perfect square you can rationalize the denominator by multiplying the expression by an appropriate form of 1 e.g.

$\sqrt{\frac{x}{y}}=\frac{\sqrt{x}}{\sqrt{y}}\cdot {\color{green} {\frac{\sqrt{y}}{\sqrt{y}}}}=\frac{\sqrt{xy}}{\sqrt{y^{2}}}=\frac{\sqrt{xy}}{y}$

Binomials like

$x\sqrt{y}+z\sqrt{w}\: \: and\: \: x\sqrt{y}-z\sqrt{w}$

are called conjugates to each other. The product of two conjugates is always a rational number which means that you can use conjugates to rationalize the denominator e.g.

$\frac{x}{4+\sqrt{x}}=\frac{x\left ( {\color{green} {4-\sqrt{x}}} \right )}{\left ( 4+\sqrt{x} \right )\left ( {\color{green}{ 4-\sqrt{x}}} \right )}=$

$=\frac{x\left ( 4-\sqrt{x} \right )}{16-\left ( \sqrt{x} \right )^{2}}=\frac{4x-x\sqrt{x}}{16-x}$

## Video lesson

$\frac{x}{5-\sqrt{x}}$