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Dividing Radical Expressions and Rationalizing Denominators

Fact of Life 1: Sometimes we need to divide radical expressions.

Fact of Life 2: Sometimes we need to put the answer in a particular form.





























Two Useful Properties (Big Facts!) of Radicals $$\sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b}$$ $$\sqrt[n]{\frac{a}{b}}=\frac{\sqrt[n]{a}}{\sqrt[n]{b}}$$





























Rationalizing Denominators

Rationalizing denominators is simply rewriting a radical expression without radicals in the denominator.

Example: $\displaystyle \frac{3}{\sqrt{2}}$

$$ \begin{array}{lll} \displaystyle \frac{3}{\sqrt{2}}&\displaystyle=\frac{3}{\sqrt{2}}\cdot \color{magenta}{1} &\mbox{multiplying by 1 doesn't change the value}\\ \displaystyle &\displaystyle=\frac{3}{\sqrt{2}}\cdot \color{magenta}{\frac{\sqrt{2}}{\sqrt{2}}} &\mbox{$\frac{\sqrt{2}}{\sqrt{2}}$ is just a fancy way to say 1}\\ \displaystyle &\displaystyle=\frac{3\sqrt{2}}{\sqrt{2}\sqrt{2}}&\mbox{}\\ \displaystyle &\displaystyle=\frac{3\sqrt{2}}{2}&\mbox{since $\sqrt{2}\sqrt{2}=2$}\\ \end{array} $$



Example: $\displaystyle \frac{\sqrt{14}}{\sqrt{7}}$

Ninja Stealth Method

$$ \begin{array}{lll} \displaystyle \frac{\sqrt{14}}{\sqrt{7}}&\displaystyle=\sqrt{\frac{14}{7}} &\mbox{}\\ \displaystyle &\displaystyle=\sqrt{2} &\mbox{}\\ \end{array} $$ Fancy One Method

$$ \begin{array}{lll} \displaystyle \frac{\sqrt{14}}{\sqrt{7}}&\displaystyle=\frac{\sqrt{14}}{\sqrt{7}}\cdot \color{magenta}{\frac{\sqrt{7}}{\sqrt{7}}} &\mbox{}\\ \displaystyle &\displaystyle=\frac{\sqrt{14}\sqrt{7}}{\sqrt{7}\sqrt{7}} &\mbox{}\\ \displaystyle &\displaystyle=\frac{\sqrt{14 \cdot 7}}{7} &\mbox{}\\ \displaystyle &\displaystyle=\frac{\sqrt{2\cdot 7 \cdot 7}}{7} &\mbox{}\\ \displaystyle &\displaystyle=\frac{7\sqrt{2}}{7} &\mbox{}\\ \displaystyle &\displaystyle=\sqrt{2} &\mbox{simplify}\\ \end{array} $$



Example: $\displaystyle \frac{15}{\sqrt{5 y}}$

$$ \begin{array}{lll} \displaystyle \frac{15}{\sqrt{5 y}}&\displaystyle=\frac{15}{\sqrt{5 y}}\cdot \color{magenta}{\frac{\sqrt{5y}}{\sqrt{5y}}} &\mbox{fancy one!}\\ \displaystyle &\displaystyle=\frac{15\sqrt{5 y}}{\sqrt{5 y}\sqrt{5 y}} &\mbox{}\\ \displaystyle &\displaystyle=\frac{15\sqrt{5 y}}{5y} &\mbox{}\\ \displaystyle &\displaystyle=\frac{3\sqrt{5 y}}{y} &\mbox{simplify}\\ \end{array} $$



