The length of the two legs of a right triangle are \begin{align*}2\sqrt{5}\end{align*} and \begin{align*}3\sqrt{4}\end{align*}. What is the length of the triangle's hypotenuse?

### Simplifying Square Roots

Before we can solve a quadratic equation using square roots, we need to review how to simplify, add, subtract, and multiply them. Recall that the **square root** is a number that, when multiplied by itself, produces another number. 4 is the square root of 16, for example. -4 is also the square root of 16 because \begin{align*}(-4)^2 = 16\end{align*}. The symbol for square root is the **radical** sign, or \begin{align*}\sqrt{}\end{align*}. The number under the radical is called the **radicand**. If the square root of an integer is not another integer, it is an irrational number.

Let's find \begin{align*}\sqrt{50}\end{align*}.

- using a calculator.

To plug the square root into your graphing calculator, typically there is a \begin{align*}\sqrt{}\end{align*} or SQRT button. Depending on your model, you may have to enter 50 before or after the square root button. Either way, your answer should be \begin{align*}\sqrt{50}=7.071067811865 \ldots\end{align*} In general, we will round to the hundredths place, so 7.07 is sufficient.

- by simplifying the square root.

To simplify the square root, the square numbers must be “pulled out.” Look for factors of 50 that are square numbers: 4, 9, 16, 25... 25 is a factor of 50, so break the factors apart.

\begin{align*}\sqrt{50}=\sqrt{25 \cdot 2} = \sqrt{25} \cdot \sqrt{2} = 5 \sqrt{2}\end{align*}. This is the most accurate answer.

**Radical Rules**

1. \begin{align*}\sqrt{ab} = \sqrt{a \cdot b} = \sqrt{a} \cdot \sqrt{b}\end{align*} Any two radicals can be multiplied together.

2. \begin{align*}x\sqrt{a} \pm y \sqrt{a} = x \pm y \sqrt{a}\end{align*} The radicands must be the same in order to add or subtract.

3. \begin{align*}\left(\sqrt{a}\right)^2 = \sqrt{a}^2 = a\end{align*} The square and square root cancel each other out.

Now, let's simplify the following expressions.

- \begin{align*}\sqrt{45}+\sqrt{80}-2\sqrt{5}\end{align*}

At first glance, it does not look like we can simplify this. But, we can simplify each radical by pulling out the perfect squares.

\begin{align*}\sqrt{45} &= \sqrt{9 \cdot 5} = 3 \sqrt{5}\\ \sqrt{80} &= \sqrt{16 \cdot 5} = 4 \sqrt{5}\end{align*}

Rewriting our expression, we have: \begin{align*}3\sqrt{5}+4\sqrt{5}-2\sqrt{5}\end{align*} and all the radicands are the same. Using the Order of Operations, our answer is \begin{align*}5 \sqrt{5}\end{align*}.

- \begin{align*}2\sqrt{35} \cdot 4 \sqrt{7}\end{align*}

Multiply across.

\begin{align*}2\sqrt{35} \cdot 4 \sqrt{7} = 2 \cdot 4 \sqrt{35 \cdot 7} = 8 \sqrt{245}\end{align*}

Now, simplify the radical. \begin{align*}8\sqrt{245} = 8 \sqrt{49 \cdot 5} = 8 \cdot 7 \sqrt{5} = 56\sqrt{5}\end{align*}

### Examples

#### Example 1

Earlier, you were asked to find the length of the triangle's hypotenuse.

We must use the Pythagorean Theorem, which states that the square of one leg of a right triangle plus the square of the other leg equals the square of the hypotenuse.

So we are looking for *c* such that \begin{align*}(2\sqrt{5})^2 + (3\sqrt{4})^2 = c^2\end{align*}.

Simplifying, we get \begin{align*}4 \cdot 5 + 9 \cdot 4 = c^2\end{align*}, or \begin{align*}20 + 36 = c^2\end{align*}.

Therefore, \begin{align*}c^2 = 56\end{align*}, so to find *c*, we must take the square root of 56.

\begin{align*}\sqrt{56} = \sqrt{4 \cdot 14} = 2\sqrt{14}\end{align*}.

