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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
6
Try it.
\(\sqrt{4}\)
Try it.
\(\sqrt{64}\)
Solution
8
Try it.
\(\sqrt{169}\)
Try it.
\(\sqrt{9}\)
Solution
3
Try it.
\(\sqrt{16}\)
Try it.
\(\sqrt{100}\)
Solution
10
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
5
Try it.
\(\sqrt{25+144}\)
Try it.
\(\sqrt{9}+\sqrt{16}\)
Solution
7
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 .Odhalte odpověď
ⓐ \(81\) ⓑ \(81\) ⓒ \(-81\)
-
Round 3.846 to the nearest hundredth.
If you missed this problem, review .Odhalte odpověď
\(3.85\)
-
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 .Odhalte odpověď
ⓐ \(-\sqrt{100}\) ⓑ \(\sqrt{-100}\).
-
Simplify: ⓐ \(\sqrt{36}\) ⓑ \(\sqrt{196}\) ⓒ \(\text{-}\sqrt{81}\) ⓓ \(\text{-}\sqrt{289}\).
Odhalte odpověď
ⓐ
ⓑ\(\sqrt{36}\) 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: ⓐ \(\text{-}\sqrt{49}\) ⓑ \(\sqrt{225}\).
Odhalte odpověď
ⓐ \(-7\) ⓑ \(15\)
-
Simplify: ⓐ \(\sqrt{64}\) ⓑ \(\text{-}\sqrt{121}\).
Odhalte odpověď
ⓐ \(8\) ⓑ \(-11\)
-
Simplify: ⓐ \(\sqrt{-169}\) ⓑ \(\text{-}\sqrt{64}\).
Odhalte odpověď
ⓐ
ⓑ\(\sqrt{-169}\) 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: ⓐ \(\sqrt{-196}\) ⓑ \(\text{-}\sqrt{81}\).
Odhalte odpověď
ⓐ not a real number ⓑ \(-9\)
-
Simplify: ⓐ \(\text{-}\sqrt{49}\) ⓑ \(\sqrt{-121}\).
Odhalte odpověď
ⓐ \(-7\) ⓑ not a real number
-
Simplify: ⓐ \(\sqrt{25}+\sqrt{144}\) ⓑ \(\sqrt{25+144}\).
Odhalte odpověď
ⓐ
ⓑ\(\sqrt{25}+\sqrt{144}\) 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: ⓐ \(\sqrt{9}+\sqrt{16}\) ⓑ \(\sqrt{9+16}\).
Odhalte odpověď
ⓐ 7 ⓑ 5
-
Simplify: ⓐ \(\sqrt{64+225}\) ⓑ \(\sqrt{64}+\sqrt{225}\).
Odhalte odpověď
ⓐ 17 ⓑ \(23\)
-
Estimate \(\sqrt{60}\) between two consecutive whole numbers.
Odhalte odpověď
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.
Odhalte odpověď
\(6<\sqrt{38}<7\)
-
Estimate the square root \(\sqrt{84}\) between two consecutive whole numbers.
Odhalte odpověď
\(9<\sqrt{84}<10\)
-
Round \(\sqrt{17}\) to two decimal places.
Odhalte odpověď
\(\sqrt{17}\) 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.
Odhalte odpověď
\(\approx 3.32\)
-
Round \(\sqrt{13}\) to two decimal places.
Odhalte odpověď
\(\approx 3.61\)
-
Simplify: ⓐ \(\sqrt{{x}^{6}}\) ⓑ \(\sqrt{{y}^{16}}\).
Odhalte odpověď
ⓐ
ⓑ\(\sqrt{{x}^{6}}\) \(\text{Since}\ {({x}^{3})}^{2}={x}^{6}.\) \({x}^{3}\) \(\sqrt{{y}^{16}}\) \(\text{Since}\ {({y}^{8})}^{2}={y}^{16}.\) \({y}^{8}\) -
Simplify: ⓐ \(\sqrt{{y}^{8}}\) ⓑ \(\sqrt{{z}^{12}}\).
Odhalte odpověď
ⓐ \({y}^{4}\) ⓑ \({z}^{6}\)
-
Simplify: ⓐ \(\sqrt{{m}^{4}}\) ⓑ \(\sqrt{{b}^{10}}\).
Odhalte odpověď
ⓐ \({m}^{2}\) ⓑ \({b}^{5}\)
-
Simplify: \(\sqrt{16{n}^{2}}\).
Odhalte odpověď
\(\sqrt{16{n}^{2}}\) \(\text{Since}\ {(4n)}^{2}=16{n}^{2}.\) \(4n\) -
Simplify: \(\sqrt{64{x}^{2}}\).
Odhalte odpověď
\(8x\)
-
Simplify: \(\sqrt{169{y}^{2}}\).
Odhalte odpověď
\(13y\)
-
Simplify: \(\text{-}\sqrt{81{c}^{2}}\).
Odhalte odpověď
\(\text{-}\sqrt{81{c}^{2}}\) \(\text{Since}\ {(9c)}^{2}=81{c}^{2}.\) \(-9c\) -
Simplify: \(\text{-}\sqrt{121{y}^{2}}\).
Odhalte odpověď
\(-11y\)
-
Simplify: \(\text{-}\sqrt{100{p}^{2}}\).
Odhalte odpověď
\(-10p\)
-
Simplify: \(\sqrt{36{x}^{2}{y}^{2}}\).
Odhalte odpověď
\(\sqrt{36{x}^{2}{y}^{2}}\) \(\text{Since}\ {(6xy)}^{2}=36{x}^{2}{y}^{2}.\) \(6xy\) -
Simplify: \(\sqrt{100{a}^{2}{b}^{2}}\).
Odhalte odpověď
\(10ab\)
-
Simplify: \(\sqrt{225{m}^{2}{n}^{2}}\).
Odhalte odpověď
\(15mn\)
-
Simplify: \(\sqrt{64{p}^{64}}\).
Odhalte odpověď
\(\sqrt{64{p}^{64}}\) \(\text{Since}\ {(8{p}^{32})}^{2}=64{p}^{64}.\) \(8{p}^{32}\) -
Simplify: \(\sqrt{49{x}^{30}}\).
Odhalte odpověď
\(7{x}^{15}\)
-
Simplify: \(\sqrt{81{w}^{36}}\).
Odhalte odpověď
\(9{w}^{18}\)
-
Simplify: \(\sqrt{121{a}^{6}{b}^{8}}\)
Odhalte odpověď
\(\sqrt{121{a}^{6}{b}^{8}}\) \(\text{Since}\ {(11{a}^{3}{b}^{4})}^{2}=121{a}^{6}{b}^{8}.\) \(11{a}^{3}{b}^{4}\) -
Simplify: \(\sqrt{169{x}^{10}{y}^{14}}\).
Odhalte odpověď
\(13{x}^{5}{y}^{7}\)
-
Simplify: \(\sqrt{144{p}^{12}{q}^{20}}\).
Odhalte odpověď
\(12{p}^{6}{q}^{10}\)
-
\(\sqrt{36}\)
Odhalte odpověď
6
-
\(\sqrt{4}\)
-
\(\sqrt{64}\)
Odhalte odpověď
8
-
\(\sqrt{169}\)
Symbols used here
The non-negative number whose square (n-th power) is x.
Equal to the precision shown, not exactly.
Inequalities that allow equality; < and > exclude it.
Both signs at once: x = 3 ± 2 means 5 and 1.
The two sides are different.
Distance from zero: |−3| = 3. For a complex number, distance from the origin.
i² = −1.
The exponent b must be raised to for x; ln uses base e.
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.
Zkuste si vlastní.
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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