ORCCA Square and nth Root Properties

Section 6.1 Square and $n$ th Root Properties

Objectives: PCC Course Content and Outcome Guide

In this section, we discuss what radical expressions like

\sqrt{50}

and

\sqrt[3]{27}

mean, and learn some properties that allow these expressions to be simplified.

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Figure 6.1.1. Alternative Video Lesson

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Subsection 6.1.1 Square Roots

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Consider the non-negative number

25 .

You can ask the question “what number multiplies by itself to make

25 ?

” We use the symbol

\sqrt{25}

to represent the answer to this question, even if you know the “answer” is

5 .

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A way to visualize this question is with a square that has

25

units of area inside it. What would a side length have to be? We use

\sqrt{25}

to represent that side length.

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Definition 6.1.2. Square Root.

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Given a non-negative number

x,

r \cdot r = x

for some positive number

r,

then

r

is called the square root of

x,

and we can write

\sqrt{x}

instead of

r .

The

\sqrt{}

symbol is called the radical or the root. We call expressions with the

\sqrt{}

symbol radical expressions. The number inside the radical is called the radicand.

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For example, if you see the expression

\sqrt{16},

you should think about the equation

r \cdot r = 16

(or if you prefer, the equation

r^{2} = 16

) and ask yourself if you know a non-negative value for

r

that solves that equation. Of course,

4

is a non-negative solution. So we can say

\sqrt{16} = 4 .

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To demonstrate vocabulary, both

\sqrt{2}

and

3 \sqrt{2}

are radical expressions. In both expressions, the number

2

is the radicand.

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The word “radical” means something like “on the fringes” when used in politics, sports, and other places. It actually has that same meaning in math, when you consider a square with area

A

as in Figure 3.

A drawing of a square labeled with area A; the side lengths are labeled the square root of A — Figure 6.1.3. “Radical” means “off to the side.”

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The one-digit multiplication times table has special numbers along the diagonal. They are known as perfect squares. And for working with square roots, it will be helpful if you can memorize these first few perfect square numbers. For example, the times table tells us that

7 \cdot 7 = 49 .

Just knowing that fact from memory lets us know that

\sqrt{49} = 7 .

It’s advisable to memorize the following:

$\sqrt{0} = 0$	$\sqrt{1} = 1$	$\sqrt{4} = 2$
$\sqrt{9} = 3$	$\sqrt{16} = 4$	$\sqrt{25} = 5$
$\sqrt{36} = 6$	$\sqrt{49} = 7$	$\sqrt{64} = 8$
$\sqrt{81} = 9$	$\sqrt{100} = 10$	$\sqrt{121} = 11$
$\sqrt{144} = 12$	$\sqrt{225} = 15$	$\sqrt{256} = 16$

$\times$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$	$9$
$1$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$	$9$
$2$	$2$	$4$	$6$	$8$	$10$	$12$	$14$	$16$	$18$
$3$	$3$	$6$	$9$	$12$	$15$	$18$	$21$	$24$	$27$
$4$	$4$	$8$	$12$	$16$	$20$	$24$	$28$	$32$	$36$
$5$	$5$	$10$	$15$	$20$	$25$	$30$	$35$	$40$	$45$
$6$	$6$	$12$	$18$	$24$	$30$	$36$	$42$	$48$	$54$
$7$	$7$	$14$	$21$	$28$	$35$	$42$	$49$	$56$	$63$
$8$	$8$	$16$	$24$	$32$	$40$	$48$	$56$	$64$	$72$
$9$	$9$	$18$	$27$	$36$	$45$	$54$	$63$	$72$	$81$

Figure 6.1.4. Multiplication table with squares

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Subsection 6.1.2 Square Root Decimal Values

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Most square roots have decimal expressions that go on forever. Take

\sqrt{5}

as an example. The number

5

is between two perfect squares,

4

and

9 .

Therefore, as demonstrated in Figure 5,

\sqrt{4} < \sqrt{5} < \sqrt{9} .

In other words,

2 < \sqrt{5} < 3

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\sqrt{5}

has a decimal value somewhere between

2

and

3 .

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With a calculator, we can see:

\sqrt{5} = 2.236 \dots

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Actually the decimal will not terminate, and we can write

\sqrt{5} \approx 2.236

with the

\approx

symbol instead of an equal sign. To get

2.236

we rounded down slightly from the true value of

\sqrt{5} .

