Simplify Expressions with Roots

By the end of this section, you will be able to: simplify expressions with roots, estimate and approximate roots, simplify variable expressions with roots.

Simplify Expressions with Roots

In Foundations, we briefly looked at square roots. Remember that when a real number n is multiplied by itself, we write $n^{2}$ and read it ‘n squared’. This number is called the square of n, and n is called the square root. For example,

\begin{matrix}13^{2}\ \text{is read ``13 squared''}\ \\ \text{169 is called the}\ \textit{square}\ \text{of 13},\text{ since}\ 13^{2} = 169\ \\ \text{13 is a}\ \textit{square root}\ \text{of 169}\end{matrix}

Square and Square Root of a number

Square

\text{If}\ n^{2} = m, \text{then}\ m\ \text{is the}\ \textbf{square}\ \text{of}\ n.

Square Root

\text{If}\ n^{2} = m, \text{then}\ n\ \text{is a}\ \textbf{square root}\ \text{of}\ m.

Notice $(-13)^2=169$ also, so $-13$ is also a square root of 169. Therefore, both 13 and $-13$ are square roots of 169.

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? We use a radical sign, and write, $\sqrt{m},$ which denotes the positive square root of m. The positive square root is also called the principal square root. This symbol, as well as other radicals to be introduced later, are grouping symbols.

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.

Square Root Notation

\begin{matrix} \\ \\ \sqrt{m}\ \text{is read ``the square root of }m\text{.''}\ \\ \text{If}\ n^{2} = m, \text{then}\ n = \sqrt{m}, \text{for}\ n \geq 0.\end{matrix}

In the expression $\sqrt{m}$ , the symbol surrounding $m$ is the radical sign, and the expression $m$ under the radical sign is the radicand.

We know that every positive number has two square roots and the radical sign indicates the positive one. We write $\sqrt{169} = 13.$ If we want to find the negative square root of a number, we place a negative in front of the radical sign. For example, $- \sqrt{169} = -13.$

Example.

Simplify: (a) $\sqrt{144}$ (b) $- \sqrt{289}.$

(a)

Step	Result
	$\sqrt{144}$
Since $12^{2} = 144.$	$12$

(b)

Step	Result
	$- \sqrt{289}$
Since $17^{2} = 289$ and the negative is in front of the radical sign.	$-17$

Simplify: $- \sqrt{64}$

Simplify: $\sqrt{225}$

Simplify: $\sqrt{100}$

Can we simplify $\sqrt{-49} ?$ Is there a number whose square is $-49 ?$

\left( \right)^{2} = -49

Any positive number squared is positive. Any negative number squared is positive. There is no real number equal to $\sqrt{-49}.$ The square root of a negative number is not a real number.

Example.

Simplify: (a) $\sqrt{-196}$ (b) $- \sqrt{64}.$

(a)

Step	Result
	$\sqrt{-196}$
There is no real number whose square is $-196.$	$\sqrt{-196}\ \text{is not a real number}.$

(b)

Step	Result
	$- \sqrt{64}$
The negative is in front of the radical.	$-8$

Simplify: $\sqrt{-169}$

Simplify: $- \sqrt{81}$

So far we have only talked about squares and square roots. Let’s now extend our work to include higher powers and higher roots.

Let’s review some vocabulary first.

$\begin{matrix}\text{We write}: & & & & & \text{We say}: \\ n^{2} & & & & & n\ \text{squared}\ \\ n^{3} & & & & & n\ \text{cubed}\ \\ n^{4} & & & & & n\ \text{to the fourth power}\ \\ n^{5} & & & & & n\ \text{to the fifth power}\end{matrix}$

The terms ‘squared’ and ‘cubed’ come from the formulas for area of a square and volume of a cube.

It will be helpful to have tables of powers of the integers from $-5$ to 5 and of a few variable expressions.

