# How do you rationalize the denominator and simplify (sqrtc-sqrtd)/(sqrtc+sqrtd)?

Oct 3, 2015

Multiply numerator and denominator by $\left(\sqrt{c} - \sqrt{d}\right)$ and simplify to find:

$\frac{\sqrt{c} - \sqrt{d}}{\sqrt{c} + \sqrt{d}} = \frac{c + d - 2 \sqrt{c d}}{c - d}$

#### Explanation:

Use the difference of squares identity: ${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$ with $a = \sqrt{c}$ and $b = \sqrt{d}$

Also use $\sqrt{a} \sqrt{b} = \sqrt{a b}$ (if $a , b \ge 0$)

$\frac{\sqrt{c} - \sqrt{d}}{\sqrt{c} + \sqrt{d}}$

$= \frac{\left(\sqrt{c} - \sqrt{d}\right) \left(\sqrt{c} - \sqrt{d}\right)}{\left(\sqrt{c} - \sqrt{d}\right) \left(\sqrt{c} + \sqrt{d}\right)}$

$= \frac{{\left(\sqrt{c}\right)}^{2} - 2 \left(\sqrt{c}\right) \left(\sqrt{d}\right) + {\left(\sqrt{d}\right)}^{2}}{{\left(\sqrt{c}\right)}^{2} - {\left(\sqrt{d}\right)}^{2}}$

$= \frac{c + d - 2 \sqrt{c d}}{c - d}$