# 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?

One-fourth of the candies in a bag of M&M’s are red. If there are 23 red candies, how many candies are in the bag?
rectangular field solutions
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Donna
the proudact of 3x^3-5×^2+3 and 2x^2+5x-4 in z7[x]/ is
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Choli
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44.19%
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40.22%
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44.2%
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if you want the discounted price subtract $725 from$1299. then divide the answer by $1299. you get 0.4419... but as percent you get 44.19... but to the nearest tenth... round .19 to .2 and you get 44.2% Orlando you could also just divide$725/$1299 and then subtract it from 1. then you get the same answer. Orlando p mulripied-5 and add 30 to it Tausif Reply p mulripied-5 and add30 Tausif p mulripied-5 and addto30 Tausif Can you explain further Monica Reply p mulripied-5 and add to 30 Tausif How do you find divisible numbers without a calculator? Jacob Reply TAKE OFF THE LAST DIGIT AND MULTIPLY IT 9. SUBTRACT IT THE DIGITS YOU HAVE LEFT. IF THE ANSWER DIVIDES BY 13(OR IS ZERO), THEN YOUR ORIGINAL NUMBER WILL ALSO DIVIDE BY 13!IS DIVISIBLE BY 13 BAINAMA When she graduates college, Linda will owe$43,000 in student loans. The interest rate on the federal loans is 4.5% and the rate on the private bank loans is 2%. The total interest she owes for one year was $1,585. What is the amount of each loan? Ariana Reply Sean took the bus from Seattle to Boise, a distance of 506 miles. If the trip took 7 2/3 hours, what was the speed of the bus? Kirisma Reply 66miles/hour snigdha How did you work it out? Esther s=mi/hr 2/3~0.67 s=506mi/7.67hr = ~66 mi/hr Orlando hello, I have algebra phobia. Subtracting negative numbers always seem to get me confused. Alicia Reply what do you need help in? Felix subtracting a negative....is adding!! Heather look at the numbers if they have different signs, it's like subtracting....but you keep the sign of the largest number... Felix for example.... -19 + 7.... different signs...subtract.... 12 keep the sign of the "largest" number 19 is bigger than 7.... 19 has the negative sign... Therefore, -12 is your answer... Felix —12 Niazmohammad Thanks Felix.l also get confused with signs. Esther Thank you for this Shatey ty Graham think about it like you lost$19 (-19), then found $7(+7). Totally you lost just$12 (-12)
Annushka
I used to struggle a lot with negative numbers and math in general what I typically do is look at it in terms of money I have -$5 in my account I then take out 5 more dollars how much do I have in my account well-$10 ... I also for a long time would draw it out on a number line to visualize it
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practicing with smaller numbers to understand then working with larger numbers helps too and the song/rhyme same sign add and keep opposite signs subtract keep the sign of the bigger # then you'll be exact
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mustee
j 17 w 11
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3
Christopher
18t+14t=96 32t=96 32/96 3
Christopher
show that a^n-b^2n is divisible by a-b By Eric Crawford By Marriyam Rana By Madison Christian By OpenStax By OpenStax By OpenStax By Edgar Delgado By Gerr Zen By Yasser Ibrahim By Richley Crapo