Number	Square	Cube	Fourth power	Fifth power
$n$	$n^2$	$n^3$	$n^4$	$n^5$
$1$	$1$	$1$	$1$	$1$
$2$	$4$	$8$	$16$	$32$
$3$	$9$	$27$	$81$	$243$
$4$	$16$	$64$	$256$	$1{,}024$
$5$	$25$	$125$	$625$	$3{,}125$
$x$	$x^2$	$x^3$	$x^4$	$x^5$
$x^2$	$x^4$	$x^6$	$x^8$	$x^{10}$

Number	Square	Cube	Fourth power	Fifth power
$n$	$n^2$	$n^3$	$n^4$	$n^5$
$-1$	$1$	$-1$	$1$	$-1$
$-2$	$4$	$-8$	$16$	$-32$
$-3$	$9$	$-27$	$81$	$-243$
$-4$	$16$	$-64$	$256$	$-1{,}024$
$-5$	$25$	$-125$	$625$	$-3{,}125$

Notice the signs in the table. All powers of positive numbers are positive, of course. But when we have a negative number, the even powers are positive and the odd powers are negative. We’ll copy the row with the powers of -2 to help you see this.

$n$	$n^2$	$n^3$	$n^4$	$n^5$
$-2$	$4$	$-8$	$16$	$-32$

The odd powers, $n$ , $n^3$ , and $n^5$ , have negative results when $n=-2$ . The even powers, $n^2$ and $n^4$ , have positive results.

We will now extend the square root definition to higher roots.

nth Root of a Number

\begin{matrix}\ \\ \\ \text{If}\ b^{n} = a, \text{then}\ b\ \text{is an}\ n^{t h}\ \text{root of}\ a. \\ \text{The principal}\ n^{t h}\ \text{root of}\ a\ \text{is written}\ \sqrt[n]{a}. \\ n\ \text{is called the}\ \textbf{index}\ \text{of the radical}.\end{matrix}

Just like we use the word ‘cubed’ for $b^3$ , we use the term ‘cube root’ for $\sqrt[3]{a}.$

We can refer to the power tables above to help find higher roots.

\begin{aligned}4^{3} & = & 64 \\ 3^{4} & = & 81 \\ \left(-2\right)^{5} & = & -32\end{aligned} \begin{aligned}\sqrt[3]{64} & = & 4 \\ \sqrt[4]{81} & = & 3 \\ \sqrt[5]{-32} & = & -2\end{aligned}

Could we have an even root of a negative number? We know that the square root of a negative number is not a real number. The same is true for any even root. Even roots of negative numbers are not real numbers. Odd roots of negative numbers are real numbers.

Properties of $\sqrt[n]{a}$

When n is an even number and

$a \geq 0,$ then $\sqrt[n]{a}$ is a real number.
$a < 0,$ then $\sqrt[n]{a}$ is not a real number.

When n is an odd number, $\sqrt[n]{a}$ is a real number for all values of a.

We will apply these properties in the next two examples.

Example.

Simplify: (a) $\sqrt[3]{64}$ (b) $\sqrt[4]{81}$ (c) $\sqrt[5]{32}.$

(a)

Step	Result
	$\sqrt[3]{64}$
Since $4^{3} = 64.$	$4$

(b)

Step	Result
	$\sqrt[4]{81}$
Since $\left(3\right)^{4} = 81.$	$3$

(c)

Step	Result
	$\sqrt[5]{32}$
Since $\left(2\right)^{5} = 32.$	$2$

Simplify: $\sqrt[3]{27}$

Simplify: $\sqrt[4]{256}$

Simplify: $\sqrt[5]{243}$

In this example be alert for the negative signs as well as even and odd powers.

Example.

Simplify: (a) $\sqrt[3]{-125}$ (b) $\sqrt[4]{- 16}$ (c) $\sqrt[5]{-243}.$

(a)

Step	Result
	$\sqrt[3]{-125}$
Since $\left(-5\right)^{3} = -125.$	$-5$

(b)

Step	Result
	$\sqrt[4]{-16}$
Think, $\left(?\right)^{4} = -16.$ No real number raised to the fourth power is negative.	Not a real number.

(c)

Step	Result
	$\sqrt[5]{-243}$
Since $\left(-3\right)^{5} = -243.$	$-3$

Simplify: $\sqrt[3]{-27}$

Simplify: $\sqrt[4]{-256}$

Simplify: $\sqrt[5]{-32}$

Estimate and Approximate Roots

When we see a number with a radical sign, we often don’t think about its numerical value. While we probably know that the $\sqrt{4} = 2,$ what is the value of $\sqrt{21}$ or $\sqrt[3]{50} ?$ In some situations a quick estimate is meaningful and in others it is convenient to have a decimal approximation.

To get a numerical estimate of a square root, we look for perfect square numbers closest to the radicand. To find an estimate of $\sqrt{11},$ we see 11 is between perfect square numbers 9 and 16, closer to 9. Its square root then will be between 3 and 4, but closer to 3.