With a calculator, we can check that

{2.236}^{2} = 4.999696,

a little less than

5,

but close.

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When the radicand is a perfect square, its square root is a rational number. If the radicand is not a perfect square, the square root is irrational. (It has a decimal that goes on forever without any repeating pattern.) We want to be able to estimate square roots without using a calculator.

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Example 6.1.6.

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To estimate

\sqrt{10}

without a calculator, we can find the nearest perfect squares that are whole numbers on either side of

10 .

Recall that the perfect squares are

1, 4, 9, 16, 25, 36, 49, 64, \dots

The perfect square that is just below

10

9

and the perfect square just above

10

16 .

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This tells us that

\sqrt{10}

is between

\sqrt{9}

and

\sqrt{16},

or between

3

and

4 .

We can also say that

\sqrt{10}

is much closer to

3

than

4

because

10

is closer to

9,

so we think

3.1

3.2

would be a good estimate.

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To check our estimates (

3.1

3.2

) we can square them and see if the result is close to

10 .

We find

{3.1}^{2} = 9.61

and

{3.2}^{2} = 10.24,

so our estimates are pretty good.

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Checkpoint 6.1.7.

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Estimate

\sqrt{19}

without a calculator.

Explanation.

The radicand,

19,

is between

16

and

25,

\sqrt{19}

is between

\sqrt{16}

and

\sqrt{25},

or between

4

and

5 .

We notice that

19

is in the middle between

16

and

25

but closer to

16 .

We estimate

\sqrt{19}

to be about

4.4 .

We can check our estimate by calculating:

{4.4}^{2} = 19.36

{4.4}^{2}

is close to

19,

and

4.4

is close to

\sqrt{19} .

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Subsection 6.1.3 Cube Root and Higher Order Roots

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The concept of a square root extends to a cube root. What if we have a number in mind, like

64,

and we would like to know what number could be multiplied with itself three times to make

64 ?

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A way to visualize this question is to imagine a cube with

64

units of volume inside it. What would an edge length have to be? We use

\sqrt[3]{64}

to represent that edge length.

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Definition 6.1.8. $n$ th Root.

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Given a number

x,

\overset{n instances}{\overset{⏞}{r \cdot r \cdot \dots \cdot r}} = x

for some number

r

(or if you prefer,

r^{n} = x

) then

r

is called an

n

th root of

x .

When $n$ is odd, there is always exactly one real number $n$ th root for any $x,$ and we write $\sqrt[n]{x}$ to mean that one $n$ th root.
When $n$ is even and $x$ is positive, there are two real number $n$ th roots, one of which is positive and the other of which is negative. We write $\sqrt[n]{x}$ to mean the positive $n$ th root.
When $n$ is even and $x$ is negative, there aren’t any real number $n$ th roots, and we say that $\sqrt[n]{x}$ is “not a real number”.
When $x$ is $0,$ then $0$ is the $n$ th root. On other words, $\sqrt[n]{0} = 0 .$

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The

\sqrt[n]{}

symbol is called the

n

th radical or the

n

th root. We call expressions with the

\sqrt[n]{}

symbol radical expressions. The number inside the radical is called the radicand. The index of a radical is the number

n

\sqrt[n]{} .

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As we noted earlier, when we have

\sqrt[3]{},

we can say “cube root” instead of “

3

rd root”. Also, when we have

\sqrt[2]{},

we usually say “square root” instead of “

2

nd root” and we usually just write

\sqrt{x}

without the “

2

”.

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Here are some examples of

n

th roots:

$\sqrt[3]{8} = 2,$ because $\overset{3 instances}{\overset{⏞}{2 \cdot 2 \cdot 2}} = 8 .$
$\sqrt[4]{81} = 3,$ because $\overset{4 instances}{\overset{⏞}{3 \cdot 3 \cdot 3 \cdot 3}} = 81 .$
$\sqrt[5]{- 32} = - 2,$ because $\overset{5 instances}{\overset{⏞}{(- 2) \cdot (- 2) \cdot (- 2) \cdot (- 2) \cdot (- 2)}} = - 32 .$

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As with square roots, in general an

n

th root’s decimal value is a decimal that goes on forever. For example,

\sqrt[3]{20} = 2.714 \dots .

