# How do you rationalize the denominator and simplify (4-sqrt3)/(6+sqrty)?

Feb 12, 2017

$\frac{4 - \sqrt{3}}{6 + \sqrt{y}} = \frac{24 - 4 \sqrt{y} - 6 \sqrt{3} + \sqrt{3 y}}{36 - y}$

#### Explanation:

To rationalize the denominator, we multiply numerator and denominator by the conjugate of its denominator .

Conjugate of an irrational number like $\sqrt{a} + \sqrt{b}$ is $\sqrt{a} - \sqrt{b}$ and that of $p + \sqrt{q}$ is $p - \sqrt{q}$.

Here in $\frac{4 - \sqrt{3}}{6 + \sqrt{y}}$, denominator is $6 + \sqrt{y}$ and assuming $y$ is rational positive number, conjugate of denominator is $6 - \sqrt{y}$. As such

$\frac{4 - \sqrt{3}}{6 + \sqrt{y}}$

= $\frac{4 - \sqrt{3}}{6 + \sqrt{y}} \times \frac{6 - \sqrt{y}}{6 - \sqrt{y}}$

= $\frac{\left(4 - \sqrt{3}\right) \left(6 - \sqrt{y}\right)}{{6}^{2} - {\left(\sqrt{y}\right)}^{2}}$

= $\frac{4 \times 6 - 4 \sqrt{y} - 6 \sqrt{3} + \sqrt{3 y}}{{6}^{2} - {\left(\sqrt{y}\right)}^{2}}$

= $\frac{24 - 4 \sqrt{y} - 6 \sqrt{3} + \sqrt{3 y}}{36 - y}$

Note - In case you have just $\sqrt{p}$ in denominator, just multiply numerator and denominator by $\sqrt{p}$.