How do you write y^(-1/2)/x^(1/2) in radical form?

Oct 18, 2015

$\sqrt{\frac{1}{x y}} \text{ }$, or $\text{ } \frac{\sqrt{x y}}{x y}$

Explanation:

Your starting expression looks like this

${y}^{- \frac{1}{2}} / {x}^{\frac{1}{2}}$

The first thing to do is rewrite the negative exponent as a positive exponent. You know that

$\textcolor{b l u e}{{n}^{- a} = \frac{1}{n} ^ a}$

${y}^{- \frac{1}{2}} = \frac{1}{y} ^ \left(\frac{1}{2}\right)$

The expression becomes

${y}^{- \frac{1}{2}} / {x}^{\frac{1}{2}} = \frac{1}{x} ^ \left(\frac{1}{2}\right) \cdot \frac{1}{y} ^ \left(\frac{1}{2}\right)$

Take a look at the denominator. You have

${x}^{\frac{1}{2}} \cdot {y}^{\frac{1}{2}} = {\left(x \cdot y\right)}^{\frac{1}{2}}$

The expression is now equaivalent to

$\frac{1}{x} ^ \left(\frac{1}{2}\right) \cdot \frac{1}{y} ^ \left(\frac{1}{2}\right) = {1}^{\frac{1}{2}} / {\left(x \cdot y\right)}^{\frac{1}{2}} = {\left(\frac{1}{x \cdot y}\right)}^{\frac{1}{2}}$

You know that

$\textcolor{b l u e}{{n}^{\frac{a}{b}} = \sqrt[b]{{n}^{a}}}$

In your case, you will have

${\left(\frac{1}{x y}\right)}^{\frac{1}{2}} = \sqrt{\frac{1}{x y}}$

Extra step

You can rationalize the denominator and simplify this expression further

$\sqrt{\frac{1}{x y}} = \frac{\sqrt{1}}{\sqrt{x y}} = \frac{1}{\sqrt{x y}} \cdot \frac{\sqrt{x y}}{\sqrt{x y}} = \frac{\sqrt{x y}}{\sqrt{x y} \cdot \sqrt{x y}} = \frac{\sqrt{x y}}{x y}$