Rationalizing Denominators

Rationalizing the denominator is when we move any fractional power from the bottom of a fraction to the top. $\frac{1}{\sqrt{2}}$, for example, has an irrational denominator. We will soon see that it equals $\frac{\sqrt{2}}{2}$. Let us look at fractions with irrational denominators.

Consider the fraction

$$\frac{a}{\sqrt{b}}.$$

It is perfectly valid to write it as

$$\begin{aligned} \frac{a}{\sqrt{b}} &=\frac{a}{\sqrt{b} } \times 1\\ &=\frac{a}{\sqrt{b} } \times \frac{\sqrt{b}}{\sqrt{b}} \\ &=\frac{a\sqrt{b}}{b}. \end{aligned}$$

We have now rationalized the denominator and most problems will depend on this simple technique.

Rationalize the denominator of $\frac{1}{\sqrt{2}}.$

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Using what we learned, we have

$$\frac{1}{\sqrt{2}}=\frac{1}{\sqrt{2} } \times \frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{2}.\ _\square$$

Rationalize the denominator of $\frac{2}{\sqrt {\sqrt {2}}}$.

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We have

$$\begin{aligned} \dfrac{2}{\sqrt {\sqrt {2}}} & = \dfrac{2 × \sqrt{\sqrt {2}}}{\sqrt{\sqrt {2}} × \sqrt{\sqrt {2}}}\\ & = \dfrac{2 \sqrt{\sqrt {2}}}{\sqrt {2}}\\ & = \dfrac{2 \sqrt{\sqrt {2}} × \sqrt{2}}{\sqrt {2} × \sqrt {2}}\\ & = \dfrac{2 \sqrt{2\sqrt{2}}}{2}\\ & = \sqrt{2\sqrt{2}}.\ _\square \end{aligned}$$

Rationalize the denominator of $\frac{\sqrt{y-1}}{\sqrt{y+1}}$.

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We have

$$\begin{aligned} \dfrac{\sqrt{y-1}}{\sqrt{y+1}} & = \dfrac{\big(\sqrt{y-1}\big)\big(\sqrt{y+1}\big)}{\big(\sqrt{y+1}\big)\big(\sqrt{y+1}\big)}\\ & =\dfrac {\sqrt{y^2-1}}{y + 1}.\ _\square \end{aligned}$$

Rationalize the expression $\frac{y^2 \sqrt{x^2 - 1}}{\sqrt{x+1}}.$

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Rationalizing the denominator, we get

$$\frac{y^2 \sqrt{x^2 - 1}}{\sqrt{x+1}} \times \frac{\sqrt{x+1}}{\sqrt{x+1}}=\frac{y^2 \sqrt{x^2 -1} \times\sqrt{x+1}}{x+1} .$$

Simplifying even further, we have

$$\begin{aligned} \frac{y^2 \sqrt{(x-1)(x+1)} \times\sqrt{x+1}}{x+1} &= \frac{y^2 \sqrt{(x-1)(x+1)^2} }{x+1} \\ &=\frac{y^2(x+1) \sqrt{x-1}}{x+1}. \end{aligned}$$

Canceling out $x+1,$ our final result is

$$y^2\sqrt{x-1}.\ _\square$$

Rationalize the denominator of $\frac {3}{\sqrt[3]{3}}$

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We have

$$\begin{aligned} \dfrac {3}{\sqrt[3]{3}} & = \dfrac {3 × \sqrt[3]{9}}{\sqrt[3]{3} × \sqrt[3]{9}}\\ & = \dfrac {3 × \sqrt[3]{9}}{\sqrt[3]{27}}\\ & = \dfrac {3 × \sqrt[3]{9}}{3}\\ & =\sqrt[3]{9}.\ _\square \end{aligned}$$

