# 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?

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