# Week 4 – Precalc 11

One of the things we learned this week in Precalc 11 is Dividing Radicals. Basically dividing a radical by a radical, the main goal is to turn the denomination into a rational number.
Ex.
$\frac{\sqrt{10}}{\sqrt{5}}$
This one is actually pretty easy, you can just divide 10 by 5 so the answer would be $\sqrt{2}$

In the case of $\frac{\sqrt{20}}{\sqrt{3}}$ you can’t divide 20 by 3
we can multiply it by $\frac{\sqrt{3}}{\sqrt{3}}$ so the denominator is a perfect square.
$=\frac{\sqrt{20}}{\sqrt{3}}$x$\frac{\sqrt{3}}{\sqrt{3}}$
$=\frac{\sqrt{60}}{\sqrt{9}}$
then we can simply it to
$=\frac{2\sqrt{15}}{3}$

For binomial radicals in the denominator, we have to use conjugates (a+b)(a-b) to square a and b term.
Ex.
$\frac{\sqrt{4}}{\sqrt{3}-\sqrt{5}}$
we have to use the conjugates of $\sqrt{3}-\sqrt{5}$ which is $\sqrt{3}+\sqrt{5}$
=$\frac{\sqrt{4}}{\sqrt{3}-\sqrt{5}}$x$\frac{\sqrt{3}+\sqrt{5}}{\sqrt{3}+\sqrt{5}}$
=$\frac{\sqrt{12}+\sqrt{20}}{\sqrt{9}-\sqrt{25}}$
=$\frac{2\sqrt{3}+2\sqrt{5}}{3-5}$
=$\frac{2\sqrt{3}+2\sqrt{5}}{-2}$

More examples