Number	Square root
$4$	$2$
$9$	$3$
$11$	$\sqrt{11}$
$16$	$4$
$25$	$5$

Thus $9<11<16$ , so $3<\sqrt{11}<4$ .

Number	Cube root
$8$	$2$
$27$	$3$
$64$	$4$
$91$	$\sqrt[3]{91}$
$125$	$5$

Thus $64<91<125$ , so $4<\sqrt[3]{91}<5$ .

Similarly, to estimate $\sqrt[3]{91},$ we see 91 is between perfect cube numbers 64 and 125. The cube root then will be between 4 and 5.

Example.

Estimate each root between two consecutive whole numbers: (a) $\sqrt{105}$ (b) $\sqrt[3]{43}.$

(a) Think of the perfect square numbers closest to 105. Make a small table of these perfect squares and their squares roots.

Step	Result
Start with the expression.	$\sqrt{105}$
The consecutive perfect squares around 105 are $10^2=100$ and $11^2=121$ .	$100\longleftrightarrow10$ and $121\longleftrightarrow11$
Locate 105 between the two consecutive perfect squares.	$100<105<121$
$\sqrt{105}$ is between their square roots.	$10<\sqrt{105}<11$

(b) Similarly we locate 43 between two perfect cube numbers.

Step	Result
Start with the expression.	$\sqrt[3]{43}$
The consecutive perfect cubes around 43 are $3^3=27$ and $4^3=64$ .	$27\longleftrightarrow3$ and $64\longleftrightarrow4$
Locate 43 between the two consecutive perfect cubes.	$27<43<64$
$\sqrt[3]{43}$ is between their cube roots.	$3<\sqrt[3]{43}<4$

Estimate each root between two consecutive whole numbers: $\sqrt{38}$

Estimate each root between two consecutive whole numbers: $\sqrt[3]{93}$

Estimate each root between two consecutive whole numbers: $\sqrt{84}$

There are mathematical methods to approximate square roots, but nowadays most people use a calculator to find square roots. To find a square root you will use the $\sqrt{x}$ key on your calculator. To find a cube root, or any root with higher index, you will use the $\sqrt[y]{x}$ key.

When you use these keys, you get an approximate value. 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

\begin{matrix}\sqrt{5} \approx 2.236067978\ \text{rounded to two decimal places is}\ \sqrt{5} \approx 2.24 \\ \sqrt[4]{93} \approx 3.105422799\ \text{rounded to two decimal places is}\ \sqrt[4]{93} \approx 3.11\end{matrix}

How do we know these values are approximations and not the exact values? Look at what happens when we square them:

\begin{aligned}\left(2.236067978\right)^{2} & = & 5.000000002 \\ \left(2.24\right)^{2} & = & 5.0176\end{aligned} \begin{aligned}\left(3.105422799\right)^{4} & = & 92.999999991 \\ \left(3.11\right)^{4} & = & 93.54951841\end{aligned}

Their squares are close to 5, but are not exactly equal to 5. The fourth powers are close to 93, but not equal to 93.

Example.

Round to two decimal places: (a) $\sqrt{17}$ (b) $\sqrt[3]{49}$ (c) $\sqrt[4]{51}.$

(a)

Step	Result
	$\sqrt{17}$
Use the calculator square root key.	$4.123105626 \ldots$
Round to two decimal places.	$4.12$
	$\sqrt{17} \approx 4.12$

(b)

Step	Result
	$\sqrt[3]{49}$
Use the calculator $\sqrt[y]{x}$ key.	$3.659305710 \ldots$
Round to two decimal places.	$3.66$
	$\sqrt[3]{49} \approx 3.66$

(c)

Step	Result
	$\sqrt[4]{51}$
Use the calculator $\sqrt[y]{x}$ key.	$2.6723451177 \ldots$
Round to two decimal places.	$2.67$
	$\sqrt[4]{51} \approx 2.67$

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

Round to two decimal places: $\sqrt[3]{71}$

Round to two decimal places: $\sqrt[4]{127}$

Simplify Variable Expressions with Roots

The odd root of a number can be either positive or negative. For example,

\begin{aligned} \sqrt[3]{4^3}&=\sqrt[3]{64}=4,\\ \sqrt[3]{(-4)^3}&=\sqrt[3]{-64}=-4. \end{aligned}

In both cases, the number being cubed and the resulting cube root have the same sign.

