# 9.1 Simplify and use square roots  (Page 2/5)

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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:

$2<\sqrt{5}<3$

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.

Estimate the square root $\sqrt{38}$ between two consecutive whole numbers.

$6<\sqrt{38}<7$

Estimate the square root $\sqrt{84}$ between two consecutive whole numbers.

$9<\sqrt{84}<10$

## 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

$\phantom{\rule{3.4em}{0ex}}\sqrt{5}\approx 2.236067978$

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:

$\begin{array}{ccc}\hfill {\left(2.236067978\right)}^{2}& =\hfill & 5.000000002\hfill \\ \hfill {\left(2.24\right)}^{2}& =\hfill & 5.0176\hfill \end{array}$

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:

$\begin{array}{ccc}\hfill \sqrt{4}& =\hfill & 2\hfill \\ \hfill \sqrt{5}& \approx \hfill & 2.24\hfill \\ \hfill \sqrt{6}& \approx \hfill & 2.45\hfill \\ \hfill \sqrt{7}& \approx \hfill & 2.65\hfill \\ \hfill \sqrt{8}& \approx \hfill & 2.83\hfill \\ \hfill \sqrt{9}& =\hfill & 3\hfill \end{array}$

Round $\sqrt{17}$ to two decimal places.

## Solution

$\begin{array}{cccc}& & & \sqrt{17}\hfill \\ \text{Use the calculator square root key.}\hfill & & & 4.123105626...\hfill \\ \text{Round to two decimal places.}\hfill & & & 4.12\hfill \\ & & & \sqrt{17}\approx 4.12\hfill \end{array}$

Round $\sqrt{11}$ to two decimal places.

$\approx 3.32$

Round $\sqrt{13}$ to two decimal places.

$\approx 3.61$

## 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 $9{x}^{2}$ ?

$\begin{array}{cccccc}\hfill {\left(?\right)}^{2}& =\hfill & 9{x}^{2}\hfill & & & \\ \hfill {\left(3x\right)}^{2}& =\hfill & 9{x}^{2},\hfill & & & \text{so}\phantom{\rule{0.2em}{0ex}}\sqrt{9{x}^{2}}=3x\hfill \end{array}$

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.

${\left({a}^{m}\right)}^{n}={a}^{m·n}$

If we square ${a}^{m}$ , the exponent will become $2m$ .

${\left({a}^{m}\right)}^{2}={a}^{2m}$

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

$\begin{array}{cc}\hfill \sqrt{25{u}^{8}}=5{u}^{4}& \text{because}\phantom{\rule{0.2em}{0ex}}{\left(5{u}^{4}\right)}^{2}=25{u}^{8}\hfill \\ \hfill \sqrt{16{r}^{20}}=4{r}^{10}& \text{because}\phantom{\rule{0.2em}{0ex}}{\left(4{r}^{10}\right)}^{2}=16{r}^{20}\hfill \\ \hfill \sqrt{196{q}^{36}}=14{q}^{18}& \text{because}\phantom{\rule{0.2em}{0ex}}{\left(14{q}^{18}\right)}^{2}=196{q}^{36}\hfill \end{array}$

Simplify: $\sqrt{{x}^{6}}$ $\sqrt{{y}^{16}}$ .

## Solution

$\begin{array}{cccc}& & & \phantom{\rule{4em}{0ex}}\sqrt{{x}^{6}}\hfill \\ \text{Since}\phantom{\rule{0.2em}{0ex}}{\left({x}^{3}\right)}^{2}={x}^{6}.\hfill & & & \phantom{\rule{4em}{0ex}}{x}^{3}\hfill \end{array}$

$\begin{array}{cccc}& & & \phantom{\rule{4em}{0ex}}\sqrt{{y}^{16}}\hfill \\ \text{Since}\phantom{\rule{0.2em}{0ex}}{\left({y}^{8}\right)}^{2}={y}^{16}.\hfill & & & \phantom{\rule{4em}{0ex}}{y}^{8}\hfill \end{array}$

Simplify: $\sqrt{{y}^{8}}$ $\sqrt{{z}^{12}}$ .

${y}^{4}$ ${z}^{6}$

Simplify: $\sqrt{{m}^{4}}$ $\sqrt{{b}^{10}}$ .

${m}^{2}$ ${b}^{5}$

Simplify: $\sqrt{16{n}^{2}}$ .

## Solution

$\begin{array}{cccc}& & & \phantom{\rule{4em}{0ex}}\sqrt{16{n}^{2}}\hfill \\ \text{Since}\phantom{\rule{0.2em}{0ex}}{\left(4n\right)}^{2}=16{n}^{2}.\hfill & & & \phantom{\rule{4em}{0ex}}4n\hfill \end{array}$

Simplify: $\sqrt{64{x}^{2}}$ .

