Dividing radicals - Kuta Software - Free Printable

Let’s solve each problem one by one. We’ll simplify the radical expressions step by step.

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1) $\frac{\sqrt{15}}{5\sqrt{20}}$

First, simplify $\sqrt{20} = \sqrt{4 \cdot 5} = 2\sqrt{5}$

So denominator becomes: $5 \cdot 2\sqrt{5} = 10\sqrt{5}$

Now we have: $\frac{\sqrt{15}}{10\sqrt{5}}$

We can write $\sqrt{15} = \sqrt{3 \cdot 5} = \sqrt{3}\sqrt{5}$

So: $\frac{\sqrt{3}\sqrt{5}}{10\sqrt{5}} = \frac{\sqrt{3}}{10}$ (cancel $\sqrt{5}$)

✔ Final Answer for #1: $\boxed{\frac{\sqrt{3}}{10}}$

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2) $\frac{\sqrt{8}}{\sqrt{100}}$

$\sqrt{8} = \sqrt{4 \cdot 2} = 2\sqrt{2}$

$\sqrt{100} = 10$

So: $\frac{2\sqrt{2}}{10} = \frac{\sqrt{2}}{5}$

✔ Final Answer for #2: $\boxed{\frac{\sqrt{2}}{5}}$

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3) $\frac{\sqrt{6}}{\sqrt{27}}$

Simplify $\sqrt{27} = \sqrt{9 \cdot 3} = 3\sqrt{3}$

So: $\frac{\sqrt{6}}{3\sqrt{3}}$

Write $\sqrt{6} = \sqrt{2 \cdot 3} = \sqrt{2}\sqrt{3}$

Then: $\frac{\sqrt{2}\sqrt{3}}{3\sqrt{3}} = \frac{\sqrt{2}}{3}$

✔ Final Answer for #3: $\boxed{\frac{\sqrt{2}}{3}}$

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4) $\frac{3\sqrt{20}}{2\sqrt{4}}$

$\sqrt{20} = 2\sqrt{5}$, so numerator: $3 \cdot 2\sqrt{5} = 6\sqrt{5}$

$\sqrt{4} = 2$, so denominator: $2 \cdot 2 = 4$

So: $\frac{6\sqrt{5}}{4} = \frac{3\sqrt{5}}{2}$

✔ Final Answer for #4: $\boxed{\frac{3\sqrt{5}}{2}}$

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5) $\frac{4}{\sqrt{5}}$

Rationalize the denominator: multiply numerator and denominator by $\sqrt{5}$

$\frac{4}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{4\sqrt{5}}{5}$

✔ Final Answer for #5: $\boxed{\frac{4\sqrt{5}}{5}}$

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6) $\frac{\sqrt{4}}{5\sqrt{5}}$

$\sqrt{4} = 2$, so: $\frac{2}{5\sqrt{5}}$

Rationalize: multiply numerator and denominator by $\sqrt{5}$

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

✔ Final Answer for #6: $\boxed{\frac{2\sqrt{5}}{25}}$

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7) $\frac{\sqrt{5}}{\sqrt{3}}$

Rationalize: multiply numerator and denominator by $\sqrt{3}$

$\frac{\sqrt{5}}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{15}}{3}$

✔ Final Answer for #7: $\boxed{\frac{\sqrt{15}}{3}}$

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8) $\frac{\sqrt{2}}{2\sqrt{3}}$

Rationalize: multiply numerator and denominator by $\sqrt{3}$

$\frac{\sqrt{2}}{2\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{6}}{2 \cdot 3} = \frac{\sqrt{6}}{6}$

✔ Final Answer for #8: $\boxed{\frac{\sqrt{6}}{6}}$

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9) $\frac{\sqrt{3x^3y^3}}{4\sqrt{5xy^3}}$

Break down radicals:

Numerator: $\sqrt{3x^3y^3} = \sqrt{3} \cdot \sqrt{x^2 \cdot x} \cdot \sqrt{y^2 \cdot y} = \sqrt{3} \cdot x\sqrt{x} \cdot y\sqrt{y} = xy\sqrt{3xy}$

Denominator: $4\sqrt{5xy^3} = 4 \cdot \sqrt{5x \cdot y^2 \cdot y} = 4 \cdot y\sqrt{5xy}$