But what about an even root? We want the principal root, so $\sqrt[4]{625} = 5.$

But notice,

\begin{aligned} \sqrt[4]{5^4}&=\sqrt[4]{625}=5,\\ \sqrt[4]{(-5)^4}&=\sqrt[4]{625}=5. \end{aligned}

The first fourth root returns the original positive number. The second returns 5, not $-5$ , because an even-indexed radical denotes the principal, nonnegative root.

How can we make sure the fourth root of -5 raised to the fourth power is 5? We can use the absolute value. $\left|-5\right| = 5.$ So we say that when n is even $\sqrt[n]{a^{n}} = \left|a\right|.$ This guarantees the principal root is positive.

Simplifying Odd and Even Roots

For any integer $n \geq 2,$

\begin{matrix}\text{when the index}\ n\ \text{is odd} & & & \sqrt[n]{a^{n}} = a\ \\ \text{when the index}\ n\ \text{is even} & & & \sqrt[n]{a^{n}} = \left|a\right|\end{matrix}

We must use the absolute value signs when we take an even root of an expression with a variable in the radical.

Example.

Simplify: (a) $\sqrt{x^{2}}$ (b) $\sqrt[3]{n^{3}}$ (c) $\sqrt[4]{p^{4}}$ (d) $\sqrt[5]{y^{5}}.$

(a) We use the absolute value to be sure to get the positive root.

Step	Result
	$\sqrt{x^{2}}$
Since the index $n$ is even, $\sqrt[n]{a^{n}} = \left	a\right

(b) This is an odd indexed root so there is no need for an absolute value sign.

Step	Result
	$\sqrt[3]{n^{3}}$
Since the index $n$ is odd, $\sqrt[n]{a^{n}} = a.$	$n$

(c)

Step	Result
	$\sqrt[4]{p^{4}}$
Since the index $n\ \text{is even}\ \sqrt[n]{a^{n}} = \left	a\right

(d)

Step	Result
	$\sqrt[5]{y^{5}}$
Since the index $n$ is odd, $\sqrt[n]{a^{n}} = a.$	$y$

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

Simplify: $\sqrt[3]{w^{3}}$

Simplify: $\sqrt[4]{m^{4}}$

What about square roots of higher powers of variables? The Power Property of Exponents says $\left(a^{m}\right)^{n} = a^{m \cdot n}.$ So if we square am, the exponent will become 2m.

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

Looking now at the square root,

$\begin{matrix} & & & \sqrt{a^{2 m}}\ \\ \text{Since}\ \left(a^{m}\right)^{2} = a^{2 m}. & & & \sqrt{\left(a^{m}\right)^{2}}\ \\ \text{Since}\ n\ \text{is even}\ \sqrt[n]{a^{n}} = \left|a\right|. & & & \left|a^{m}\right| \\ & & & \text{So}\ \sqrt{a^{2 m}} = \left|a^{m}\right|.\end{matrix}$

We apply this concept in the next example.

Example.

Simplify: (a) $\sqrt{x^{6}}$ (b) $\sqrt{y^{16}}.$

(a)

Step	Result
	$\sqrt{x^{6}}$
Since $\left(x^{3}\right)^{2} = x^{6}.$	$\sqrt{\left(x^{3}\right)^{2}}$
Since the index $n$ is even $\sqrt{a^{n}} = \left	a\right

(b)

Step	Result
	$\sqrt{y^{16}}$
Since $\left(y^{8}\right)^{2} = y^{16}.$	$\sqrt{\left(y^{8}\right)^{2}}$
Since the index $n$ is even $\sqrt[n]{a^{n}} = \left	a\right
In this case the absolute value sign is not needed as $y^{8}$ is positive.

Simplify: $\sqrt{y^{18}}$

Simplify: $\sqrt{z^{12}}$

Simplify: $\sqrt{m^{4}}$

The next example uses the same idea for higher roots.

Example.