For practical applications, we may want to use a calculator to find a decimal approximation to an

n

th root. Some calculators will do this for you directly, and some will not.

Maybe your calculator has a button that looks like $\sqrt[x]{y} .$ Then you should be able to type something like 3 $\sqrt[x]{y}$ 20 to get $\sqrt[3]{20} \approx 2.714 .$
Maybe your calculator has a button that looks like $\sqrt[n]{} .$ Then you should be able to type something like 3 $\sqrt[n]{}$ 20 to get $\sqrt[3]{20} \approx 2.714 .$
Maybe your calculator allows you to type letters, parentheses, and commas, and you can type root(3,20) to get $\sqrt[3]{20} \approx 2.714 .$
Maybe your calculator allows you to type letters, parentheses, and commas, but the syntax for an $n$ th root is reversed from the last example, and you can type root(20,3) to get $\sqrt[3]{20} \approx 2.714 .$
If your calculator has none of the above options, then you should be able to type 20^(1/3) as a way to get $\sqrt[3]{20} \approx 2.714 \dots .$ This is technically using mathematics we will learn later in Section 3.

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Try using your own calculator to calculate

\sqrt[3]{20}

so that you can become familiar with whatever method it uses.

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Fact 6.1.9.

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Since

\sqrt[n]{a}

is can be used to fill in the blanks in the equation

\overset{n times}{\overset{⏞}{() \cdot () \cdot \dots \cdot ()}} = a

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it means that

\sqrt[n]{a}

is a solution to the equation

x^{n} = a .

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Example 6.1.10.

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A solution to

x^{5} = 48

\sqrt[5]{48} .

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Example 6.1.11.

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A pyramid has a square base, and its height is equal to one side length of the square at its base. In this situation, the volume

V

of the pyramid, in in³, is given by

V = \frac{1}{3} s^{3},

where

s

is the pyramid’s base side length in in. Archimedes dropped the pyramid in a bathtub, and judging by how high the water level rose, the volume of the pyramid is 243 in³ (a little more than 1 gal). How tall is the pyramid?

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The equation tells us that:

243 = \frac{1}{3} s^{3}

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We can multiply on both sides by

3

and:

729 = s^{3}

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So we are looking for a solution to the equation

s^{3} = 729 .

This means that

s

\sqrt[3]{729} .

A calculator tells us that this equals

9 .

So the pyramid’s height is

9

inches.

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Subsection 6.1.4 Roots of Negative Numbers

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Can we find the square root of a negative number, like

\sqrt{- 64} ?

How about its cube root,

\sqrt[3]{- 64} ?

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As noted in Definition 8, when the index of an

n

th root is odd, there will always be a real number

n

th root even when the radicand is negative. For example,

\sqrt[3]{- 64}

- 4,

because

(- 4) (- 4) (- 4) = - 64 .

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When the index of an

n

th root is even, it is a problem to have a negative radicand. For example, to find

\sqrt{- 64},

you would need to find a value

r

so that

r \cdot r = - 64 .

But whether

r

is positive or negative, multiplying

r

by itself will give a positive result. It could never be

- 64 .

So there is no way to have a real number square root of a negative number. And the same thing is true for any even index

n

th root with a negative radicand, such as

\sqrt[4]{- 64} .

An even-indexed root of a negative number is not a real number.

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If you are confronted with an expression like

\sqrt{- 25}

\sqrt[4]{- 16}

(any square root or even-indexed root of a negative number), you can state that the expression “is not real” or that it is “not defined” (as a real number). Don’t get carried away though. Expressions like

\sqrt[3]{- 27}

and

\sqrt[5]{- 1}

are defined, because the index is odd.

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Subsection 6.1.5 Radical Properties and Exponent Properties

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In Chapter 5, we learned some algebra properties of exponents. A summary of these properties is in List 5.6.13. There are some similar properties for radicals, presented here without explanation. (These properties are easier to explain once we are in Section 3.)