Now consider fractions of the form $$\frac{a}{b + \sqrt{c}},$$ where $a$,$b$ and $c$ are integers. If we multiply by $\frac{b+\sqrt{c}}{b+\sqrt{c}},$ we will obtain $$\frac{ab +a\sqrt{c}}{b^2 + 2b\sqrt{c} + c} ,$$ which as you can see is a more complicated form of the fraction. Therefore, whenever we obtain such type of fractions, we utilize the difference of squares formula $$(x-y)(x+y)=x^2 - y^2.$$ This is particularly useful because if $x = \sqrt{m}$ and $y=\sqrt{n},$ then $$\big(\sqrt{m}-\sqrt{n}\big)\big(\sqrt{m}+\sqrt{n}\big) = m - n .$$ We are now ready to rationalize the denominators of our original expression by multiplying by $\frac{b-\sqrt{c}}{b-\sqrt{c}}:$ $$\frac{a}{b + \sqrt{c}} \times \frac{b-\sqrt{c}}{b - \sqrt{c} }=\frac{ab - a\sqrt{c}}{b^2 - c} .$$

Simplify the following expression:

$$\frac{3}{\sqrt{3}} + \frac{2}{\sqrt{2} + \sqrt{5} }.$$

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Let us approach the two components separately. $\frac{3}{\sqrt{3}}$ can be simplified by rationalizing with $\frac{\sqrt{3}}{\sqrt{3}}$ and the right part $\frac{2}{\sqrt{2} + \sqrt{5} }$ can be simplified by multiplying by $\frac{\sqrt{2}-\sqrt{5}}{\sqrt{2}-\sqrt{5}} :$ $$\frac{3}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}+ \frac{2}{\sqrt{2} + \sqrt{5} } \times \frac{\sqrt{2}-\sqrt{5}}{\sqrt{2}-\sqrt{5}} =\frac{3\sqrt{3}}{3} + \frac{2(\sqrt{2}-\sqrt{5})}{-3}.$$ From this we obtain our answer $$\frac{3\sqrt{3}-2\sqrt{2}+2\sqrt{5}}{3}.\ _\square$$

Simplify the following expression: $$\frac{7\sqrt{6} + \sqrt{303} }{ \sqrt{3} } \times (\sqrt{101} - \sqrt{98} ).$$

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We begin by writing $\sqrt{98}$ as $7\sqrt{2}:$ $$\frac{7\sqrt{6} + \sqrt{303} }{ \sqrt{3} } \times (\sqrt{101} - 7\sqrt{2} ).$$ Rationalizing the denominator by $\sqrt{3}$ gives $$\begin{aligned} \frac{7\sqrt{6} \times \sqrt{3} + \sqrt{303} \sqrt{3} }{ 3 } \times (\sqrt{101} - 7\sqrt{2} ) = &\frac{7\sqrt{3\times 2 \times 3} + \sqrt{3\times101\times 3} }{ 3 } \times (\sqrt{101} - 7\sqrt{2} )\\ =& \frac{7\sqrt{3^2 \times 2 } + \sqrt{3^2 \times101 } }{ 3 } \times (\sqrt{101} - 7\sqrt{2} )\\ =& \frac{3\times 7\sqrt{2 } + 3\sqrt{101 } }{ 3 } \times (\sqrt{101} - 7\sqrt{2} ). \end{aligned}$$ Cancelling out the $3$'s on the left side gives $$\begin{aligned} (7\sqrt{2 } + \sqrt{101 } ) \times (\sqrt{101} - 7\sqrt{2} ) &=7\sqrt{202} - 98 + 101 - 7\sqrt{202}\\ &=3.\ _\square \end{aligned}$$

Simplify

$$\dfrac{\big({\sqrt{5}\big)}^3 - {\big(\sqrt{4}\big)}^3}{{\big(\sqrt{5}\big)}^3 + {\big(\sqrt{4}\big)}^3}.$$

