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# 5.5: Simplifying Square Roots

Difficulty Level: Advanced Created by: CK-12
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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?

### Guidance

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 a number is not an integer, it is an irrational number.

#### Example A

Find \begin{align*}\sqrt{50}\end{align*} using:

a) A calculator.

b) By simplifying the square root.

Solution:

a) 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.

b) 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.

#### Example B

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

Solution: 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*}.

#### Example C

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

Solution: 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*}

Intro Problem Revisit 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*}.

### Guided Practice

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

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

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

1. 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*}

2. 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.

3. 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.

### Vocabulary

Square Root
A number, that when multiplied by itself, produces another number.
Perfect Square
A number that has an integer for a square root.
The \begin{align*}\sqrt{}\end{align*}, or square root, sign.

### Practice

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*}

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### Vocabulary Language: English

TermDefinition
Perfect Square A perfect square is a number whose square root is an integer.
Rationalize the denominator To rationalize the denominator means to rewrite the fraction so that the denominator no longer contains a radical.
Square Root The square root of a term is a value that must be multiplied by itself to equal the specified term. The square root of 9 is 3, since 3 * 3 = 9.
Variable Expression A variable expression is a mathematical phrase that contains at least one variable or unknown quantity.

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