So expression becomes: $\frac{xy\sqrt{3xy}}{4y\sqrt{5xy}}$

Cancel $y$ and $\sqrt{xy}$ from top and bottom:

$\frac{x\sqrt{3}}{4\sqrt{5}}$

Now rationalize: multiply numerator and denominator by $\sqrt{5}$

$\frac{x\sqrt{3}}{4\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{x\sqrt{15}}{20}$

✔ Final Answer for #9: $\boxed{\frac{x\sqrt{15}}{20}}$

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10) $\frac{\sqrt{15xy}}{3\sqrt{10xy^3}}$

Break down denominator: $\sqrt{10xy^3} = \sqrt{10x \cdot y^2 \cdot y} = y\sqrt{10xy}$

So denominator: $3y\sqrt{10xy}$

Expression: $\frac{\sqrt{15xy}}{3y\sqrt{10xy}}$

Combine under one square root: $\frac{1}{3y} \cdot \sqrt{\frac{15xy}{10xy}} = \frac{1}{3y} \cdot \sqrt{\frac{15}{10}} = \frac{1}{3y} \cdot \sqrt{\frac{3}{2}}$

But better to keep as:

$\frac{\sqrt{15xy}}{3y\sqrt{10xy}} = \frac{1}{3y} \cdot \sqrt{\frac{15}{10}} = \frac{1}{3y} \cdot \sqrt{\frac{3}{2}}$

Rationalize $\sqrt{\frac{3}{2}} = \frac{\sqrt{6}}{2}$

So: $\frac{1}{3y} \cdot \frac{\sqrt{6}}{2} = \frac{\sqrt{6}}{6y}$

Wait — let me double-check that.

Actually, better approach:

$\frac{\sqrt{15xy}}{3y\sqrt{10xy}} = \frac{1}{3y} \cdot \sqrt{\frac{15xy}{10xy}} = \frac{1}{3y} \cdot \sqrt{\frac{3}{2}}$

Now rationalize $\sqrt{\frac{3}{2}} = \frac{\sqrt{6}}{2}$

So: $\frac{1}{3y} \cdot \frac{\sqrt{6}}{2} = \frac{\sqrt{6}}{6y}$

✔ Final Answer for #10: $\boxed{\frac{\sqrt{6}}{6y}}$

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11) $\frac{3 - 3\sqrt{3a}}{4\sqrt{8a}}$

First, factor numerator: $3(1 - \sqrt{3a})$

Denominator: $\sqrt{8a} = \sqrt{4 \cdot 2a} = 2\sqrt{2a}$, so denominator is $4 \cdot 2\sqrt{2a} = 8\sqrt{2a}$? Wait no:

Wait: denominator is $4\sqrt{8a}$, not $4 \times \sqrt{8a}$? Actually yes it is.

$\sqrt{8a} = 2\sqrt{2a}$, so $4 \cdot 2\sqrt{2a} = 8\sqrt{2a}$

So expression: $\frac{3(1 - \sqrt{3a})}{8\sqrt{2a}}$

This doesn’t simplify nicely unless we rationalize or see if terms cancel — but they don’t.

Alternatively, maybe leave as is? But usually we rationalize denominators.

So multiply numerator and denominator by $\sqrt{2a}$:

Numerator: $3(1 - \sqrt{3a}) \cdot \sqrt{2a} = 3[\sqrt{2a} - \sqrt{3a} \cdot \sqrt{2a}] = 3[\sqrt{2a} - \sqrt{6a^2}] = 3[\sqrt{2a} - a\sqrt{6}]$

Because $\sqrt{6a^2} = a\sqrt{6}$

Denominator: $8\sqrt{2a} \cdot \sqrt{2a} = 8 \cdot 2a = 16a$

So overall: $\frac{3(\sqrt{2a} - a\sqrt{6})}{16a}$

We can split: $\frac{3\sqrt{2a}}{16a} - \frac{3a\sqrt{6}}{16a} = \frac{3\sqrt{2a}}{16a} - \frac{3\sqrt{6}}{16}$

But perhaps better to leave factored.

Actually, original form might be acceptable, but since instruction is to simplify, and we rationalized, this is simplified.

But let me check if there's a better way.

