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Simplify and Use Square Roots

Simplify expressions with square roots

Simplify Expressions with Square Roots

Remember that when a number \(n\) is multiplied by itself, we write \({n}^{2}\) and read it “n squared.” For example, \({15}^{2}\) reads as “15 squared,” and 225 is called the square of 15, since \({15}^{2}=225\).

Sometimes we will need to look at the relationship between numbers and their squares in reverse. Because 225 is the square of 15, we can also say that 15 is a square root of 225. A number whose square is \(m\) is called a square root of \(m\).

Notice \({(-15)}^{2}=225\) also, so \(-15\) is also a square root of 225. Therefore, both 15 and \(-15\) are square roots of 225.

So, every positive number has two square roots—one positive and one negative. What if we only wanted the positive square root of a positive number? The radical sign, \(\sqrt{m}\), denotes the positive square root. The positive square root is also called the principal square root.

We also use the radical sign for the square root of zero. Because \({0}^{2}=0\), \(\sqrt{0}=0\). Notice that zero has only one square root.

Since 15 is the positive square root of 225, we write \(\sqrt{225}=15\). Fill in to make a table of square roots you can refer to as you work this chapter.

We know that every positive number has two square roots and the radical sign indicates the positive one. We write \(\sqrt{225}=15\). If we want to find the negative square root of a number, we place a negative in front of the radical sign. For example, \(\text{-}\sqrt{225}=-15\).

Example

Try it.

Simplify: ⓐ \(\sqrt{36}\) ⓑ \(\sqrt{196}\) ⓒ \(\text{-}\sqrt{81}\) ⓓ \(\text{-}\sqrt{289}\).

Solution

ⓐ

	\(\sqrt{36}\)
Since \({6}^{2}=36\)	6

ⓑ

	\(\sqrt{196}\)
Since \({14}^{2}=196\)	14

ⓒ

	\(\text{-}\sqrt{81}\)
The negative is in front of the radical sign.	−9

ⓓ

	\(\text{-}\sqrt{289}\)
The negative is in front of the radical sign.	−17

Example

Try it.

Simplify: ⓐ \(\sqrt{-169}\) ⓑ \(\text{-}\sqrt{64}\).

Solution

ⓐ

	\(\sqrt{-169}\)
There is no real number whose square is −169.	\(\sqrt{-169}\) is not a real number.

ⓑ

	\(\text{-}\sqrt{64}\)
The negative is in front of the radical.	−8

Condensed — the full section is in OpenStax Elementary Algebra 2e.

Estimate Square Roots

So far we have only considered square roots of perfect square numbers. The square roots of other numbers are not whole numbers. Look at below.

Number	Square Root
4	\(\sqrt{4}\) = 2
5	\(\sqrt{5}\)
6	\(\sqrt{6}\)
7	\(\sqrt{7}\)
8	\(\sqrt{8}\)
9	\(\sqrt{9}\) = 3

The square roots of numbers between 4 and 9 must be between the two consecutive whole numbers 2 and 3, and they are not whole numbers. Based on the pattern in the table above, we could say that \(\sqrt{5}\) must be between 2 and 3. Using inequality symbols, we write:

\[2<\sqrt{5}<3\]

Example

Try it.

Estimate \(\sqrt{60}\) between two consecutive whole numbers.

Solution

Think of the perfect square numbers closest to 60. Make a small table of these perfect squares and their squares roots.


Locate 60 between two consecutive perfect squares.
\(\sqrt{60}\) is between their square roots.

Approximate Square Roots

There are mathematical methods to approximate square roots, but nowadays most people use a calculator to find them. Find the \(\sqrt{x}\) key on your calculator. You will use this key to approximate square roots.

When you use your calculator to find the square root of a number that is not a perfect square, the answer that you see is not the exact square root. It is an approximation, accurate to the number of digits shown on your calculator’s display. The symbol for an approximation is \(\approx\) and it is read ‘approximately.’

