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Rationalizing the Denominator

In mathematics, it is standard practice to write expressions without radicals in the denominator. The process of removing a radical from the denominator is called rationalizing the denominator. This does not change the value of the expression but converts it to a standard form which is often easier to work with.
Method Rationalizing a Monomial Denominator
To rationalize a denominator of the form \(\sqrt{a}\) (where \(a>0\)), multiply the numerator and the denominator by \(\sqrt{a}\).$$ \frac{b}{\sqrt{a}} = \frac{b}{\sqrt{a}} \times \frac{\sqrt{a}}{\sqrt{a}} = \frac{b\sqrt{a}}{a} $$This works because multiplying by \(\frac{\sqrt{a}}{\sqrt{a}}\) is equivalent to multiplying by 1, which does not change the value.
Example
Rationalize the denominator of \(\frac{6}{\sqrt{2}}\).

We multiply the numerator and denominator by \(\sqrt{2}\).$$\begin{aligned}\frac{6}{\sqrt{2}} &= \frac{6}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} \\ &= \frac{6\sqrt{2}}{2} \\ &= \boldsymbol{3\sqrt{2}}\end{aligned}$$

When the denominator is a binomial containing a square root, such as \(a+\sqrt{b}\), we use a special tool called the conjugate to rationalize it.
Definition Conjugate
The conjugate of a binomial expression \(x+y\) is \(x-y\).
Example
  • The conjugate of \(a+\sqrt{b}\) is \(\boldsymbol{a-\sqrt{b}}\).
  • The conjugate of \(a-\sqrt{b}\) is \(\boldsymbol{a+\sqrt{b}}\).
The power of the conjugate lies in its product, which follows the difference of squares identity: \((x+y)(x-y) = x^2 - y^2\).$$ (a+\sqrt{b})(a-\sqrt{b}) = a^2 - (\sqrt{b})^2 = a^2 - b $$The result is a rational number, which achieves our goal.
Method Rationalizing a Binomial Denominator
To rationalize a binomial denominator, multiply the numerator and the denominator by the conjugate of the denominator.
Example
Rationalize the denominator of \(\frac{4}{3+\sqrt{5}}\).

The denominator is \(3+\sqrt{5}\). Its conjugate is \(3-\sqrt{5}\). We multiply the numerator and denominator by this conjugate.$$\begin{aligned}\frac{4}{3+\sqrt{5}} &= \frac{4}{(3+\sqrt{5})} \times \frac{(3-\sqrt{5})}{(3-\sqrt{5})} \\ &= \frac{4(3 - \sqrt{5})}{3^2 - (\sqrt{5})^2} &&\color{gray}{\text{(Using }(x+y)(x-y)=x^2-y^2)}\\ &= \frac{12 - 4\sqrt{5}}{9 - 5} \\ &= \frac{12 - 4\sqrt{5}}{4} \\ &= \frac{4(3 - \sqrt{5})}{4} &&\color{gray}{\text{(Factor out 4 to simplify)}} \\ &= \boldsymbol{3 - \sqrt{5}}\end{aligned}$$