Example: $\displaystyle \frac{\sqrt[3]{9}}{\sqrt[3]{16}}$

$$ \begin{array}{lll} \displaystyle \frac{\sqrt[3]{9}}{\sqrt[3]{16}}&\displaystyle=\frac{\sqrt[3]{9}}{\sqrt[3]{\underline{ 2 \cdot 2 \cdot 2}\cdot 2}} &\mbox{}\\ \displaystyle &\displaystyle=\frac{\sqrt[3]{9}}{2\sqrt[3]{2}} &\mbox{simplify radical}\\ \displaystyle &\displaystyle=\frac{\sqrt[3]{9}}{2\sqrt[3]{2}}\cdot \color{magenta}{\frac{\sqrt[3]{2}}{\sqrt[3]{2}}}\cdot \color{magenta}{\frac{\sqrt[3]{2}}{\sqrt[3]{2}}} &\mbox{}\\ \displaystyle &\displaystyle=\frac{\sqrt[3]{9}\sqrt[3]{2}\sqrt[3]{2}}{2\sqrt[3]{2}\sqrt[3]{2}\sqrt[3]{2}} &\mbox{}\\ \displaystyle &\displaystyle=\frac{\sqrt[3]{9\cdot 2\cdot 2}}{2\cdot 2} &\mbox{}\\ \displaystyle &\displaystyle=\frac{\sqrt[3]{36}}{4} &\mbox{}\\ \end{array} $$ Brute Force Alternative $$ \begin{array}{lll} \displaystyle \frac{\sqrt[3]{9}}{\sqrt[3]{16}}&\displaystyle=\frac{\sqrt[3]{9}}{\sqrt[3]{16}}\cdot \color{magenta}{\frac{\sqrt[3]{16}}{\sqrt[3]{16}}}\cdot \color{magenta}{\frac{\sqrt[3]{16}}{\sqrt[3]{16}}} &\mbox{}\\ \displaystyle &\displaystyle=\frac{\sqrt[3]{9}\sqrt[3]{16}\sqrt[3]{16}}{\sqrt[3]{16}\sqrt[3]{16}\sqrt[3]{16}} &\mbox{}\\ \displaystyle &\displaystyle=\frac{\sqrt[3]{9\cdot 16\cdot 16}}{16} &\mbox{}\\ \displaystyle &\displaystyle=\frac{\sqrt[3]{9 \underline{\cdot 2 \cdot 2 \cdot 2} \cdot \underline{2 \cdot 2 \cdot 2} \cdot 2 \cdot 2}}{16} &\mbox{}\\ \displaystyle &\displaystyle=\frac{2\cdot 2\sqrt[3]{9 \cdot 2 \cdot 2}}{16} &\mbox{}\\ \displaystyle &\displaystyle=\frac{4\sqrt[3]{36}}{16} &\mbox{}\\ \displaystyle &\displaystyle=\frac{\sqrt[3]{36}}{4} &\mbox{}\\ \end{array} $$































Process: To rationalize a denominator, we multiply the top and bottom by the same quantity (a fancy one!) which gets rid of the radical in the bottom.

Sometimes, other shenanigans work too.





























More Examples



Example: $\displaystyle \frac{\sqrt[3]{12}}{\sqrt[3]{36 n}}$

Ninja Stealth Method

$$ \begin{array}{lll} \displaystyle \frac{\sqrt[3]{12}}{\sqrt[3]{36 n}}&=\displaystyle \sqrt[3]{\frac{12}{36 n}} &\mbox{ninja stealth move}\\ &=\displaystyle \sqrt[3]{\frac{1}{3 n}} &\mbox{simplify fraction}\\ &=\displaystyle \frac{\sqrt[3]{1}}{\sqrt[3]{3 n}} &\mbox{re-express as quotient of radicals}\\ &=\displaystyle \frac{1}{\sqrt[3]{3 n}} &\mbox{simplify}\\ &=\displaystyle \frac{1}{\sqrt[3]{3 n}}\cdot \color{magenta}{\frac{\sqrt[3]{3 n}}{\sqrt[3]{3 n}}}\cdot \color{magenta}{\frac{\sqrt[3]{3 n}}{\sqrt[3]{3 n}}} &\mbox{}\\ &=\displaystyle \frac{\sqrt[3]{3 n}\sqrt[3]{3 n}}{3 n} &\mbox{}\\ &=\displaystyle \frac{\sqrt[3]{3 n\cdot 3 n}}{3 n} &\mbox{}\\ &=\displaystyle \frac{\sqrt[3]{9n^2}}{3 n} &\mbox{}\\ \end{array} $$ Brute Force Method

$$ \begin{array}{lll} \displaystyle \frac{\sqrt[3]{12}}{\sqrt[3]{36 n}}&\displaystyle=\frac{\sqrt[3]{12}}{\sqrt[3]{36 n}}\cdot \color{magenta}{\frac{\sqrt[3]{36 n}}{\sqrt[3]{36 n}}}\cdot \color{magenta}{\frac{\sqrt[3]{36 n}}{\sqrt[3]{36 n}}} &\mbox{}\\ \displaystyle &\displaystyle=\frac{\sqrt[3]{12}\sqrt[3]{36 n}\sqrt[3]{36 n}}{\sqrt[3]{36 n}\sqrt[3]{36 n}\sqrt[3]{36 n}} &\mbox{}\\ \displaystyle &\displaystyle=\frac{\sqrt[3]{12\cdot 36 n\cdot 36 n}}{36 n} &\mbox{}\\ \displaystyle &\displaystyle=\frac{\sqrt[3]{2\cdot 2 \cdot 3 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot n^2}}{36 n} &\mbox{}\\ \displaystyle &\displaystyle=\frac{\sqrt[3]{\underline{2\cdot 2 \cdot 2} \cdot \underline{2 \cdot 2 \cdot 2} \cdot \underline{3 \cdot 3 \cdot 3} \cdot 3 \cdot 3\cdot n^2}}{36 n} &\mbox{}\\ \displaystyle &\displaystyle=\frac{2\cdot 2 \cdot 3 \sqrt[3]{ 3 \cdot 3\cdot n^2}}{36 n} &\mbox{}\\ \displaystyle &\displaystyle=\frac{12 \sqrt[3]{ 9 n^2}}{36 n} &\mbox{}\\ \displaystyle &\displaystyle=\frac{\sqrt[3]{ 9 n^2}}{3 n} &\mbox{}\\ \end{array} $$