Suppose your calculator has a 10-digit display. You would see that

\[\ \sqrt{5}\approx 2.236067978\]

If we wanted to round \(\sqrt{5}\) to two decimal places, we would say

\[\sqrt{5}\approx 2.24\]

How do we know these values are approximations and not the exact values? Look at what happens when we square them:

\[\begin{array}{lll}{(2.236067978)}^{2} & = & 5.000000002 \\ {(2.24)}^{2} & = & 5.0176\end{array}\]

Their squares are close to 5, but are not exactly equal to 5.

Using the square root key on a calculator and then rounding to two decimal places, we can find:

\[\begin{array}{lll}\sqrt{4} & = & 2 \\ \sqrt{5} & \approx & 2.24 \\ \sqrt{6} & \approx & 2.45 \\ \sqrt{7} & \approx & 2.65 \\ \sqrt{8} & \approx & 2.83 \\ \sqrt{9} & = & 3\end{array}\]

Example

Try it.

Round \(\sqrt{17}\) to two decimal places.

Solution

	\(\sqrt{17}\)
Use the calculator square root key.	4.123105626...
Round to two decimal places.	4.12
	\(\sqrt{17}\approx 4.12\)

Simplify Variable Expressions with Square Roots

What if we have to find a square root of an expression with a variable? Consider \(\sqrt{9{x}^{2}}\). Can you think of an expression whose square is \(9{x}^{2}\)?

\[\begin{array}{llllll}{(?)}^{2} & = & 9{x}^{2} & & & \\ {(3x)}^{2} & = & 9{x}^{2}, & & & \text{so}\ \sqrt{9{x}^{2}}=3x\end{array}\]

When we use the radical sign to take the square root of a variable expression, we should specify that \(x\ge 0\) to make sure we get the principal square root.

However, in this chapter we will assume that each variable in a square-root expression represents a non-negative number and so we will not write \(x\ge 0\) next to every radical.

What about square roots of higher powers of variables? Think about the Power Property of Exponents we used in Chapter 6.

\[{({a}^{m})}^{n}={a}^{m\cdot n}\]

If we square \({a}^{m}\), the exponent will become \(2m\).

\[{({a}^{m})}^{2}={a}^{2m}\]

How does this help us take square roots? Let’s look at a few:

\[\begin{array}{ll}\sqrt{25{u}^{8}}=5{u}^{4} & \text{because}\ {(5{u}^{4})}^{2}=25{u}^{8} \\ \sqrt{16{r}^{20}}=4{r}^{10} & \text{because}\ {(4{r}^{10})}^{2}=16{r}^{20} \\ \sqrt{196{q}^{36}}=14{q}^{18} & \text{because}\ {(14{q}^{18})}^{2}=196{q}^{36}\end{array}\]

Example

Try it.

Simplify: ⓐ \(\sqrt{{x}^{6}}\) ⓑ \(\sqrt{{y}^{16}}\).

Solution

ⓐ

	\(\sqrt{{x}^{6}}\)
\(\text{Since}\ {({x}^{3})}^{2}={x}^{6}.\)	\({x}^{3}\)

ⓑ

	\(\sqrt{{y}^{16}}\)
\(\text{Since}\ {({y}^{8})}^{2}={y}^{16}.\)	\({y}^{8}\)

Example

Try it.

Simplify: \(\sqrt{16{n}^{2}}\).

Solution

	\(\sqrt{16{n}^{2}}\)
\(\text{Since}\ {(4n)}^{2}=16{n}^{2}.\)	\(4n\)

Example

Try it.

Simplify: \(\text{-}\sqrt{81{c}^{2}}\).

Solution

	\(\text{-}\sqrt{81{c}^{2}}\)
\(\text{Since}\ {(9c)}^{2}=81{c}^{2}.\)	\(-9c\)

Example

Try it.

Simplify: \(\sqrt{36{x}^{2}{y}^{2}}\).

