Rationalizing Denominators Practice worksheet featuring eight problems to simplify expressions with radicals in the denominator.

Let’s solve each problem step by step. The goal is to rationalize the denominator — that means get rid of any square roots in the bottom (denominator) of the fraction. We do this by multiplying numerator and denominator by something that makes the denominator a perfect square (or just removes the radical).

We’ll go one by one.

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1. $\frac{5}{\sqrt{7}}$

Multiply top and bottom by $\sqrt{7}$:

$\frac{5}{\sqrt{7}} \cdot \frac{\sqrt{7}}{\sqrt{7}} = \frac{5\sqrt{7}}{7}$

✔ Done. Denominator is now rational.

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2. $\frac{6^2}{\sqrt{5}}$ → Wait, $6^2 = 36$, so it’s $\frac{36}{\sqrt{5}}$

Multiply top and bottom by $\sqrt{5}$:

$\frac{36}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{36\sqrt{5}}{5}$

✔ Done.

*(Note: The original says “6²” — we assume that’s 36, not 6 times sqrt(2) or anything else.)*

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3. $\frac{9x}{8\sqrt{x}}$

We can simplify before rationalizing. Notice $\frac{x}{\sqrt{x}} = \sqrt{x}$, because $x = (\sqrt{x})^2$, so $\frac{(\sqrt{x})^2}{\sqrt{x}} = \sqrt{x}$.

So:

$\frac{9x}{8\sqrt{x}} = \frac{9}{8} \cdot \frac{x}{\sqrt{x}} = \frac{9}{8} \cdot \sqrt{x} = \frac{9\sqrt{x}}{8}$

Alternatively, multiply numerator and denominator by $\sqrt{x}$:

$\frac{9x}{8\sqrt{x}} \cdot \frac{\sqrt{x}}{\sqrt{x}} = \frac{9x\sqrt{x}}{8x} = \frac{9\sqrt{x}}{8}$ (cancel x)

✔ Same answer.

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4. $\frac{y^2}{\sqrt{y}}$

Again, $\frac{y^2}{\sqrt{y}} = y^{2 - 0.5} = y^{1.5} = y \cdot \sqrt{y}$

Or: Multiply numerator and denominator by $\sqrt{y}$:

$\frac{y^2}{\sqrt{y}} \cdot \frac{\sqrt{y}}{\sqrt{y}} = \frac{y^2 \sqrt{y}}{y} = y \sqrt{y}$

✔ Answer: $y\sqrt{y}$

*(The note says “Write the trick as an odd for you won’t be penalized again.” — probably typo, ignore.)*

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5. $\frac{2}{\sqrt[3]{2}}$

This is cube root! So we need to make the denominator a perfect cube.

Denominator: $\sqrt[3]{2} = 2^{1/3}$

To make it $2^1 = 2$, we multiply by $2^{2/3} = \sqrt[3]{4}$

So:

$\frac{2}{\sqrt[3]{2}} \cdot \frac{\sqrt[3]{4}}{\sqrt[3]{4}} = \frac{2 \cdot \sqrt[3]{4}}{\sqrt[3]{8}} = \frac{2 \sqrt[3]{4}}{2} = \sqrt[3]{4}$

✔ Answer: $\sqrt[3]{4}$

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6. $\frac{4}{\sqrt[3]{3}}$

Same idea: multiply numerator and denominator by $\sqrt[3]{9}$ to make denominator $\sqrt[3]{27} = 3$

Because: $\sqrt[3]{3} \cdot \sqrt[3]{9} = \sqrt[3]{27} = 3$

So:

$\frac{4}{\sqrt[3]{3}} \cdot \frac{\sqrt[3]{9}}{\sqrt[3]{9}} = \frac{4 \sqrt[3]{9}}{3}$

✔ Answer: $\frac{4 \sqrt[3]{9}}{3}$

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7. $\frac{3a}{12\sqrt[3]{a^2}}$

First, simplify coefficients: $\frac{3}{12} = \frac{1}{4}$, so:

$\frac{a}{4 \sqrt[3]{a^2}}$

Now, to rationalize denominator: $\sqrt[3]{a^2} = a^{2/3}$, so we want to multiply by $a^{1/3} = \sqrt[3]{a}$ to make it $a^{1} = a$

So multiply numerator and denominator by $\sqrt[3]{a}$:

$\frac{a}{4 \sqrt[3]{a^2}} \cdot \frac{\sqrt[3]{a}}{\sqrt[3]{a}} = \frac{a \cdot \sqrt[3]{a}}{4 \cdot \sqrt[3]{a^3}} = \frac{a \sqrt[3]{a}}{4a} = \frac{\sqrt[3]{a}}{4}$

✔ Answer: $\frac{\sqrt[3]{a}}{4}$

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8. $\frac{c^2 \sqrt{c}}{c^2 \sqrt{c}}$

Wait — numerator and denominator are identical!

So this is just 1.

But let’s check: Is there a typo? Original says:

> 8. $\frac{c^2 \sqrt{c}}{c^2 \sqrt{c}}$

Yes — same on top and bottom. So unless c=0 (which would be undefined), this equals 1.

Even if we try to rationalize — no need, denominator already has no radical? Wait — actually, denominator has $\sqrt{c}$, but since numerator also has it, they cancel.

So:

$\frac{c^2 \sqrt{c}}{c^2 \sqrt{c}} = 1$

✔ Answer: 1

*(Unless the problem meant something else like $\frac{c^2}{\sqrt{c}}$ — but as written, it's 1.)*

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Final Answers:

1. $\frac{5\sqrt{7}}{7}$
2. $\frac{36\sqrt{5}}{5}$
3. $\frac{9\sqrt{x}}{8}$
4. $y\sqrt{y}$
5. $\sqrt[3]{4}$
6. $\frac{4 \sqrt[3]{9}}{3}$
7. $\frac{\sqrt[3]{a}}{4}$
8. $1$