Bonus Example (if time permits): $\displaystyle \frac{\sqrt[4]{55s^{7}}}{\sqrt[4]{12}}$

$$ \begin{array}{lll} \displaystyle \frac{\sqrt[4]{55s^{7}}}{\sqrt[4]{12}}&\displaystyle=\frac{\sqrt[4]{55s^{7}}}{\sqrt[4]{12}}\cdot \color{magenta}{\frac{\sqrt[4]{12}}{\sqrt[4]{12}}}\cdot \color{magenta}{\frac{\sqrt[4]{12}}{\sqrt[4]{12}}}\cdot \color{magenta}{\frac{\sqrt[4]{12}}{\sqrt[4]{12}}}&\mbox{}\\ \displaystyle &\displaystyle=\frac{\sqrt[4]{55s^{7}}\sqrt[4]{12}\sqrt[4]{12}\sqrt[4]{12}}{\sqrt[4]{12}\sqrt[4]{12}\sqrt[4]{12}\sqrt[4]{12}} &\mbox{}\\ \displaystyle &\displaystyle=\frac{\sqrt[4]{55s^{7}\cdot 12\cdot 12\cdot 12}}{12} &\mbox{}\\ \displaystyle &\displaystyle=\frac{\sqrt[4]{55 \cdot s \cdot s \cdot s \cdot s \cdot s \cdot s \cdot s\cdot 2 \cdot 2 \cdot 3 \cdot 2 \cdot 2 \cdot 3\cdot 2 \cdot 2 \cdot 3}}{12} &\mbox{}\\ \displaystyle &\displaystyle=\frac{\sqrt[4]{55 \cdot \underline{2 \cdot 2 \cdot 2 \cdot 2} \cdot 2 \cdot 2 \cdot 3 \cdot 3\cdot 3 \cdot \underline{s \cdot s \cdot s \cdot s} \cdot s \cdot s \cdot s}}{12} &\mbox{}\\ \displaystyle &\displaystyle=\frac{2s\sqrt[4]{55 \cdot 2 \cdot 2 \cdot 3 \cdot 3\cdot 3 \cdot s \cdot s \cdot s}}{12} &\mbox{}\\ \displaystyle &\displaystyle=\frac{s\sqrt[4]{5940 s^3}}{6} &\mbox{}\\ \end{array} $$

































Rationalizing More Complicated Denominators

Fact of Life: sometimes denominators are a bit more complicated.

Example: Simplfy. $\displaystyle \frac{2}{5+\sqrt{7}}$

$$ \begin{array}{lll} \displaystyle \frac{2}{5+\sqrt{7}}&\displaystyle=\frac{2}{5+\sqrt{7}} \cdot \color{magenta}{\frac{5-\sqrt{7}}{5-\sqrt{7}}}&\mbox{a special fancy one}\\ \displaystyle &\displaystyle=\frac{2(5-\sqrt{7})}{(5+\sqrt{7})(5-\sqrt{7})} &\mbox{}\\ \displaystyle &\displaystyle=\frac{10-2\sqrt{7}}{25-5\sqrt{7}+5\sqrt{7}-7} &\mbox{FOIL out denominator}\\ \displaystyle &\displaystyle=\frac{10-2\sqrt{7}}{18} &\mbox{}\\ \displaystyle &\displaystyle=\frac{2(5-\sqrt{7})}{2\cdot 9} &\mbox{}\\ \displaystyle &\displaystyle=\frac{5-\sqrt{7}}{9} &\mbox{}\\ \end{array} $$



Conjugates: The conjugate of $5+\sqrt{7}$ is $5-\sqrt{7}.$ To find a conjugate, switch the middle sign.

Fact: to rationalize a denominator like the above, multiply top and bottom by the conjugate of the denominator.





























More Examples: Rationalize the denominators.