Solution

	\(\sqrt{36{x}^{2}{y}^{2}}\)
\(\text{Since}\ {(6xy)}^{2}=36{x}^{2}{y}^{2}.\)	\(6xy\)

Example

Try it.

Simplify: \(\sqrt{64{p}^{64}}\).

Solution

	\(\sqrt{64{p}^{64}}\)
\(\text{Since}\ {(8{p}^{32})}^{2}=64{p}^{64}.\)	\(8{p}^{32}\)

Example

Try it.

Simplify: \(\sqrt{121{a}^{6}{b}^{8}}\)

Solution

	\(\sqrt{121{a}^{6}{b}^{8}}\)
\(\text{Since}\ {(11{a}^{3}{b}^{4})}^{2}=121{a}^{6}{b}^{8}.\)	\(11{a}^{3}{b}^{4}\)

Condensed — the full section is in OpenStax Elementary Algebra 2e.

Key Concepts

Note that the square root of a negative number is not a real number.
Every positive number has two square roots, one positive and one negative. The positive square root of a positive number is the principal square root.
We can estimate square roots using nearby perfect squares.
We can approximate square roots using a calculator.
When we use the radical sign to take the square root of a variable expression, we should specify that \(x\ge 0\) to make sure we get the principal square root.

Simplify and Use Square Roots

Simplify Expressions with Square Roots

In the following exercises, simplify.

Try it.

\(\sqrt{36}\)

Solution

Try it.

\(\sqrt{4}\)

Try it.

\(\sqrt{64}\)

Solution

Try it.

\(\sqrt{169}\)

Try it.

\(\sqrt{9}\)

Solution

Try it.

\(\sqrt{16}\)

Try it.

\(\sqrt{100}\)

Solution

Try it.

\(\sqrt{144}\)

Try it.

\(\text{-}\sqrt{4}\)

Solution

\(-2\)

Try it.

\(\text{-}\sqrt{100}\)

Try it.

\(\text{-}\sqrt{1}\)

Solution

\(-1\)

Try it.

\(\text{-}\sqrt{121}\)

Try it.

\(\sqrt{-121}\)

Solution

not a real number

Try it.

\(\sqrt{-36}\)

Try it.

\(\sqrt{-9}\)

Solution

not a real number

Try it.

\(\sqrt{-49}\)

Try it.

\(\sqrt{9+16}\)

Solution

Try it.

\(\sqrt{25+144}\)

Try it.

\(\sqrt{9}+\sqrt{16}\)

Solution

Try it.

\(\sqrt{25}+\sqrt{144}\)

Estimate Square Roots

In the following exercises, estimate each square root between two consecutive whole numbers.

Try it.

\(\sqrt{70}\)

Solution

\(8<\sqrt{70}<9\)

Try it.

\(\sqrt{55}\)

Try it.

\(\sqrt{200}\)

Solution

\(14<\sqrt{200}<15\)

Try it.

\(\sqrt{172}\)

Approximate Square Roots

In the following exercises, approximate each square root and round to two decimal places.

Try it.

\(\sqrt{19}\)

Solution

4.36

Try it.

\(\sqrt{21}\)

Try it.

\(\sqrt{53}\)

Solution

7.28

Try it.

\(\sqrt{47}\)

Simplify Variable Expressions with Square Roots

In the following exercises, simplify.

Try it.

\(\sqrt{{y}^{2}}\)

Solution

\(y\)

Try it.

\(\sqrt{{b}^{2}}\)

Try it.

\(\sqrt{{a}^{14}}\)

Solution

\({a}^{7}\)

Try it.

\(\sqrt{{w}^{24}}\)

Try it.

\(\sqrt{49{x}^{2}}\)

Solution

\(7x\)

Try it.

\(\sqrt{100{y}^{2}}\)

Try it.

\(\sqrt{121{m}^{20}}\)

Solution

\(11{m}^{10}\)

Try it.

\(\sqrt{25{h}^{44}}\)

Try it.

\(\sqrt{81{x}^{36}}\)

Solution

\(9{x}^{18}\)

Try it.

