# How do you simplify (sqrta- sqrtb)/(sqrta+sqrtb)?

Mar 14, 2018

See a solution process below:

#### Explanation:

To simplify we need to rationalize the denominator by multiplying by the appropriate form of $1$:

$\frac{\textcolor{red}{\sqrt{a}} - \textcolor{red}{\sqrt{b}}}{\textcolor{red}{\sqrt{a}} - \textcolor{red}{\sqrt{b}}} \times \frac{\sqrt{a} - \sqrt{b}}{\sqrt{a} + \sqrt{b}} \implies$

${\left(\sqrt{a} - \sqrt{b}\right)}^{2} / \left(\textcolor{red}{\sqrt{a}} \sqrt{a} + \textcolor{red}{\sqrt{a}} \sqrt{b} - \textcolor{red}{\sqrt{b}} \sqrt{a} - \textcolor{red}{\sqrt{b}} \sqrt{b}\right) \implies$

${\left(\sqrt{a} - \sqrt{b}\right)}^{2} / \left(a - b\right)$

Mar 14, 2018

$\frac{\sqrt{a} - \sqrt{b}}{\sqrt{a} + \sqrt{b}}$

Multiply both the numerator and denominator by $\sqrt{a} - \sqrt{b}$

$\frac{\sqrt{a} - \sqrt{b}}{\sqrt{a} + \sqrt{b}} \cdot \frac{\sqrt{a} - \sqrt{b}}{\sqrt{a} - \sqrt{b}}$

${\left(\sqrt{a} - \sqrt{b}\right)}^{2} / \left({\left(\sqrt{a}\right)}^{2} - {\left(\sqrt{b}\right)}^{2}\right)$

$\frac{a + b - 2 \sqrt{a b}}{a - b}$

Hope this helps :)

Mar 14, 2018

$\frac{a - 2 \sqrt{a b} + b}{a - b}$

#### Explanation:

$\text{multiply the numerator/denominator by the "color(blue)"conjugate}$
$\text{of the denominator}$

$\text{the conjugate of "sqrta+sqrtb" is } \sqrt{a} \textcolor{red}{-} \sqrt{b}$

$\text{this ensures the denominator is a rational value}$

•color(white)(x)sqrtaxxsqrta=a

•color(white)(x)(sqrta+sqrtb)(sqrta-sqrtb)=a-b

$\Rightarrow \frac{\sqrt{a} - \sqrt{b}}{\sqrt{a} + \sqrt{b}} \times \frac{\sqrt{a} - \sqrt{b}}{\sqrt{a} - \sqrt{b}}$

$= \frac{\left(\sqrt{a} - \sqrt{b}\right) \left(\sqrt{a} - \sqrt{b}\right)}{\left(\sqrt{a} + \sqrt{b}\right) \left(\sqrt{a} - \sqrt{b}\right)}$

$= \frac{a - \sqrt{a} b - \sqrt{a} b + b}{a - b}$

$= \frac{a - 2 \sqrt{a b} + b}{a - b}$