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We have

$$\begin{aligned} \dfrac{{(\sqrt{5})}^3 - {(\sqrt{4})}^3}{{(\sqrt{5})}^3 + {(\sqrt{4})}^3} & = \dfrac{(\sqrt 5 - \sqrt 4)(5 + \sqrt{20} + 4)}{(\sqrt 5 + \sqrt 4)(5 - \sqrt{20} + 4)}\\ & = \dfrac{\big(\sqrt{5} - \sqrt{4}\big)\big(9 + 2\sqrt{5}\big)}{\big(\sqrt{5} + \sqrt{4}\big)\big(9 - 2\sqrt{5}\big)}\\ & = \dfrac{\big(\sqrt{5} - \sqrt{4}\big)\big(9 + 2\sqrt{5}\big)\big(9 + 2\sqrt{5}\big)}{\big(\sqrt{5} + \sqrt{4}\big)\big(9 - 2\sqrt{5}\big)\big(9+ 2\sqrt{5}\big)}\\ & = \dfrac{\big(\sqrt{5} - 2\big)\big(81 + 20 + 36\sqrt{5}\big)}{\big(\sqrt{5} + 2\big)(81-20)}\\ & = \dfrac{\big(\sqrt 5 - 2\big)\big(101 + 36\sqrt {5}\big)}{\big(\sqrt 5 + 2\big)(61)}\\ & = \dfrac{\big(101 + 36\sqrt {5}\big)\big(\sqrt 5 - 2\big)}{(61)\big(\sqrt 5 + 2\big)}\\ & = \dfrac{\big(101 + 36\sqrt {5}\big)\big(\sqrt 5 - 2\big)\big(\sqrt 5 - 2\big)}{(61)\big(\sqrt 5 + 2\big)\big(\sqrt 5 - 2\big)}\\ & = \dfrac{\big(101 + 36\sqrt {5}\big)\big(5 + 4 - 4\sqrt 5\big)}{(61)(5-4)}\\ & = \dfrac{\big(101 + 36\sqrt {5}\big)\big(9 - 4\sqrt 5\big)}{61}\\ & = \dfrac {189 - 80\sqrt 5}{61}.\ _\square \end{aligned}$$

Rationalize the denominator of $\dfrac {{\sqrt{\sqrt[3]{2}}}}{{\sqrt[4]{\sqrt{5}}}-{\sqrt[4]{\sqrt{3}}}}.$

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We have

$$\begin{aligned} \dfrac{{\sqrt{\sqrt[3]{2}}}}{{\sqrt[4]{\sqrt{5}}}-{\sqrt[4]{\sqrt{3}}}} & = \dfrac {\left(2^{\frac{1}{3}}\right)^{\frac{1}{2}}}{{\left(5^{\frac {1}{4}}\right)^{\frac{1}{2}}-\left(3^{\frac {1}{4}}\right)^{\frac{1}{2}}}}\\ & = \dfrac{2^{\frac{1}{6}}}{5^{\frac{1}{8}} - 3^{\frac{1}{8}}}\\ & = \dfrac{2^{\frac{1}{6}}\left(5^{\frac{1}{8}} + 3^{\frac{1}{8}}\right)}{\left(5^{\frac{1}{8}} - 3^{\frac{1}{8}}\right)\left(5^{\frac{1}{8}} + 3^{\frac{1}{8}}\right)}\\ & = \dfrac{2^{\frac{1}{6}}\left(5^{\frac{1}{8}} + 3^{\frac{1}{8}}\right)\left(5^{\frac{1}{4}} + 3^{\frac{1}{4}}\right)}{\left(5^{\frac{1}{4}} - 3^{\frac{1}{4}}\right) \left(5^{\frac{1}{4}} + 3^{\frac{1}{4}}\right)}\\ & = \dfrac{2^{\frac{1}{6}}\left(5^{\frac{1}{8}} + 3^{\frac{1}{8}}\right)\left(5^{\frac{1}{4}} + 3^{\frac{1}{4}}\right)}{\left(5^{\frac{1}{2}} - 3^{\frac{1}{2}}\right)}\\ & = \dfrac{2^{\frac{1}{6}}\left(5^{\frac{1}{8}} + 3^{\frac{1}{8}}\right)\left(5^{\frac{1}{4}} + 3^{\frac{1}{4}}\right) \left(5^{\frac{1}{2}} + 3^{\frac{1}{2}}\right)}{\left(5^{\frac{1}{2}} - 3^{\frac{1}{2}}\right)\left(5^{\frac{1}{2}} + 3^{\frac{1}{2}}\right)}\\ & = \dfrac{2^{\frac{1}{6}}\left(5^{\frac{1}{8}} + 3^{\frac{1}{8}}\right)\left(5^{\frac{1}{4}} + 3^{\frac{1}{4}}\right) \left(5^{\frac{1}{2}} + 3^{\frac{1}{2}}\right)}{5-3}\\ & = \dfrac{2^{\frac{1}{6}}\left(5^{\frac{1}{8}} + 3^{\frac{1}{8}}\right)\left(5^{\frac{1}{4}} + 3^{\frac{1}{4}}\right) \left(5^{\frac{1}{2}} + 3^{\frac{1}{2}}\right)}{2}.\ _\square \end{aligned}$$