$8x$

Simplify: $\sqrt{169{y}^{2}}$ .

$13y$

Simplify: $\text{−}\sqrt{81{c}^{2}}$ .

## Solution

$\begin{array}{cccc}& & & \phantom{\rule{4em}{0ex}}\text{−}\sqrt{81{c}^{2}}\hfill \\ \text{Since}\phantom{\rule{0.2em}{0ex}}{\left(9c\right)}^{2}=81{c}^{2}.\hfill & & & \phantom{\rule{4em}{0ex}}-9c\hfill \end{array}$

Simplify: $\text{−}\sqrt{121{y}^{2}}$ .

$-11y$

Simplify: $\text{−}\sqrt{100{p}^{2}}$ .

$-10p$

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

## Solution

$\begin{array}{cccc}& & & \phantom{\rule{4em}{0ex}}\sqrt{36{x}^{2}{y}^{2}}\hfill \\ \text{Since}\phantom{\rule{0.2em}{0ex}}{\left(6xy\right)}^{2}=36{x}^{2}{y}^{2}.\hfill & & & \phantom{\rule{4em}{0ex}}6xy\hfill \end{array}$

Simplify: $\sqrt{100{a}^{2}{b}^{2}}$ .

$10ab$

Simplify: $\sqrt{225{m}^{2}{n}^{2}}$ .

$15mn$

Simplify: $\sqrt{64{p}^{64}}$ .

## Solution

$\begin{array}{cccc}& & & \phantom{\rule{4em}{0ex}}\sqrt{64{p}^{64}}\hfill \\ \text{Since}\phantom{\rule{0.2em}{0ex}}{\left(8{p}^{32}\right)}^{2}=64{p}^{64}.\hfill & & & \phantom{\rule{4em}{0ex}}8{p}^{32}\hfill \end{array}$

Simplify: $\sqrt{49{x}^{30}}$ .

$7{x}^{15}$

Simplify: $\sqrt{81{w}^{36}}$ .

$9{w}^{18}$

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

## Solution

$\begin{array}{cccc}& & & \phantom{\rule{4em}{0ex}}\sqrt{121{a}^{6}{b}^{8}}\hfill \\ \text{Since}\phantom{\rule{0.2em}{0ex}}{\left(11{a}^{3}{b}^{4}\right)}^{2}=121{a}^{6}{b}^{8}.\hfill & & & \phantom{\rule{4em}{0ex}}11{a}^{3}{b}^{4}\hfill \end{array}$

Simplify: $\sqrt{169{x}^{10}{y}^{14}}$ .

$13{x}^{5}{y}^{7}$

Simplify: $\sqrt{144{p}^{12}{q}^{20}}$ .

$12{p}^{6}{q}^{10}$

Access this online resource for additional instruction and practice with square roots.

## 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.

## Practice makes perfect

Simplify Expressions with Square Roots

In the following exercises, simplify.

$\sqrt{36}$

6

$\sqrt{4}$

$\sqrt{64}$

8

$\sqrt{169}$

$\sqrt{9}$

3

$\sqrt{16}$

$\sqrt{100}$

10

$\sqrt{144}$

$\text{−}\sqrt{4}$

$-2$

$\text{−}\sqrt{100}$

$\text{−}\sqrt{1}$

$-1$

$\text{−}\sqrt{121}$

$\sqrt{-121}$

not a real number

$\sqrt{-36}$

$\sqrt{-9}$

not a real number

$\sqrt{-49}$

$\sqrt{9+16}$

5

$\sqrt{25+144}$

$\sqrt{9}+\sqrt{16}$

7

$\sqrt{25}+\sqrt{144}$

Estimate Square Roots

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

$\sqrt{70}$

$8<\sqrt{70}<9$

$\sqrt{55}$

$\sqrt{200}$

$14<\sqrt{200}<15$

$\sqrt{172}$

Approximate Square Roots

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

$\sqrt{19}$

4.36

$\sqrt{21}$

$\sqrt{53}$

7.28

$\sqrt{47}$

Simplify Variable Expressions with Square Roots

In the following exercises, simplify.