\(\sqrt{144{z}^{84}}\)

Try it.

\(\text{-}\sqrt{81{x}^{18}}\)

Solution

\(-9{x}^{9}\)

Try it.

\(\text{-}\sqrt{100{m}^{32}}\)

Try it.

\(\text{-}\sqrt{64{a}^{2}}\)

Solution

\(-8a\)

Try it.

\(\text{-}\sqrt{25{x}^{2}}\)

Try it.

\(\sqrt{144{x}^{2}{y}^{2}}\)

Solution

\(12xy\)

Try it.

\(\sqrt{196{a}^{2}{b}^{2}}\)

Try it.

\(\sqrt{169{w}^{8}{y}^{10}}\)

Solution

\(13{w}^{4}{y}^{5}\)

Try it.

\(\sqrt{81{p}^{24}{q}^{6}}\)

Try it.

\(\sqrt{9{c}^{8}{d}^{12}}\)

Solution

\(3{c}^{4}{d}^{6}\)

Try it.

\(\sqrt{36{r}^{6}{s}^{20}}\)

Try it.

Decorating Denise wants to have a square accent of designer tiles in her new shower. She can afford to buy 625 square centimeters of the designer tiles. How long can a side of the accent be?

Solution

25 centimeters

Try it.

Decorating Morris wants to have a square mosaic inlaid in his new patio. His budget allows for 2025 square inch tiles. How long can a side of the mosaic be?

Try it.

Why is there no real number equal to \(\sqrt{-64}\)?

Solution

Answers will vary.

Try it.

What is the difference between \({9}^{2}\) and \(\sqrt{9}\)?

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ On a scale of 1–10, how would you rate your mastery of this section in light of your responses on the checklist? How can you improve this?

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Simplify: ⓐ \({9}^{2}\) ⓑ \({(-9)}^{2}\) ⓒ \(\text{-}{9}^{2}\).
If you missed this problem, review .

答えを明らかにしろ

ⓐ \(81\) ⓑ \(81\) ⓒ \(-81\)
Round 3.846 to the nearest hundredth.
If you missed this problem, review .

答えを明らかにしろ

\(3.85\)
For each number, identify whether it is a real number or not a real number:
ⓐ \(\text{-}\sqrt{100}\) ⓑ \(\sqrt{-100}\).
If you missed this problem, review .

答えを明らかにしろ

ⓐ \(-\sqrt{100}\) ⓑ \(\sqrt{-100}\).
Simplify: ⓐ \(\sqrt{36}\) ⓑ \(\sqrt{196}\) ⓒ \(\text{-}\sqrt{81}\) ⓓ \(\text{-}\sqrt{289}\).

答えを明らかにしろ

ⓐ

\(\sqrt{36}\)
Since \({6}^{2}=36\) 6
ⓑ

\(\sqrt{196}\)
Since \({14}^{2}=196\) 14
ⓒ

\(\text{-}\sqrt{81}\)
The negative is in front of the radical sign. −9
ⓓ

\(\text{-}\sqrt{289}\)
The negative is in front of the radical sign. −17
Simplify: ⓐ \(\text{-}\sqrt{49}\) ⓑ \(\sqrt{225}\).

答えを明らかにしろ

ⓐ \(-7\) ⓑ \(15\)
Simplify: ⓐ \(\sqrt{64}\) ⓑ \(\text{-}\sqrt{121}\).

答えを明らかにしろ

ⓐ \(8\) ⓑ \(-11\)
Simplify: ⓐ \(\sqrt{-169}\) ⓑ \(\text{-}\sqrt{64}\).

答えを明らかにしろ

ⓐ

\(\sqrt{-169}\)
There is no real number whose square is −169. \(\sqrt{-169}\) is not a real number.
ⓑ

\(\text{-}\sqrt{64}\)
The negative is in front of the radical. −8
Simplify: ⓐ \(\sqrt{-196}\) ⓑ \(\text{-}\sqrt{81}\).