Rationalize the denominator of $\dfrac{1}{\sqrt[5]{\sqrt[4]{\sqrt[3]{\sqrt{2}}}}}.$

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We have

$$\begin{aligned} \dfrac{1}{\sqrt[5]{\sqrt[4]{\sqrt[3]{\sqrt{2}}}}} & = \dfrac{1 × {\left(\sqrt[5]{\sqrt[4]{\sqrt[3]{\sqrt{2}}}}\right)}^{4}}{\sqrt[5]{\sqrt[4]{\sqrt[3]{\sqrt{2}}}} × {\left(\sqrt[5]{\sqrt[4]{\sqrt[3]{\sqrt{2}}}}\right)}^{4}}\\ & = \dfrac{\sqrt[5]{\sqrt[3]{\sqrt{2}}}}{\sqrt[4]{\sqrt[3]{\sqrt{2}}}}\\ & = \dfrac{\sqrt[5]{\sqrt[3]{\sqrt{2}}} × {\left(\sqrt[4]{\sqrt[3]{\sqrt{2}}}\right)}^{3}}{\sqrt[4]{\sqrt[3]{\sqrt{2}}} × {\left(\sqrt[4]{\sqrt[3]{\sqrt{2}}}\right)}^{3}}\\ & = \dfrac{\sqrt[5]{\sqrt[3]{\sqrt{2}}} × \sqrt[4]{\sqrt{2}}}{\sqrt[3]{\sqrt{2}}}\\ & = \dfrac{\sqrt[5]{\sqrt[3]{\sqrt{2}}} × \sqrt[4]{\sqrt{2}} × {\left(\sqrt[3]{\sqrt{2}}\right)}^{2}}{\sqrt[3]{\sqrt{2}} × {\left(\sqrt[3]{\sqrt{2}}\right)}^{2}}\\ & = \dfrac{\sqrt[5]{\sqrt[3]{\sqrt{2}}} × \sqrt[4]{\sqrt{2}} × \sqrt[3]{2}}{\sqrt{2}}\\ & = \dfrac{\sqrt[5]{\sqrt[3]{\sqrt{2}}} × \sqrt[4]{\sqrt{2}} × \sqrt[3]{2} × \sqrt{2}}{\sqrt{2} × \sqrt{2}}\\ & =\dfrac{\sqrt[5]{\sqrt[3]{\sqrt{2}}} × \sqrt[4]{\sqrt{2}} × \sqrt[3]{2} × \sqrt{2}}{2}.\ _\square \end{aligned}$$

• Radical Equations
• Simplifying Expressions