Dividing radicals - Kuta Software - Free Printable

Let's solve each of these radical expression simplification problems step by step. We'll use the properties of radicals, including:

- $\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}$
- $\sqrt{ab} = \sqrt{a} \cdot \sqrt{b}$
- Rationalizing the denominator (if needed)
- Simplifying square roots by factoring perfect squares

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

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

So:
$$
\frac{\sqrt{15}}{5 \cdot 2\sqrt{5}} = \frac{\sqrt{15}}{10\sqrt{5}}
$$

Now rationalize or simplify:
$$
= \frac{\sqrt{15}}{10\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{\sqrt{75}}{10 \cdot 5} = \frac{\sqrt{75}}{50}
$$

But $\sqrt{75} = \sqrt{25 \cdot 3} = 5\sqrt{3}$, so:
$$
\frac{5\sqrt{3}}{50} = \frac{\sqrt{3}}{10}
$$

✔ Answer: $\boxed{\frac{\sqrt{3}}{10}}$

---

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

$\sqrt{8} = \sqrt{4 \cdot 2} = 2\sqrt{2}$, and $\sqrt{100} = 10$

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

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

---

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

We can write this as:
$$
\sqrt{\frac{6}{27}} = \sqrt{\frac{2}{9}} = \frac{\sqrt{2}}{3}
$$

Alternatively:
$\sqrt{27} = \sqrt{9 \cdot 3} = 3\sqrt{3}$, so:
$$
\frac{\sqrt{6}}{3\sqrt{3}} = \frac{\sqrt{6}}{3\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{18}}{9} = \frac{3\sqrt{2}}{9} = \frac{\sqrt{2}}{3}
$$

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

---

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

$\sqrt{20} = 2\sqrt{5}$, $\sqrt{4} = 2$

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

✔ Answer: $\boxed{\frac{3\sqrt{5}}{2}}$

---

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

Rationalize the denominator:
$$
\frac{4}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{4\sqrt{5}}{5}
$$

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

---

6) $\frac{\sqrt{4}}{5\sqrt{9}}$

$\sqrt{4} = 2$, $\sqrt{9} = 3$

So:
$$
\frac{2}{5 \cdot 3} = \frac{2}{15}
$$

✔ Answer: $\boxed{\frac{2}{15}}$

---

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

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

✔ Answer: $\boxed{\frac{\sqrt{15}}{3}}$

---

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

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

✔ Answer: $\boxed{\frac{\sqrt{6}}{6}}$

---

9) $\frac{\sqrt{3x^2y^3}}{4\sqrt{5xy^5}}$

Break down both radicals:

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

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

So:
$$
\frac{x y \sqrt{3y}}{4 y^2 \sqrt{5xy}} = \frac{x \sqrt{3y}}{4 y \sqrt{5xy}}
$$

Now combine:
$$
= \frac{x}{4y} \cdot \frac{\sqrt{3y}}{\sqrt{5xy}} = \frac{x}{4y} \cdot \sqrt{\frac{3y}{5xy}} = \frac{x}{4y} \cdot \sqrt{\frac{3}{5x}}
$$

Now simplify:
$$
= \frac{x}{4y} \cdot \frac{\sqrt{3}}{\sqrt{5x}} = \frac{\sqrt{3}}{4y} \cdot \frac{x}{\sqrt{5x}} = \frac{\sqrt{3}}{4y} \cdot \sqrt{\frac{x}{5x}} = \frac{\sqrt{3}}{4y} \cdot \sqrt{\frac{1}{5}} = \frac{\sqrt{3}}{4y\sqrt{5}}
$$

Now rationalize:
$$
= \frac{\sqrt{3}}{4y\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{\sqrt{15}}{4y \cdot 5} = \frac{\sqrt{15}}{20y}
$$

✔ Answer: $\boxed{\frac{\sqrt{15}}{20y}}$

---

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

Simplify numerator and denominator:

Numerator: $\sqrt{15xy}$

Denominator: $\sqrt{10xy^3} = \sqrt{10xy \cdot y^2} = y \sqrt{10xy}$

So:
$$
\frac{\sqrt{15xy}}{3 y \sqrt{10xy}} = \frac{1}{3y} \cdot \frac{\sqrt{15xy}}{\sqrt{10xy}} = \frac{1}{3y} \cdot \sqrt{\frac{15xy}{10xy}} = \frac{1}{3y} \cdot \sqrt{\frac{3}{2}} = \frac{1}{3y} \cdot \frac{\sqrt{6}}{2}
$$

Wait — let’s do it carefully.

$$
\sqrt{\frac{15}{10}} = \sqrt{\frac{3}{2}} = \frac{\sqrt{6}}{2}
$$

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

✔ Answer: $\boxed{\frac{\sqrt{6}}{6y}}$

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

First simplify denominator:
$\sqrt{8a} = \sqrt{4 \cdot 2a} = 2\sqrt{2a}$

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

Now rationalize denominator:
Multiply numerator and denominator by $\sqrt{2a}$:

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

Wait — better to keep it symbolic:

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

But maybe we should just leave it in simplified form before rationalizing.

Alternatively, factor numerator and see if anything cancels.

But there is no common factor. So best approach:

Leave as:
$$
\frac{3(1 - \sqrt{3a})}{8\sqrt{2a}}
$$

Then rationalize:
Multiply numerator and denominator by $\sqrt{2a}$:

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

Denominator: $8 \cdot 2a = 16a$

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

This is messy. Let's check if original expression can be simplified differently.

Wait — perhaps the problem expects us to factor and simplify before rationalizing.

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

Now, unless further simplification is possible, we rationalize:

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

Now distribute:
$$
= \frac{3\sqrt{2a} - 3\sqrt{3a} \cdot \sqrt{2a}}{16a} = \frac{3\sqrt{2a} - 3\sqrt{6a^2}}{16a}
$$

But $\sqrt{6a^2} = a\sqrt{6}$, so:
$$
= \frac{3\sqrt{2a} - 3a\sqrt{6}}{16a} = \frac{3}{16a}\left(\sqrt{2a} - a\sqrt{6}\right)
$$

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

Simplify first term:
$$
\frac{3\sqrt{2a}}{16a} = \frac{3}{16} \cdot \frac{\sqrt{2a}}{a} = \frac{3}{16} \cdot \sqrt{\frac{2}{a}}
$$

So final answer:
$$
\frac{3}{16} \sqrt{\frac{2}{a}} - \frac{3\sqrt{6}}{16}
$$

But this seems overly complicated. Perhaps the intended simplification was just:

$$
\frac{3(1 - \sqrt{3a})}{8\sqrt{2a}} \quad \text{(not fully simplified)}
$$

But likely, they want rationalized form:

✔ Answer: $\boxed{\frac{3\sqrt{2a} - 3a\sqrt{6}}{16a}}$

Or better: $\boxed{\frac{3}{16a}(\sqrt{2a} - a\sqrt{6})}$

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

First simplify numerator:
$\sqrt{2n^2} = \sqrt{2} \cdot \sqrt{n^2} = n\sqrt{2}$

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

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

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

Now write $n = \sqrt{n^2}$, so:
$$
= \frac{\sqrt{n^2}(3n + \sqrt{2})}{\sqrt{10n}} = (3n + \sqrt{2}) \cdot \frac{\sqrt{n^2}}{\sqrt{10n}} = (3n + \sqrt{2}) \cdot \sqrt{\frac{n^2}{10n}} = (3n + \sqrt{2}) \cdot \sqrt{\frac{n}{10}}
$$

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

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

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

✔ Answer: $\boxed{\frac{(3n + \sqrt{2})\sqrt{10n}}{10}}$

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✔ Final Answers Summary:

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

Let me know if you'd like any explanation expanded!