$\displaystyle \frac{3}{\sqrt{7}-\sqrt{2}}$

$$ \begin{array}{lll} \displaystyle \frac{3}{\sqrt{7}-\sqrt{2}}&\displaystyle=\frac{3}{\sqrt{7}-\sqrt{2}} \cdot \color{magenta}{\frac{\sqrt{7}+\sqrt{2}}{\sqrt{7}+\sqrt{2}}}&\mbox{conjugate fancy one}\\ \displaystyle &\displaystyle=\frac{3(\sqrt{7}+\sqrt{2})}{(\sqrt{7}-\sqrt{2})(\sqrt{7}+\sqrt{2})} &\mbox{}\\ \displaystyle &\displaystyle=\frac{3\sqrt{7}+3\sqrt{2}}{7+\sqrt{7}\sqrt{2}-\sqrt{2}\sqrt{7}-2} &\mbox{}\\ \displaystyle &\displaystyle=\frac{3\sqrt{7}+3\sqrt{2}}{7-2} &\mbox{}\\ \displaystyle &\displaystyle=\frac{3\sqrt{7}+3\sqrt{2}}{5} &\mbox{}\\ \end{array} $$



$\displaystyle \frac{\sqrt{p}}{\sqrt{q}-\sqrt{p}}$

$$ \begin{array}{lll} \displaystyle \frac{\sqrt{p}}{\sqrt{q}-\sqrt{p}}&\displaystyle=\frac{\sqrt{p}}{\sqrt{q}-\sqrt{p}} \cdot \color{magenta}{\frac{\sqrt{q}+\sqrt{p}}{\sqrt{q}+\sqrt{p}}}&\mbox{a conjugate fancy one}\\ \displaystyle &\displaystyle=\frac{\sqrt{p}(\sqrt{q}+\sqrt{p})}{(\sqrt{q}-\sqrt{p})(\sqrt{q}+\sqrt{p})} &\mbox{}\\ \displaystyle &\displaystyle=\frac{\sqrt{p}\sqrt{q}+\sqrt{p}\sqrt{p}}{q+\sqrt{q}\sqrt{p}-\sqrt{p}\sqrt{q}-p} &\mbox{}\\ \displaystyle &\displaystyle=\frac{\sqrt{pq}+p}{q-p} &\mbox{}\\ \end{array} $$



$\displaystyle \frac{11}{\sqrt{2 u}+\sqrt{7 j}}$

$$ \begin{array}{lll} \displaystyle \frac{11}{\sqrt{2 u}+\sqrt{7 j}}&\displaystyle=\frac{11}{\sqrt{2 u}+\sqrt{7 j}}\cdot \color{magenta}{\frac{\sqrt{2 u}-\sqrt{7 j}}{\sqrt{2 u}-\sqrt{7 j}}}&\mbox{}\\ \displaystyle &\displaystyle=\frac{11(\sqrt{2 u}-\sqrt{7 j})}{(\sqrt{2 u}+\sqrt{7 j})(\sqrt{2 u}-\sqrt{7 j})} &\mbox{}\\ \displaystyle &\displaystyle=\frac{11\sqrt{2 u}-11\sqrt{7 j}}{2 u-\sqrt{2 u}\sqrt{7j} +\sqrt{7j} \sqrt{2 u}-7 j} &\mbox{}\\ \displaystyle &\displaystyle=\frac{11\sqrt{2 u}-11\sqrt{7 j}}{2 u-7 j} &\mbox{}\\ \end{array} $$



$\displaystyle \frac{\sqrt{q}-\sqrt{10}}{\sqrt{q}+\sqrt{10}}$

$$ \begin{array}{lll} \displaystyle \frac{\sqrt{q}-\sqrt{10}}{\sqrt{q}+\sqrt{10}}&\displaystyle=\frac{\sqrt{q}-\sqrt{10}}{\sqrt{q}+\sqrt{10}}\cdot \color{magenta}{\frac{\sqrt{q}-\sqrt{10}}{\sqrt{q}-\sqrt{10}}}&\mbox{}\\ \displaystyle &\displaystyle=\frac{(\sqrt{q}-\sqrt{10})(\sqrt{q}-\sqrt{10})}{(\sqrt{q}+\sqrt{10})(\sqrt{q}-\sqrt{10})} &\mbox{}\\ \displaystyle &\displaystyle=\frac{q-\sqrt{q}\sqrt{10}-\sqrt{10}\sqrt{q}+10}{q-\sqrt{q}\sqrt{10}+\sqrt{10}\sqrt{q}-10} &\mbox{}\\ \displaystyle &\displaystyle=\frac{q-2\sqrt{10q}+10}{q-10} &\mbox{}\\ \end{array} $$