Therefore, \begin{align*}c = 2\sqrt{14}\end{align*}.

**Simplify the following radicals.**

#### Example 2

\begin{align*}\sqrt{150}\end{align*}

Pull out all the square numbers.

\begin{align*}\sqrt{150}=\sqrt{25 \cdot 6}= 5 \sqrt{6}\end{align*}

Alternate Method: Write out the prime factorization of 150.

\begin{align*}\sqrt{150} = \sqrt{2 \cdot 3 \cdot 5 \cdot 5}\end{align*}

Now, pull out any number that has a pair. Write it *once* in front of the radical and multiply together what is left over under the radical.

\begin{align*}\sqrt{150} = \sqrt{2 \cdot 3 \cdot {\color{red}5 \cdot 5}}={\color{red}5}\sqrt{6}\end{align*}

#### Example 3

\begin{align*}2\sqrt{3}-\sqrt{6}+\sqrt{96}\end{align*}

Simplify \begin{align*}\sqrt{96}\end{align*} to see if anything can be combined. We will use the alternate method above.

\begin{align*}\sqrt{96}= \sqrt{{\color{red}2 \cdot 2} \cdot {\color{blue}2 \cdot 2} \cdot 2 \cdot 3} = {\color{red}2} \cdot {\color{blue}2} \sqrt{6} = 4 \sqrt{6}\end{align*}

Rewrite the expression: \begin{align*}2 \sqrt{3} - \sqrt{6} + 4\sqrt{6} = 2\sqrt{3}+3\sqrt{6}\end{align*}. This is fully simplified. \begin{align*}\sqrt{3}\end{align*} and \begin{align*}\sqrt{6}\end{align*} cannot be combined because they do not have the same value under the radical.

#### Example 4

\begin{align*}\sqrt{8} \cdot \sqrt{20}\end{align*}

This problem can be done two different ways.

**First Method**: Multiply radicals, then simplify the answer.

\begin{align*}\sqrt{8} \cdot \sqrt{20} = \sqrt{160} = \sqrt{16 \cdot 10} = 4\sqrt{10}\end{align*}

**Second Method**: Simplify radicals, then multiply.

\begin{align*}\sqrt{8} \cdot \sqrt{20} = \left(\sqrt{4 \cdot 2}\right) \cdot \left(\sqrt{4 \cdot 5}\right)= 2 \sqrt{2} \cdot 2 \sqrt{5}=2 \cdot 2 \sqrt{2 \cdot 5} = 4 \sqrt{10}\end{align*}

Depending on the complexity of the problem, either method will work. Pick whichever method you prefer.

### Review

Find the square root of each number by using the calculator. Round your answer to the nearest hundredth.

- 56
- 12
- 92

Simplify the following radicals. If it cannot be simplified further, write "*cannot be simplified*".

- \begin{align*}\sqrt{18}\end{align*}
- \begin{align*}\sqrt{75}\end{align*}
- \begin{align*}\sqrt{605}\end{align*}
- \begin{align*}\sqrt{48}\end{align*}
- \begin{align*}\sqrt{50} \cdot \sqrt{2}\end{align*}
- \begin{align*}4\sqrt{3} \cdot \sqrt{21}\end{align*}
- \begin{align*}\sqrt{6} \cdot \sqrt{20}\end{align*}
- \begin{align*}\left(4\sqrt{5}\right)^2\end{align*}
- \begin{align*}\sqrt{24} \cdot \sqrt{27}\end{align*}
- \begin{align*}\sqrt{16} + 2\sqrt{8}\end{align*}
- \begin{align*}\sqrt{28} + \sqrt{7}\end{align*}
- \begin{align*}-8 \sqrt{3} - \sqrt{12}\end{align*}
- \begin{align*}\sqrt{72} - \sqrt{50}\end{align*}
- \begin{align*}\sqrt{6} +7 \sqrt{6} - \sqrt{54}\end{align*}
- \begin{align*}8 \sqrt{10} - \sqrt{90}+7\sqrt{5}\end{align*}

### Answers for Review Problems

To see the Review answers, open this PDF file and look for section 5.5.