Radicals and rationalizing the denominator

If the denominator is just $\sqrt{a}$, multiply the top and bottom by $\sqrt{a}$: $\frac{2}{\sqrt{7}}=\frac{2\sqrt{7}}{7}$.

If the denominator is $\sqrt{a}+b$ or $\sqrt{a}-b$, multiply by the expression with the middle sign flipped. The difference of squares removes the root.

The denominator is $(\sqrt{6})^2-1^2=5$, which you can get straight from the difference of squares.

Examples

1. 01Rationalize the denominator:$\displaystyle \frac{2}{\sqrt{6}-1}$AnswerHide

   Answer$\displaystyle \frac{2(\sqrt{6}+1)}{5}$

   Steps

   1. $\displaystyle \frac{2}{\sqrt{6}-1}$$\displaystyle = \frac{2(\sqrt{6}+1)}{(\sqrt{6}-1)(\sqrt{6}+1)}$$\displaystyle = \frac{2(\sqrt{6}+1)}{6-1}$$\displaystyle = \frac{2(\sqrt{6}+1)}{5}$
2. 02After rationalizing, what is the denominator?$\displaystyle \frac{2}{\sqrt{10}-1}$AnswerHide

   Answer$\displaystyle 9$

   Steps

   1. $\displaystyle (\sqrt{10}-1)(\sqrt{10}+1)$$\displaystyle = 10 - 1$$\displaystyle = 9$
3. 03Rationalize the numerator:$\displaystyle \frac{\sqrt{x+9}-3}{x}$AnswerHide

   Answer$\displaystyle \frac{1}{\sqrt{x+9}+3}$

   Steps

   1. $\displaystyle \frac{\sqrt{x+9}-3}{x} \cdot \frac{\sqrt{x+9}+3}{\sqrt{x+9}+3}$$\displaystyle = \frac{(x+9) - 9}{x(\sqrt{x+9}+3)}$$\displaystyle = \frac{x}{x(\sqrt{x+9}+3)}$$\displaystyle = \frac{1}{\sqrt{x+9}+3}$
4. 04To rationalize the denominator, what do you multiply the top and bottom by?$\displaystyle \frac{1}{\sqrt{3}+1}$AnswerHide

   Answer$\displaystyle \sqrt{3}-1$
5. 05Rationalize the denominator:$\displaystyle \frac{2}{\sqrt{7}}$AnswerHide

   Answer$\displaystyle \frac{2\sqrt{7}}{7}$

   Steps

   1. $\displaystyle \frac{2}{\sqrt{7}}$$\displaystyle = \frac{2\sqrt{7}}{\sqrt{7}\sqrt{7}}$$\displaystyle = \frac{2\sqrt{7}}{7}$
6. 06After rationalizing, what is the denominator?$\displaystyle \frac{1}{\sqrt{3}-1}$AnswerHide

   Answer$\displaystyle 2$

   Steps

   1. $\displaystyle (\sqrt{3}-1)(\sqrt{3}+1)$$\displaystyle = 3 - 1$$\displaystyle = 2$

FAQ

Q1Can I multiply $\frac{1}{\sqrt{3}+1}$ by $\sqrt{3}+1$?AnswerHide

A

Then the denominator becomes $(\sqrt{3}+1)^2=4+2\sqrt{3}$, and the root stays. Multiply by $\sqrt{3}-1$, with the sign flipped, and the denominator becomes $3-1=2$.

Q2When is rationalizing used in limits and derivatives?AnswerHide

A

For an expression like $\frac{\sqrt{x+4}-2}{x}$, which becomes $\frac{0}{0}$ as $x\to 0$, multiply the top and bottom by $\sqrt{x+4}+2$ to rationalize the numerator. It simplifies to $\frac{1}{\sqrt{x+4}+2}$, and the limit can be found.