# How do you rationalize the denominator and simplify (x-3)/(sqrtx-sqrt3)?

Sep 26, 2016

To rationalize a denominator in the form of $\sqrt{a} - \sqrt{b}$, you multiply the fraction by 1 in the form $\frac{\sqrt{a} + \sqrt{b}}{\sqrt{a} + \sqrt{b}}$

#### Explanation:

The reason for doing this practice comes from general form for factoring binomials that contain the difference two squares:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

Returning to the given fraction, we multiply by 1 in form $\frac{\sqrt{x} + \sqrt{3}}{\sqrt{x} + \sqrt{3}}$

$\frac{x - 3}{\sqrt{x} - \sqrt{3}} \frac{\sqrt{x} + \sqrt{3}}{\sqrt{x} + \sqrt{3}} =$

$\frac{\left(x - 3\right) \left(\sqrt{x} + \sqrt{3}\right)}{x - 3} =$

$\sqrt{x} + \sqrt{3}$

Sep 26, 2016

$\sqrt{x} + \sqrt{3}$
divide the Numerator and denominator by $\sqrt{x} + \sqrt{3}$.
we get, $\frac{x - 3}{\sqrt{x} - \sqrt{3}} \cdot \frac{\sqrt{x} + \sqrt{3}}{\sqrt{x} + \sqrt{3}}$
= [(x - 3)(sqrt x + sqrt 3)]/[(sqrt x)^2 - (sqrt 3)^2] = [(x - 3)(sqrt x + sqrt 3)]/(x - 3) = sqrt x + sqrt 3