答えを明らかにしろ

ⓐ not a real number ⓑ \(-9\)
Simplify: ⓐ \(\text{-}\sqrt{49}\) ⓑ \(\sqrt{-121}\).

答えを明らかにしろ

ⓐ \(-7\) ⓑ not a real number
Simplify: ⓐ \(\sqrt{25}+\sqrt{144}\) ⓑ \(\sqrt{25+144}\).

答えを明らかにしろ

ⓐ
\(\sqrt{25}+\sqrt{144}\)
Use the order of operations. \(5+12\)
Simplify. 17
ⓑ
\(\sqrt{25+144}\)
Simplify under the radical sign. \(\sqrt{169}\)
Simplify. 13
Notice the different answers in parts ⓐ and ⓑ !
Simplify: ⓐ \(\sqrt{9}+\sqrt{16}\) ⓑ \(\sqrt{9+16}\).

答えを明らかにしろ

ⓐ 7 ⓑ 5
Simplify: ⓐ \(\sqrt{64+225}\) ⓑ \(\sqrt{64}+\sqrt{225}\).

答えを明らかにしろ

ⓐ 17 ⓑ \(23\)
Estimate \(\sqrt{60}\) between two consecutive whole numbers.

答えを明らかにしろ

Think of the perfect square numbers closest to 60. Make a small table of these perfect squares and their squares roots.
Locate 60 between two consecutive perfect squares.
\(\sqrt{60}\) is between their square roots.
Estimate the square root \(\sqrt{38}\) between two consecutive whole numbers.

答えを明らかにしろ

\(6<\sqrt{38}<7\)
Estimate the square root \(\sqrt{84}\) between two consecutive whole numbers.

答えを明らかにしろ

\(9<\sqrt{84}<10\)
Round \(\sqrt{17}\) to two decimal places.

答えを明らかにしろ

\(\sqrt{17}\)
Use the calculator square root key. 4.123105626...
Round to two decimal places. 4.12
\(\sqrt{17}\approx 4.12\)
Round \(\sqrt{11}\) to two decimal places.

答えを明らかにしろ

\(\approx 3.32\)
Round \(\sqrt{13}\) to two decimal places.

答えを明らかにしろ

\(\approx 3.61\)
Simplify: ⓐ \(\sqrt{{x}^{6}}\) ⓑ \(\sqrt{{y}^{16}}\).

答えを明らかにしろ

ⓐ
\(\sqrt{{x}^{6}}\)
\(\text{Since}\ {({x}^{3})}^{2}={x}^{6}.\) \({x}^{3}\)
ⓑ
\(\sqrt{{y}^{16}}\)
\(\text{Since}\ {({y}^{8})}^{2}={y}^{16}.\) \({y}^{8}\)
Simplify: ⓐ \(\sqrt{{y}^{8}}\) ⓑ \(\sqrt{{z}^{12}}\).

答えを明らかにしろ

ⓐ \({y}^{4}\) ⓑ \({z}^{6}\)
Simplify: ⓐ \(\sqrt{{m}^{4}}\) ⓑ \(\sqrt{{b}^{10}}\).

答えを明らかにしろ

ⓐ \({m}^{2}\) ⓑ \({b}^{5}\)
Simplify: \(\sqrt{16{n}^{2}}\).

答えを明らかにしろ

\(\sqrt{16{n}^{2}}\)
\(\text{Since}\ {(4n)}^{2}=16{n}^{2}.\) \(4n\)
Simplify: \(\sqrt{64{x}^{2}}\).

答えを明らかにしろ

\(8x\)
Simplify: \(\sqrt{169{y}^{2}}\).

答えを明らかにしろ

\(13y\)
Simplify: \(\text{-}\sqrt{81{c}^{2}}\).

答えを明らかにしろ

\(\text{-}\sqrt{81{c}^{2}}\)
\(\text{Since}\ {(9c)}^{2}=81{c}^{2}.\) \(-9c\)
Simplify: \(\text{-}\sqrt{121{y}^{2}}\).

