Dividing radicals in Math comes down to rationalizing the denominator on the bottom line of a fraction.

While not entirely wrong, it’s more Mathematically ‘correct’ to only have rational or whole numbers in the denominator, rather than an irrational radical.

We will see how this works on this page in some rationalizing the denominator examples.

How to Rationalize a Denominator:

A standard fraction with an irrational denominator is    \bf{\frac{1}{\sqrt{5}}}.

To rationalize, we want to try and get rid of the radical sign in the denominator, her multiplying both the numerator and denominator by  \sqrt{5}  will achieve this.

\bf{\frac{1}{\sqrt{5}}} × \bf{\frac{\sqrt{5}}{\sqrt{5}}}   =   \bf{\frac{\sqrt{5}}{\sqrt{5}\sqrt{5}}}   =   \bf{\frac{\sqrt{5}}{5}}

As we are multiplying the top and bottom both by the same number/amount, this will not change the overall value of the fraction.

Also, with rationalizing the denominator examples we are ok with having an irrational number or term as the numerator on the top line, it’s just the denominator that we don’t want to be irrational.

Rationalizing the Denominator Examples

(1.1)

Rationalize   {\frac{2}{\sqrt{r}}}.

Solution

The approach of multiplying the top and bottom by a root is also just fine with variables involved as well as numbers.

{\frac{2}{\sqrt{r}}} × {\frac{\sqrt{r}}{\sqrt{r}}}   =   {\frac{2\sqrt{r}}{\sqrt{r}\sqrt{r}}}   =   \frac{2\sqrt{r}}{r}

(1.2)
Rationalize   \sqrt{\frac{4}{y}}.

Solution

\sqrt{\frac{4}{y}}   =   \frac{\sqrt{4}}{\sqrt{y}}

\frac{\sqrt{4}}{\sqrt{y}}   =   \frac{\sqrt{4}}{\sqrt{y}} × \frac{\sqrt{y}}{\sqrt{y}}   =   \frac{\sqrt{4}\sqrt{y}}{\sqrt{y}\sqrt{y}}   =   \frac{\sqrt{4y}}{y}

(1.3)

Rationalize   \frac{1}{5 \space {\text{--}} \space \sqrt{a}}.

Solution

This is a division of  ‘1’  by  ‘5 \space {\text{--}} \space \sqrt{a}‘.

A situation like this requires us to multiply top and bottom by the conjugate of the denominator, where the operation of plus or minus is reversed.
As multiplying the denominator by its conjugate will eliminate the radicals present.

Conjugate of   5 \space {\text{--}} \space \sqrt{a}   is   5 + \sqrt{a}.

\frac{1}{5 \space {\text{--}} \space \sqrt{a}} × \frac{5 \space + \space \sqrt{a}}{5 \space + \space \sqrt{a}}   =   \frac{5 \space + \space \sqrt{a}}{(5 \space {\text{--}} \space \sqrt{a})(5 \space + \space \sqrt{a})}

=   \frac{5 \space + \space \sqrt{a}}{25 \space + \space 5\sqrt{a} \space {\text{--}} \space 5\sqrt{a} \space {\text{--}} \space (\sqrt{a})^2}   =   \frac{5 \space + \space \sqrt{a}}{25 \space {\text{--}} \space a}

(1.4)

Rationalize   \frac{6}{3\sqrt{v} \space + \space w}.

Solution

Like in example (1.3), we take this division on by making use of the conjugate of the denominator.

\frac{6}{3\sqrt{v} \space + \space w} × \frac{3\sqrt{v} \space {\text{--}} \space w}{3\sqrt{v} \space {\text{--}} \space w}   =   \frac{6(3\sqrt{v} \space {\text{--}} \space w)}{(3\sqrt{v} \space + \space w)(3\sqrt{v} \space {\text{--}} \space w)}

=   \frac{18\sqrt{v} \space {\text{--}} \space 6w}{(3\sqrt{v})^2 \space {\text{--}} \space 3\sqrt{v}w \space + \space 3\sqrt{v}w \space {\text{--}} \space w^2}

=   \frac{18\sqrt{v} \space {\text{--}} \space 6w}{9v \space {\text{--}} \space w^2}

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