## Evaluate and estimate numerical square and cube roots

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Simplifying Square Roots

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*}.

1. 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.

1. 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.

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.

1. \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*}.

1. \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*}.

#### 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.

\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.

1. 56
2. 12
3. 92

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

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

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

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### Vocabulary Language: English

TermDefinition
Perfect Square A perfect square is a number whose square root is an integer.