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List 6.1.12. Properties of Radicals for Multiplication and Division

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a

and

b

are positive real numbers, and

m

is a positive [cross-reference to target(s) "item-integer-definition" missing or not unique], then we have the following properties:

Root of a Product Property: 🔗
$\sqrt[m]{a \cdot b} = \sqrt[m]{a} \cdot \sqrt[m]{b}$
Root of a Quotient Property: 🔗
$\sqrt[m]{\frac{a}{b}} = \frac{\sqrt[m]{a}}{\sqrt[m]{b}}$ as long as $b \neq 0$

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Knowing these algebra properties helps to make complicated radical expressions look simpler.

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Example 6.1.13.

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Simplify

\sqrt{18} .

Anything we can do to make the radicand a smaller simpler number is helpful. Note that

18 = 9 \cdot 2,

so we can write

\begin{aligned} \sqrt{18} & = \sqrt{9 \cdot 2} \\ = \sqrt{9} \cdot \sqrt{2} & according to the Root of a Product Property \\ = 3 \sqrt{2} \end{aligned}

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This expression

3 \sqrt{2}

is considered “simpler” than

\sqrt{18}

because the radicand is so much smaller.

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Checkpoint 6.1.14.

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Simplify

\sqrt{72} .

Explanation.

As with the previous example, it will help if

72

can be written as a product of a perfect square. In this case,

4

divides

72,

and

72 = 4 \cdot 18 .

\begin{aligned} \sqrt{72} & = \sqrt{4 \cdot 18} \\ = \sqrt{4} \cdot \sqrt{18} \\ = 2 \sqrt{18} \end{aligned}

But we aren’t done. Can

18

can be written as a product of a perfect square? Yes, because

18 = 9 \cdot 2 .

\begin{aligned} \sqrt{72} & = 2 \sqrt{18} \\ = 2 \sqrt{9 \cdot 2} \\ = 2 \sqrt{9} \cdot \sqrt{2} \\ = 2 \cdot 3 \sqrt{2} \\ = 6 \sqrt{2} \end{aligned}

This is as simple as we can make this expression.

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Example 6.1.15.

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Simplify

\sqrt[3]{80} .

Anything we can do to make the radicand a smaller simpler number is helpful. With lessons learned from the previous examples, maybe there is a way to rewrite

80

as the product of two numbers in a helpful way. Since we have a cube root, writing

80

as a product of a perfect cube would be helpful. We can write

80 = 8 \cdot 10,

where

8

is a perfect cube.

\begin{aligned} \sqrt[3]{80} & = \sqrt[3]{8 \cdot 10} \\ = \sqrt[3]{8} \cdot \sqrt[3]{10} & according to the Root of a Product Property \\ = 2 \sqrt[3]{10} \end{aligned}

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This expression

2 \sqrt[3]{10}

is considered “simpler” than

\sqrt[3]{80}

because the radicand is so much smaller.

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Checkpoint 6.1.16.

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Simplify

\sqrt[4]{48} .

Explanation.

As with the previous example, it will help if

48

can be written as a product of a

4

th power. In this case,

16

divides

48,

and

48 = 16 \cdot 3 .

\begin{aligned} \sqrt[4]{48} & = \sqrt[4]{16 \cdot 3} \\ = \sqrt[4]{16} \sqrt[4]{3} \\ = 2 \sqrt[4]{3} \end{aligned}

This is as simple as we can make this expression.

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When a radical is applied to a fraction, the Root of a Quotient Property is useful.

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Example 6.1.17.

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Simplify

\sqrt{\frac{8}{25}} .

According to the Root of a Quotient Property,

\begin{aligned} \sqrt{\frac{8}{25}} & = \frac{\sqrt{8}}{\sqrt{25}} \\ = \frac{\sqrt{8}}{5} \\ = \frac{\sqrt{4 \cdot 2}}{5} \\ = \frac{\sqrt{4} \cdot \sqrt{2}}{5} \\ = \frac{2 \sqrt{2}}{5} \end{aligned}

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This is as simple as we can make this expression, unless you prefer to write it as

\frac{2}{5} \sqrt{2} .

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Checkpoint 6.1.18.

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Simplify each of the following.