Simplify: (a) $\sqrt[3]{y^{18}}$ (b) $\sqrt[4]{z^{8}}.$

(a)

Step	Result
	$\sqrt[3]{y^{18}}$
Since $\left(y^{6}\right)^{3} = y^{18}.$	$\sqrt[3]{\left(y^{6}\right)^{3}}$
Since $n$ is odd, $\sqrt[n]{a^{n}} = a.$	$y^{6}$

(b)

Step	Result
	$\sqrt[4]{z^{8}}$
Since $\left(z^{2}\right)^{4} = z^{8}.$	$\sqrt[4]{\left(z^{2}\right)^{4}}$
Since $z^{2}$ is positive, we do not need an absolute value sign.	$z^{2}$

Simplify: $\sqrt[4]{u^{12}}$

Simplify: $\sqrt[3]{v^{15}}$

Simplify: $\sqrt[5]{c^{20}}$

In the next example, we now have a coefficient in front of the variable. The concept $\sqrt{a^{2 m}} = \left|a^{m}\right|$ works in much the same way.

\sqrt{16 r^{22}} = 4 \left|r^{11}\right| \text{because}\ \left(4 r^{11}\right)^{2} = 16 r^{22}.

But notice $\sqrt{25 u^{8}} = 5 u^{4}$ and no absolute value sign is needed as $u^4$ is always positive.

Example.

Simplify: (a) $\sqrt{16 n^{2}}$ (b) $- \sqrt{81 c^{2}}.$

(a)

Step	Result
	$\sqrt{16 n^{2}}$
Since $\left(4 n\right)^{2} = 16 n^{2}.$	$\sqrt{\left(4 n\right)^{2}}$
Since the index $n$ is even $\sqrt[n]{a^{n}} = \left	a\right

(b)

Step	Result
	$- \sqrt{81 c^{2}}$
Since $\left(9 c\right)^{2} = 81 c^{2}.$	$- \sqrt{\left(9 c\right)^{2}}$
Since the index $n$ is even $\sqrt[n]{a^{n}} = \left	a\right

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

Simplify: $- \sqrt{100 p^{2}}$

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

This example just takes the idea farther as it has roots of higher index.

Example.

Simplify: (a) $\sqrt[3]{64 p^{6}}$ (b) $\sqrt[4]{16 q^{12}}.$

(a)

Step	Result
	$\sqrt[3]{64 p^{6}}$
Rewrite $64 p^{6}$ as $\left(4 p^{2}\right)^{3}.$	$\sqrt[3]{\left(4 p^{2}\right)^{3}}$
Take the cube root.	$4 p^{2}$

(b)

Step	Result
	$\sqrt[4]{16 q^{12}}$
Rewrite the radicand as a fourth power.	$\sqrt[4]{\left(2 q^{3}\right)^{4}}$
Take the fourth root.	$2 \left

Simplify: $\sqrt[3]{27 x^{27}}$

Simplify: $\sqrt[4]{81 q^{28}}$

Simplify: $\sqrt[3]{125 q^{9}}$

The next examples have two variables.

Example.

Simplify: (a) $\sqrt{36 x^{2} y^{2}}$ (b) $\sqrt{121 a^{6} b^{8}}$ (c) $\sqrt[3]{64 p^{63} q^{9}}.$

(a)

Step	Result
	$\sqrt{36 x^{2} y^{2}}$
Since $\left(6 x y\right)^{2} = 36 x^{2} y^{2}$	$\sqrt{\left(6 x y\right)^{2}}$
Take the square root.	$6 \left

(b)

Step	Result
	$\sqrt{121 a^{6} b^{8}}$
Since $\left(11 a^{3} b^{4}\right)^{2} = 121 a^{6} b^{8}$	$\sqrt{\left(11 a^{3} b^{4}\right)^{2}}$
Take the square root.	$11 \left

(c)

Step	Result
	$\sqrt[3]{64 p^{63} q^{9}}$
Since $\left(4 p^{21} q^{3}\right)^{3} = 64 p^{63} q^{9}$	$\sqrt[3]{\left(4 p^{21} q^{3}\right)^{3}}$
Take the cube root.	$4 p^{21} q^{3}$

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

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

Simplify: $\sqrt[3]{8 x^{30} y^{12}}$

This section is adapted from Intermediate Algebra 2e, Section 8.1: Simplify Expressions with Roots by Lynn Marecek and Andrea Honeycutt Mathis, © OpenStax, licensed under CC BY-NC-SA 4.0. Access the original for free at openstax.org. Changes: reformatted worked solutions for the web; omitted the Be Prepared quiz, media links, self-check reflection, and end-of-section exercise bank; and converted the source Try It practice into interactive exercises.

Roots and Radicals Simplify Radical Expressions

Simplify Expressions with Roots