$\sqrt{{y}^{2}}$

$y$

$\sqrt{{b}^{2}}$

$\sqrt{{a}^{14}}$

${a}^{7}$

$\sqrt{{w}^{24}}$

$\sqrt{49{x}^{2}}$

$7x$

$\sqrt{100{y}^{2}}$

$\sqrt{121{m}^{20}}$

$11{m}^{10}$

$\sqrt{25{h}^{44}}$

$\sqrt{81{x}^{36}}$

$9{x}^{18}$

$\sqrt{144{z}^{84}}$

$\text{−}\sqrt{81{x}^{18}}$

$-9{x}^{9}$

$\text{−}\sqrt{100{m}^{32}}$

$\text{−}\sqrt{64{a}^{2}}$

$-8a$

$\text{−}\sqrt{25{x}^{2}}$

$\sqrt{144{x}^{2}{y}^{2}}$

$12xy$

$\sqrt{196{a}^{2}{b}^{2}}$

$\sqrt{169{w}^{8}{y}^{10}}$

$13{w}^{4}{y}^{5}$

$\sqrt{81{p}^{24}{q}^{6}}$

$\sqrt{9{c}^{8}{d}^{12}}$

$3{c}^{4}{d}^{6}$

$\sqrt{36{r}^{6}{s}^{20}}$

## Everyday math

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?

25 centimeters

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?

## Writing exercises

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

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

## Self check

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?

how many typos can we find...?
5
Joseph
In the LCM Prime Factors exercises, the LCM of 28 and 40 is 280. Not 420!
4x+7y=29,x+3y=11 substitute method of linear equation
substitute method of linear equation
Srinu
Solve one equation for one variable. Using the 2nd equation, x=11-3y. Substitute that for x in first equation. this will find y. then use the value for y to find the value for x.
bruce
I want to learn
Elizebeth
help
Elizebeth
I want to learn. Please teach me?
Wayne
1) Use any equation, and solve for any of the variables. Since the coefficient of x (the number in front of the x) in the second equation is 1 (it actually isn't shown, but 1 * x = x), use that equation. Subtract 3y from both sides (this isolates the x on the left side of the equal sign).
bruce
2) This results in x=11-3y. x is note in terms of y. Use that as the value of x and substitute for all x in the first equation. The first equation becomes 4(11-3y)+7y =29. Note that the only variable left in the first equation is the y. If you have multiple variable, then something is wrong.
bruce
3) Distribute (multiply) the 4 across 11-3y to get 44-12y. Add this to the 7y. So, the equation is now 44-5y=29.
bruce
4) Solve 44-5y=29 for y. Isolate the y by subtracting 44 from birth sides, resulting in -5y=-15. Now, divide birth sides by -5 (since you have -5y). This results in y=3. You now have the value of one variable.
bruce
5) The last step is to take the value of y from Step 4) and substitute into the 2nd equation. Therefore: x+3y=11 becomes x+3(3)=11. Then multiplying, x+9=11. Finally, solve for x by subtracting 9 from both sides. Therefore, x=2.
bruce
6) The ordered pair of (2, 3) is the proposed solution. To check, substitute those values into either equation. If the result is true, then the solution is correct. 4(2)+7(3)=8+21=29. TRUE! Finished.
bruce
At 1:30 Marlon left his house to go to the beach, a distance of 5.625 miles. He rose his skateboard until 2:15, and then walked the rest of the way. He arrived at the beach at 3:00. Marlon's speed on his skateboard is 1.5 times his walking speed. Find his speed when skateboarding and when walking.
divide 3x⁴-4x³-3x-1 by x-3
how to multiply the monomial
Two sisters like to compete on their bike rides. Tamara can go 4 mph faster than her sister, Samantha. If it takes Samantha 1 hours longer than Tamara to go 80 miles, how fast can Samantha ride her bike? Got questions? Get instant answers now!
how do u solve that question
Seera
Two sisters like to compete on their bike rides. Tamara can go 4 mph faster than her sister, Samantha. If it takes Samantha 1 hours longer than Tamara to go 80 miles, how fast can Samantha ride her bike?
Seera
Speed=distance ÷ time
Tremayne
x-3y =1; 3x-2y+4=0 graph
Brandon has a cup of quarters and dimes with a total of 5.55\$. The number of quarters is five less than three times the number of dimes
app is wrong how can 350 be divisible by 3.
June needs 48 gallons of punch for a party and has two different coolers to carry it in. The bigger cooler is five times as large as the smaller cooler. How many gallons can each cooler hold?
Susanna if the first cooler holds five times the gallons then the other cooler. The big cooler holda 40 gallons and the 2nd will hold 8 gallons is that correct?
Georgie
@Susanna that person is correct if you divide 40 by 8 you can see it's 5 it's simple
Ashley
@Geogie my bad that was meant for u
Ashley
Hi everyone, I'm glad to be connected with you all. from France.
I'm getting "math processing error" on math problems. Anyone know why?
Can you all help me I don't get any of this
4^×=9