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(a)

🔗

\sqrt{\frac{1}{36}}

Explanation.

\begin{aligned} \sqrt{\frac{1}{36}} & = \frac{\sqrt{1}}{\sqrt{36}} \\ = \frac{1}{6} \end{aligned}

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(b)

🔗

\sqrt[3]{\frac{8}{27}}

Explanation.

\begin{aligned} \sqrt[3]{\frac{8}{27}} & = \frac{\sqrt[3]{8}}{\sqrt[3]{27}} \\ = \frac{2}{3} \end{aligned}

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(c)

🔗

\sqrt[3]{\frac{81}{125}}

Explanation.

\begin{aligned} \sqrt[3]{\frac{81}{125}} & = \frac{\sqrt[3]{81}}{\sqrt[3]{125}} \\ = \frac{\sqrt[3]{81}}{5} \\ = \frac{\sqrt[3]{27 \cdot 3}}{5} \\ = \frac{\sqrt[3]{27} \sqrt[3]{3}}{5} \\ = \frac{3 \sqrt[3]{3}}{5} \end{aligned}

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Subsection 6.1.6 Multiplying Square Root Expressions

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We can use the Root of a Product Property and the Root of a Quotient Property to multiply and divide square root expressions. We want to simplify each radical first to keep the radicands as small as possible.

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Example 6.1.19.

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Multiply

\sqrt{8} \cdot \sqrt{54} .

Explanation.

We will simplify each radical first, and then multiply them together. We do not want to multiply

8 \cdot 54

because we would end up with a larger number that is harder to factor.

\begin{aligned} \sqrt{8} \cdot \sqrt{54} & = \sqrt{4 \cdot 2} \cdot \sqrt{9 \cdot 6} \\ = \sqrt{4} \cdot \sqrt{2} \cdot \sqrt{9} \cdot \sqrt{6} \\ = 2 \sqrt{2} \cdot 3 \sqrt{6} \\ = 6 \sqrt{2} \cdot \sqrt{6} \\ = 6 \sqrt{12} \\ = 6 \sqrt{4 \cdot 3} \\ = 6 \sqrt{4} \cdot \sqrt{3} \\ = 6 \cdot 2 \sqrt{3} \\ = 12 \sqrt{3} \end{aligned}

It is worth noting that this is considered as simple as we can make it, because the radicand of

3

is so small.

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Checkpoint 6.1.20.

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Multiply

2 \sqrt{7} \cdot 3 \sqrt{21} .

Explanation.

First multiply the non-radical factors together and the radical factors together. Then look for further simplifications.

\begin{aligned} 2 \sqrt{7} \cdot 3 \sqrt{21} & = 2 \cdot 3 \cdot \sqrt{7} \cdot \sqrt{21} \\ = 6 \cdot \sqrt{7} \cdot \sqrt{7 \cdot 3} \\ = 6 \sqrt{49 \cdot 3} \\ = 6 \cdot 7 \cdot \sqrt{3} \\ = 42 \sqrt{3} \end{aligned}

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Example 6.1.21.

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Multiply

\sqrt{\frac{6}{5}} \cdot \sqrt{\frac{3}{5}} .

Explanation.

First multiply the fractions together under the radical. Then look for further simplifications.

\begin{aligned} \sqrt{\frac{6}{5}} \cdot \sqrt{\frac{3}{5}} & = \sqrt{\frac{6}{5} \cdot \frac{3}{5}} \\ = \sqrt{\frac{18}{25}} \\ = \frac{\sqrt{18}}{\sqrt{25}} \\ = \frac{\sqrt{9 \cdot 2}}{5} \\ = \frac{3 \sqrt{2}}{5} \end{aligned}

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Subsection 6.1.7 Adding and Subtracting Square Root Expressions

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We learned the Root of a Product Property previously and applied this to multiplication of square roots, but we cannot apply this property to the operations of addition or subtraction. Here are two examples to demonstrate why not.

\begin{aligned} \sqrt{9} + \sqrt{16} & \overset{?}{=} \sqrt{9 + 16} & \sqrt{169} - \sqrt{25} & \overset{?}{=} \sqrt{169 - 25} \\ 3 + 4 & \overset{?}{=} \sqrt{25} & 13 - 5 & \overset{?}{=} \sqrt{144} \\ 7 & \overset{no}{=} 5 & 8 & \overset{no}{=} 12 \end{aligned}

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We do not get the same result if we combine radical sums and differences in the same way we can combine radical products and quotiens.

