# How do I rationalize Cubic root?

Apr 2, 2015

Use the fact that $\sqrt{{a}^{3}} = a$
Together with $\sqrt{a} \sqrt{b} = \sqrt{a b}$
(And, of course, the fact that multiplying by one may change the way a number is written, but it does not change the (value of) the number.)
Example 1
Rationalize the denominator: $\frac{2}{\sqrt{5}}$

We'll use the facts mentioned above to write:

$\frac{2}{\sqrt{5}} = \frac{2}{\sqrt{5}} \cdot \frac{\sqrt{{5}^{2}}}{\sqrt{{5}^{2}}} = \frac{2 \sqrt{25}}{\sqrt{{5}^{3}}} = \frac{2 \sqrt{25}}{5}$

Example 2
Rationalize the denominator: $\frac{7}{\sqrt{4}}$ .
We could multiply by $\frac{\sqrt{{4}^{2}}}{\sqrt{{4}^{2}}}$, but $\sqrt{16}$ is reducible!

We'll take a more direct path to the solution if we Realize that what we have is:$\frac{7}{\sqrt{{2}^{2}}}$ so we only need to multiply by $\frac{\sqrt{2}}{\sqrt{2}}$,

$\frac{7}{\sqrt{4}} = \frac{7}{\sqrt{4}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{7 \sqrt{2}}{\sqrt{{2}^{3}}} = \frac{7 \sqrt{2}}{2}$

Example 3 (last)
If the denominator is $\sqrt{20}$, the similar path to rationalizing would be:

$\sqrt{20} = \sqrt{{2}^{2} \cdot 5}$, so we would multiply by

think . . . (what is missing to make a perfect cube?

think some more, you can get it. . .

.

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multiply numerator and denominator by $\sqrt{2 \cdot {5}^{2}}$.