# Rationalizing the Denominator

Rationalizing the denominator is pretty easy. Harder problem in rationalizing requires basic technique and knowledge in algebraic identities. In rule of math, all expressions with radical denominator are not on their simplest form. This topic is advisable if you are at least grade 8.

Application and Extension

Worked Problem 1:

Rationalize the denominator  $\displaystyle\frac{3}{\sqrt{2}}$

Solution:

$\displaystyle\frac{3}{\sqrt{2}}\cdot\displaystyle\frac{\sqrt{2}}{\sqrt{2}}$

The purpose of multiplying $\displaystyle\frac{\sqrt{2}}{\sqrt{2}}$ is to make the denominator a rational. $\displaystyle\frac{\sqrt{2}}{\sqrt{2}}=1$, multiplying 1 in the whole expression won’t change anything on the original expression.

$=\displaystyle\frac{3\sqrt{2}}{\sqrt{4}}$

$=\boxed{\displaystyle\frac{3\sqrt{2}}{2}}$

Worked Problem 2:

Simplify:  $\displaystyle\frac{5}{1-\sqrt{2}}$

Solution:

If the denominator is binomial, we need to use its conjugate. The conjugate of $2-\sqrt{2}$ is $2+\sqrt{2}$.  The conjugate of  $5-\sqrt{3}$  is  $5+\sqrt{3}$. In our given, $1+\sqrt{2}$ is the required conjugate that we are looking for.

$\displaystyle\frac{5}{1-\sqrt{2}}\cdot\displaystyle\frac{1+\sqrt{2}}{1+\sqrt{2}}$

Notice that $(1+\sqrt{2})(1-\sqrt{2})$ is in the form of $(a+b)(a-b)=a^2-b^2$, a sum and difference of two squares.

$=\displaystyle\frac{5(1+\sqrt{2})}{1^2-(\sqrt{2})^2}$

$=\boxed{\displaystyle\frac{5+5\sqrt{2}}{-3}}$

Worked Problem 3:

Rationalize the denominator:  $\displaystyle\frac{1}{\root 3\of{2}+\root 3\of{3}}$

Solution:

Recall that  $a^3+b^3=(a+b)(a^2-ab+a^2)$.

$\displaystyle\frac{1}{\root 3\of{2}+\root 3\of{3}}$

$=\displaystyle\frac{1}{\root 3\of{2}+\root 3\of{3}}\cdot\displaystyle\frac{(\root 3\of{2})^2-\root 3\of{2}\root 3\of{3}+(\root 3\of{3})^2}{(\root 3\of{2})^2-\root 3\of{2}\root 3\of{3}+(\root 3\of{3})^2}$

$=\displaystyle\frac{(\root 3\of{2})^2-\root 3\of{6}+(\root 3\of{3})^2}{(\root 3\of{2})^3+(\root 3\of{2})^3}$

$=\displaystyle\frac{(\root 3\of{2})^2-\root 3\of{6}+(\root 3\of{3})^2}{2+3}$

$=\boxed{\displaystyle\frac{\root 3\of{4}-\root 3\of{6}+\root 3\of{9}}{5}}$