🔗

To add and subtract radical expressions, we need to recognize that we can only add and subtract like terms. In this case, we will call them like radicals. Adding like radicals will work just like adding like terms. In the same way that

x + 3 x = 4 x

combines two like terms,

\sqrt{5} + 3 \sqrt{5} = 4 \sqrt{5}

combines two like radicals.

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Example 6.1.22.

🔗

Simplify

\sqrt{2} + \sqrt{8} .

Explanation.

First, simplify each radical. Simplifying is the best way to understand whether or not we even have two like radicals that could be combined.

\begin{aligned} \sqrt{2} + \sqrt{8} & = \sqrt{2} + \sqrt{4 \cdot 2} \\ = \sqrt{2} + 2 \sqrt{2} \\ = 3 \sqrt{2} \end{aligned}

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Checkpoint 6.1.23.

🔗

Simplify

2 \sqrt{3} - 3 \sqrt{48} .

Explanation.

First we will simplify the radical term where

48

is the radicand, and we may see that we then have like radicals.

\begin{aligned} 2 \sqrt{3} - 3 \sqrt{48} & = 2 \sqrt{3} - 3 \sqrt{16 \cdot 3} \\ = 2 \sqrt{3} - 3 \cdot 4 \sqrt{3} \\ = 2 \sqrt{3} - 12 \sqrt{3} \\ = - 10 \sqrt{3} \end{aligned}

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Example 6.1.24.

🔗

Simplify

\sqrt{2} + \sqrt{27} .

Explanation.

\begin{aligned} \sqrt{2} + \sqrt{27} & = \sqrt{2} + \sqrt{9 \cdot 3} \\ = \sqrt{2} + 3 \sqrt{3} \end{aligned}

We cannot simplify the expression further because

\sqrt{2}

and

\sqrt{3}

are not like radicals.

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Example 6.1.25.

🔗

Simplify

\sqrt{6} - \sqrt{18} \cdot \sqrt{12} .

Explanation.

In this example, we should multiply the latter two square roots first (after simplifying them) and then see if we have like radicals.

\begin{aligned} \sqrt{6} - \sqrt{18} \cdot \sqrt{12} & = \sqrt{6} - \sqrt{9 \cdot 2} \cdot \sqrt{4 \cdot 3} \\ = \sqrt{6} - 3 \sqrt{2} \cdot 2 \sqrt{3} \\ = \sqrt{6} - 6 \sqrt{2} \cdot \sqrt{3} \\ = \sqrt{6} - 6 \sqrt{6} \\ = - 5 \sqrt{6} \end{aligned}

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Subsection 6.1.8 Distributing with Square Roots

🔗

In Section 5.4, we learned how to multiply polynomials like

2 (x + 3)

and

(x + 2) (x + 3) .

All the methods we learned there apply when we multiply square root expressions. We will look at a few examples done with different methods.

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Example 6.1.26.

🔗

Multiply

\sqrt{5} (\sqrt{3} - \sqrt{2}) .

Explanation.

We will use the distributive property to do this problem:

\begin{aligned} \sqrt{5} (\sqrt{3} - \sqrt{2}) & = \sqrt{5} \cdot \sqrt{3} - \sqrt{5} \cdot \sqrt{2} \\ = \sqrt{15} - \sqrt{10} \end{aligned}

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Example 6.1.27.

🔗

Multiply

(\sqrt{6} + \sqrt{12}) (\sqrt{3} - \sqrt{2}) .

Explanation.

We will use the FOIL Method to expand the product. There is an opportunity to simplify some of the radicals after multiplying.

\begin{aligned} (\sqrt{6} + \sqrt{12}) (\sqrt{3} - \sqrt{2}) & = \sqrt{6} \cdot \sqrt{3} - \sqrt{6} \cdot \sqrt{2} + \sqrt{12} \cdot \sqrt{3} - \sqrt{12} \cdot \sqrt{2} \\ = \sqrt{18} - \sqrt{12} + \sqrt{36} - \sqrt{24} \\ = 3 \sqrt{2} - 2 \sqrt{3} + 6 - 2 \sqrt{6} \end{aligned}

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Example 6.1.28.

🔗

Expand

{(\sqrt{3} - \sqrt{2})}^{2} .

Explanation.