Original: $\frac{3 - 3\sqrt{3a}}{4\sqrt{8a}} = \frac{3(1 - \sqrt{3a})}{4 \cdot 2\sqrt{2a}} = \frac{3(1 - \sqrt{3a})}{8\sqrt{2a}}$

After rationalizing: $\frac{3(1 - \sqrt{3a})\sqrt{2a}}{8 \cdot 2a} = \frac{3(1 - \sqrt{3a})\sqrt{2a}}{16a}$

Which expands to: $\frac{3\sqrt{2a} - 3\sqrt{6a^2}}{16a} = \frac{3\sqrt{2a} - 3a\sqrt{6}}{16a}$

We can factor 3: $\frac{3(\sqrt{2a} - a\sqrt{6})}{16a}$

That’s simplified.

✔ Final Answer for #11: $\boxed{\frac{3(\sqrt{2a} - a\sqrt{6})}{16a}}$

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12) $\frac{3n^2 + \sqrt{2n^2}}{\sqrt{10n}}$

First, simplify $\sqrt{2n^2} = n\sqrt{2}$ (assuming $n \geq 0$)

So numerator: $3n^2 + n\sqrt{2}$

Denominator: $\sqrt{10n}$

So expression: $\frac{3n^2 + n\sqrt{2}}{\sqrt{10n}}$

Factor numerator: $n(3n + \sqrt{2})$

So: $\frac{n(3n + \sqrt{2})}{\sqrt{10n}}$

We can write as: $n \cdot \frac{3n + \sqrt{2}}{\sqrt{10n}}$

Or split into two fractions:

$\frac{3n^2}{\sqrt{10n}} + \frac{n\sqrt{2}}{\sqrt{10n}}$

Simplify each term:

First term: $\frac{3n^2}{\sqrt{10n}} = 3n^2 \cdot (10n)^{-1/2} = 3n^{2 - 1/2} / \sqrt{10} = 3n^{3/2}/\sqrt{10}$ → messy.

Better to rationalize entire expression.

Multiply numerator and denominator by $\sqrt{10n}$:

Numerator: $(3n^2 + n\sqrt{2}) \cdot \sqrt{10n} = 3n^2\sqrt{10n} + n\sqrt{2} \cdot \sqrt{10n} = 3n^2\sqrt{10n} + n\sqrt{20n}$

Simplify $\sqrt{20n} = \sqrt{4 \cdot 5n} = 2\sqrt{5n}$

So numerator: $3n^2\sqrt{10n} + n \cdot 2\sqrt{5n} = 3n^2\sqrt{10n} + 2n\sqrt{5n}$

Denominator: $\sqrt{10n} \cdot \sqrt{10n} = 10n$

So overall: $\frac{3n^2\sqrt{10n} + 2n\sqrt{5n}}{10n}$

Factor numerator: $n(3n\sqrt{10n} + 2\sqrt{5n})$

Divide by $10n$: $\frac{3n\sqrt{10n} + 2\sqrt{5n}}{10}$

We can factor $\sqrt{n}$ out? Let’s see:

$\sqrt{10n} = \sqrt{10}\sqrt{n}$, $\sqrt{5n} = \sqrt{5}\sqrt{n}$

So: $\frac{3n \cdot \sqrt{10} \sqrt{n} + 2 \cdot \sqrt{5} \sqrt{n}}{10} = \frac{\sqrt{n}(3n\sqrt{10} + 2\sqrt{5})}{10}$

That’s simplified.

✔ Final Answer for #12: $\boxed{\frac{\sqrt{n}(3n\sqrt{10} + 2\sqrt{5})}{10}}$

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## Final Answers:

1) $\boxed{\frac{\sqrt{3}}{10}}$
2) $\boxed{\frac{\sqrt{2}}{5}}$
3) $\boxed{\frac{\sqrt{2}}{3}}$
4) $\boxed{\frac{3\sqrt{5}}{2}}$
5) $\boxed{\frac{4\sqrt{5}}{5}}$
6) $\boxed{\frac{2\sqrt{5}}{25}}$
7) $\boxed{\frac{\sqrt{15}}{3}}$
8) $\boxed{\frac{\sqrt{6}}{6}}$
9) $\boxed{\frac{x\sqrt{15}}{20}}$
10) $\boxed{\frac{\sqrt{6}}{6y}}$
11) $\boxed{\frac{3(\sqrt{2a} - a\sqrt{6})}{16a}}$
12) $\boxed{\frac{\sqrt{n}(3n\sqrt{10} + 2\sqrt{5})}{10}}$