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Section P.6 Rational Exponents and Radicals

Before Starting This Section, Review

1 Absolute value (Section P.1 , page 9)
2 Integer exponents (Section P.2 , page 20)
3 Special-product formulas (Section P.3 , page 37)
4 Properties of fractions (Section P.1 , page 13)

Objectives

1 Define and evaluate square roots.
2 Simplify square roots.
3 Define and evaluate nth roots.
4 Combine like radicals.
5 Rationalize denominators or numerators.
6 Define and evaluate rational exponents.

Friction and Tap-Water Flow

Suppose your water company had a large reservoir or water tower 80 meters or more above the level of your cold-water faucet. If there was no friction in the pipes and you turned on your faucet, the water would come pouring out of the faucet at nearly 90 miles per hour! This fact follows from a result known as Bernoulli’s equation, which predicts the velocity of the water (ignoring friction and related turbulence) in this situation to be equal to 2gh−−−√, $\sqrt{2 g h},$ where g is the acceleration due to gravity and h is the height of the water level above a hole from which the water escapes.

In Example 3, we use Bernoulli’s equation to find the velocity of water.

Square Roots

1 Define and evaluate square roots.

When we raise the number 5 to the power 2, we write $5^{2} = 25 .$ The reverse of this squaring process is called finding a square root. In the case of 25, we say that 5 is a square root of 25 and write $\sqrt{25} = 5 .$ In general, we say that b is a square root of a if $b^{2} = a .$ We can’t call 5 the square root of 25 because it is also true that ${(- 5)}^{2} = 25,$ so that $- 5$ is another square root of 25.

The symbol $\sqrt{},$ called a radical sign, is used to distinguish between 5 and $- 5,$ the two square roots of 25. We write $5 = \sqrt{25}$ and $- 5 = - \sqrt{25}$ and call $\sqrt{25}$ the principal square root of 25.

If $\sqrt{a} = b,$ then $a = b^{2} \geq 0 .$ Consequently, the symbol $\sqrt{a}$ denotes a real number only when $a \geq 0 .$ The number or expression under the radical sign is called the radicand, and the radicand together with the radical sign is called a radical.

According to the definition of the square root, the number b must pass two tests for the statement $\sqrt{a} = b$ to be correct: (1) $b^{2} = a$ and (2) $b \geq 0 .$ These tests are used to decide whether the statements below are true or false.

Statement $\sqrt{a} = b$	Test (1) $b^{2} = a ?$	Test (2) $b \geq 0 ?$	Conclusion True when `b` passes both tests
$\sqrt{4} = 2$	$2^{2} = 4$ ✓	$2 \geq 0^{2}$ ✓	True; 2 passed both tests.
$\sqrt{4} = - 2$	${(- 2)}^{2} = 4$ ✓	$- 2 \geq 0$ ✗	False; $- 2$ failed Test (2).
$\sqrt{\frac{1}{9}} = \frac{1}{3}$	${(\frac{1}{3})}^{2} = \frac{1}{9}$ ✓	$\frac{1}{3} \geq 0$ ✓	True; $\frac{1}{3}$ passed both tests.
$\sqrt{7^{2}} = 7$	$7^{2} = 7^{2}$ ✓	$7 \geq 0$ ✓	True; 7 passed both tests.
$\sqrt{{(- 7)}^{2}} = - 7$	${(- 7)}^{2} = {(- 7)}^{2}$ ✓	$- 7 \geq 0$ ✗	False; $- 7$ failed Test (2).

Side Note

Either $| x | = x or | x | = - x .$ Then $| x |^{2} = | x | \cdot | x | = x \cdot x = x^{2}$ or $| x |^{2} = | x | \cdot | x | = (- x) \cdot (- x) = x^{2}$ so $| x |^{2} = x^{2}$

The last Example (letting x represent $- 7$ ) shows that $\sqrt{x^{2}}$ is not always equal to x. Now apply Test (1) and Test (2) to the statement $\sqrt{x^{2}} = | x |,$ where x represents any real number.

Statement	Test (1)	Test (2)	Conclusion
$\sqrt{x^{2}} = \| x \|$	$\| x \|^{2} = x^{2}$ ✓	$\| x \| \geq 0$ ✓	True; $\| x \|$ passed both tests.

If $\sqrt{x} = b,$ then by Test (1), $b^{2} = x$ and by Test (2), b is nonnegative. Replacing b with $\sqrt{x}$ in the equation $b^{2} = x$ yields the following result.

Example 1 Finding the Square Roots of Perfect Squares and Their Ratios

Find.