We will use the FOIL method to expand this expression:

\begin{aligned} {(\sqrt{3} - \sqrt{2})}^{2} & = (\sqrt{3} - \sqrt{2}) (\sqrt{3} - \sqrt{2}) \\ = {(\sqrt{3})}^{2} - \sqrt{3} \cdot \sqrt{2} - \sqrt{2} \cdot \sqrt{3} + {(\sqrt{2})}^{2} \\ = 3 - \sqrt{6} - \sqrt{6} + 2 \\ = 5 - 2 \sqrt{6} \end{aligned}

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Example 6.1.29.

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Multiply

(\sqrt{5} - \sqrt{7}) (\sqrt{5} + \sqrt{7}) .

Explanation.

We can once again use the FOIL method to expand this expression. (But it is worth noting that this expression is in the special form

(a - b) (a + b)

and will simplify to

a^{2} - b^{2} .

)

\begin{aligned} (\sqrt{5} - \sqrt{7}) (\sqrt{5} + \sqrt{7}) & = {(\sqrt{5})}^{2} + \sqrt{5} \cdot \sqrt{7} - \sqrt{7} \cdot \sqrt{5} - {(\sqrt{7})}^{2} \\ = 5 + \sqrt{35} - \sqrt{35} - 7 \\ = - 2 \end{aligned}

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Reading Questions 6.1.9 Reading Questions

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1.

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Is there a difference between

\sqrt[3]{2}

and

3 \sqrt{2} ?

Explain.

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2.

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Choose one of the radical properties from List 12. Then find its counterpart in the exponent properties from List 5.6.13.

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3.

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Describe a way you can visualize

\sqrt{81}

in a geometric shape. Describe a way you can visualize

\sqrt[3]{27}

in a geometric shape.

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Exercises 6.1.10 Exercises

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Exercise Group.

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Which of the following are square numbers? There may be more than one correct answer.

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Evaluate the expression without using a calculator.

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\sqrt{121}

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(b)

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\sqrt{1.21}

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(c)

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\sqrt{0.0121}

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Exercise Group.

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Without using a calculator...

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13.

🔗

Estimate the value of

\sqrt{99} :

🔗

10.95
10.05
9.95
9.05

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14.

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Estimate the value of

\sqrt{19} :

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4.64
3.36
3.64
4.36

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24.

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- \sqrt{\frac{9}{16}}

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Section 6.1 Square and nth Root Properties

Objectives: PCC Course Content and Outcome Guide

Subsection 6.1.1 Square Roots

Definition 6.1.2. Square Root.

Subsection 6.1.2 Square Root Decimal Values

Example 6.1.6.

Checkpoint 6.1.7.

Subsection 6.1.3 Cube Root and Higher Order Roots

Definition 6.1.8. nth Root.

Fact 6.1.9.

Example 6.1.10.

Example 6.1.11.

Subsection 6.1.4 Roots of Negative Numbers

Subsection 6.1.5 Radical Properties and Exponent Properties

Example 6.1.13.

Checkpoint 6.1.14.

Example 6.1.15.

Checkpoint 6.1.16.

Example 6.1.17.

Checkpoint 6.1.18.

(a)

(b)

(c)

Subsection 6.1.6 Multiplying Square Root Expressions

Example 6.1.19.

Checkpoint 6.1.20.

Example 6.1.21.

Subsection 6.1.7 Adding and Subtracting Square Root Expressions

Example 6.1.22.

Checkpoint 6.1.23.

Example 6.1.24.

Example 6.1.25.

Subsection 6.1.8 Distributing with Square Roots

Example 6.1.26.

Example 6.1.27.

Example 6.1.28.

Example 6.1.29.

Reading Questions 6.1.9 Reading Questions

1.

2.

3.

Exercises 6.1.10 Exercises

Exercise Group.

1.

2.

Exercise Group.

3.

(a)

(b)

(c)

4.

(a)

(b)

(c)

5.

(a)

(b)

6.

(a)

(b)

7.

(a)

(b)

(c)

8.

(a)

(b)

(c)

9.

(a)

(b)

(c)

10.

(a)

(b)

(c)

11.

(a)

(b)

(c)

Section 6.1 Square and $n$ th Root Properties

Definition 6.1.8. $n$ th Root.