答えを明らかにしろ

\(-11y\)
Simplify: \(\text{-}\sqrt{100{p}^{2}}\).

答えを明らかにしろ

\(-10p\)
Simplify: \(\sqrt{36{x}^{2}{y}^{2}}\).

答えを明らかにしろ

\(\sqrt{36{x}^{2}{y}^{2}}\)
\(\text{Since}\ {(6xy)}^{2}=36{x}^{2}{y}^{2}.\) \(6xy\)
Simplify: \(\sqrt{100{a}^{2}{b}^{2}}\).

答えを明らかにしろ

\(10ab\)
Simplify: \(\sqrt{225{m}^{2}{n}^{2}}\).

答えを明らかにしろ

\(15mn\)
Simplify: \(\sqrt{64{p}^{64}}\).

答えを明らかにしろ

\(\sqrt{64{p}^{64}}\)
\(\text{Since}\ {(8{p}^{32})}^{2}=64{p}^{64}.\) \(8{p}^{32}\)
Simplify: \(\sqrt{49{x}^{30}}\).

答えを明らかにしろ

\(7{x}^{15}\)
Simplify: \(\sqrt{81{w}^{36}}\).

答えを明らかにしろ

\(9{w}^{18}\)
Simplify: \(\sqrt{121{a}^{6}{b}^{8}}\)

答えを明らかにしろ

\(\sqrt{121{a}^{6}{b}^{8}}\)
\(\text{Since}\ {(11{a}^{3}{b}^{4})}^{2}=121{a}^{6}{b}^{8}.\) \(11{a}^{3}{b}^{4}\)
Simplify: \(\sqrt{169{x}^{10}{y}^{14}}\).

答えを明らかにしろ

\(13{x}^{5}{y}^{7}\)
Simplify: \(\sqrt{144{p}^{12}{q}^{20}}\).

答えを明らかにしろ

\(12{p}^{6}{q}^{10}\)
\(\sqrt{36}\)

答えを明らかにしろ

6
\(\sqrt{4}\)
\(\sqrt{64}\)

答えを明らかにしろ

8
\(\sqrt{169}\)

Symbols used here

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

\approx

approximately equal
Equal to the precision shown, not exactly.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\pm

plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.

\neq

not equal
The two sides are different.

|x|

absolute value / modulus
Distance from zero: |−3| = 3. For a complex number, distance from the origin.

imaginary unit
i² = −1.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\mathbb{N},\ \mathbb{Z},\ \mathbb{Q},\ \mathbb{R},\ \mathbb{C}

number sets
Naturals, integers, rationals, reals, complex numbers.

How to: Simplify and Use Square Roots

Simplify expressions with square roots
Estimate square roots
Approximate square roots
Simplify variable expressions with square roots
Note that the square root of a negative number is not a real number.
Every positive number has two square roots, one positive and one negative. The positive square root of a positive number is the principal square root.
We can estimate square roots using nearby perfect squares.
We can approximate square roots using a calculator.

Questions people ask

What does it mean to solve an equation?

To find every value of the unknown that makes both sides equal. Each step is an operation applied to both sides that keeps the solution set the same, until the unknown stands alone.

Why do I sometimes get two answers?

A quadratic can cross the axis twice, so it can have two solutions. A degree-n polynomial has up to n. The graph shows where each one comes from.

How do I know whether to factor or use the quadratic formula?

Try factoring for a few seconds: look for two numbers that multiply to a·c and add to b. If nothing obvious appears, the discriminant b² − 4ac tells you how many real roots there are, and the formula finds them without guessing.

あなた自身を試してみてください

Parts of this page are adapted from OpenStax Elementary Algebra 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

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	\(\sqrt{25}+\sqrt{144}\)
Use the order of operations.	\(5+12\)
Simplify.	17

	\(\sqrt{25+144}\)
Simplify under the radical sign.	\(\sqrt{169}\)
Simplify.	13
Notice the different answers in parts ⓐ and ⓑ !