$\sqrt{121}$
$\sqrt{\frac{25}{81}}$
$\sqrt{\frac{16}{169}}$

Solution

Write each radicand as a perfect square.

$\sqrt{121} = \sqrt{11^{2}} = 11$
$\sqrt{\frac{25}{81}} = \sqrt{\frac{5^{2}}{9^{2}}} = \sqrt{{(\frac{5}{9})}^{2}} = \frac{5}{9}$
$\sqrt{\frac{16}{169}} = \sqrt{\frac{4^{2}}{13^{2}}} = \sqrt{{(\frac{4}{13})}^{2}} = \frac{4}{13}$

Practice Problem 1

Find.
1. $\sqrt{144}$
2. $\sqrt{\frac{1}{49}}$
3. $\sqrt{\frac{4}{64}}$

Simplifying Square Roots

2 Simplify square roots.

When working with square roots, we can simplify a radical expression by removing from under the radical sign perfect squares that are factors of the radicand. For example, to simplify $\sqrt{8},$ we write $\sqrt{8} = \sqrt{4 \cdot 2} = \sqrt{4} \sqrt{2} = 2 \sqrt{2} .$ We rely on the following rules.

Example 2 Simplifying Square Roots

Simplify:

$\sqrt{117}$
$\sqrt{6} \sqrt{12}$
$\sqrt{48 x^{2}}$
$\frac{\sqrt{36}}{\sqrt{3}}$
$\sqrt{\frac{25 y^{3}}{9 x^{2}}}, x \neq 0, y \geq 0$

Solution

$\sqrt{117} = \sqrt{9 \cdot 13} = \sqrt{9} \sqrt{13} = 3 \sqrt{13}$
$\sqrt{6} \sqrt{12} = \sqrt{6 \cdot 12} = \sqrt{6 \cdot 6 \cdot 2} = \sqrt{6^{2} \cdot 2} = \sqrt{6^{2}} \sqrt{2} = 6 \sqrt{2}$
$\sqrt{48 x^{2}} = \sqrt{48} \sqrt{x^{2}} = \sqrt{16 \cdot 3} \sqrt{x^{2}} = \sqrt{16} \sqrt{3} \sqrt{x^{2}} = 4 \sqrt{3} | x |$
$\frac{\sqrt{36}}{\sqrt{3}} = \sqrt{\frac{36}{3}} = \sqrt{12} = \sqrt{4 \cdot 3} = \sqrt{4} \sqrt{3} = 2 \sqrt{3}$
$\begin{array}{rcll} \sqrt{\frac{25 y^{3}}{9 x^{2}}} = \frac{\sqrt{25 y^{3}}}{\sqrt{9 x^{2}}} = \frac{\sqrt{25 \cdot y^{2} \cdot y}}{\sqrt{9 \cdot x^{2}}} = \frac{\sqrt{25} \sqrt{y^{2}} \sqrt{y}}{\sqrt{9} \sqrt{x^{2}}} & = & \frac{5 | y | \sqrt{y}}{3 | x |} & x \neq 0, y \geq 0 \\ = & \frac{5 y \sqrt{y}}{3 | x |} & | y | = y \end{array}$

Practice Problem 2

Simplify.
1. $\sqrt{20}$
2. $\sqrt{6} \sqrt{8}$
3. $\sqrt{12 x^{2}}$
4. $\sqrt{\frac{20 y^{3}}{27 x^{2}}}, x \neq 0, y \geq 0$

Warning

It is important to note that $\sqrt{a + b} \neq \sqrt{a} + \sqrt{b} .$ For example, with $a = 1$ and $b = 4$ , $\sqrt{1 + 4} \neq \sqrt{1} + \sqrt{4}$ because $\sqrt{1 + 4} = \sqrt{5},$ whereas $\sqrt{1} + \sqrt{4} = 1 + 2 = 3$ and $\sqrt{5} \neq 3 .$ Similarly, $\sqrt{a - b} \neq \sqrt{a} - \sqrt{b} .$

Example 3 Calculating Water Flow from a Punctured Water Tower

From the introduction to this section, verify the claim that the water would come pouring out of the faucet at nearly 90 miles per hour.

Solution

We can use Bernoulli’s equation to find the velocity of the water from the faucet. If we ignore friction, the velocity of water will be $\sqrt{2 g h},$ where g is the acceleration due to gravity and h is the height of the water level above the faucet.

We replace h with 80 meters and roughly approximate g by 10 meters per second per second $(g \approx 10 m / s^{2})$ in the expression $\sqrt{2 g h} .$

Thus, the velocity of the water is

$\begin{array}{l} = & \sqrt{2 g h} \\ \approx & \sqrt{2 (10 m / s^{2}) (80 m)} = \sqrt{1600 m^{2} / s^{2}} = 40 meters per second \\ = & 40 \times 60 \times 60 meters per hour 60 \times 60 seconds each hour \\ = & 144 \times 1000 meters per hour \\ \approx & 144 \times 0.6214 mph 1000 m \approx 0.6214 miles \\ = & 89.4816 mph \\ \approx & 90 miles per hour \end{array}$

Practice Problem 3

Repeat Example 3 if the water tower is 45 meters high.

Other Roots

3 Define and evaluate nth roots.

If n is a positive integer, we say that b is an nth root of a if $b^{n} = a .$ We next define the principal nth root of a real number a, denoted by the symbol $\sqrt[n]{a}$ .

We use the same vocabulary for nth roots that we used for square roots. That is, $\sqrt[n]{a}$ is called a radical expression, $\sqrt{}$ is the radical sign, and a is the radicand. The positive integer n is called the index. The index 2 is not written; we write $\sqrt{a}$ rather than $\sqrt[2]{a}$ for the principal square root of a. It is also common practice to call $\sqrt[3]{a}$ the cube root of a.

Example 4 Finding Principal `n`th Roots

Find each root.

$\sqrt[3]{27}$
$\sqrt[3]{- 64}$
$\sqrt[4]{16}$
$\sqrt[4]{{(- 3)}^{4}}$
$\sqrt[8]{- 46}$

Solution

The radicand, 27, is positive; so $\sqrt[3]{27} = 3$ because $3^{3} = 27$ and $3 > 0 .$
The radicand, $- 64,$ is negative and 3 is odd; so $\sqrt[3]{- 64} = - 4$ because ${(- 4)}^{3} = - 64 .$
The radicand, 16, is positive; so $\sqrt[4]{16} = 2$ because $2^{4} = 16$ and $2 > 0 .$
The radicand, ${(- 3)}^{4},$ is positive; so $\sqrt[4]{{(- 3)}^{4}} = 3$ because $3^{4} = {(- 3)}^{4}$ and $3 > 0 .$
The radicand, $- 46,$ is negative and the index, 8, is even; so $\sqrt[8]{- 46}$ is not a real number.

Practice Problem 4

Find each root.
1. $\sqrt[3]{- 8}$
2. $\sqrt[5]{32}$
3. $\sqrt[4]{81}$
4. $\sqrt[6]{- 4}$

The fact that the definition of the principal nth root depends on whether n is even or odd results in two rules to replace the square root rule $\sqrt{x^{2}} = | x | .$

The product and quotient rules for square roots can be extended to apply to nth roots as well.

Example 5 Simplifying `n`th Roots

Simplify.

$\sqrt[3]{135}$
$\sqrt[4]{162 a^{4}}$
$\sqrt[3]{\frac{5}{64}}$
$\sqrt[4]{\sqrt[3]{7}}$

Solution

We find a factor of 135 that is a perfect cube. Here $135 = 27 \cdot 5$ and 27 is a perfect cube $(3^{3} = 27) .$ Then

$\sqrt[3]{135} = \sqrt[3]{27 \cdot 5} = \sqrt[3]{27} \sqrt[3]{5} = 3 \sqrt[3]{5} .$
We find a factor of 162 that is a perfect fourth power. Here $162 = 81 \cdot 2$ and 81 is a perfect fourth power: $3^{4} = 81 .$ Thus,

$\sqrt[4]{162 a^{4}} = \sqrt[4]{162} \sqrt[4]{a^{4}} = \sqrt[4]{81 \cdot 2} | a | = \sqrt[4]{81} \sqrt[4]{2} | a | = 3 \sqrt[4]{2} | a | .$
$\begin{array}{rcll} \sqrt[3]{\frac{5}{64}} & = & \frac{\sqrt[3]{5}}{\sqrt[3]{64}} & Quotient rule \\ = & \frac{\sqrt[3]{5}}{4} & 64 = 4^{3}, so \sqrt[3]{64} = 4 \end{array}$
$\begin{array}{rcll} \sqrt[4]{\sqrt[3]{7}} & = & \sqrt[4 \cdot 3]{7} & Power rule \\ = & \sqrt[12]{7} \end{array}$

Practice Problem 5

Simplify.
1. $\sqrt[3]{72}$
2. $\sqrt[4]{48 a^{2}}$
3. $\sqrt[3]{\sqrt{64}}$

Like Radicals

4 Combine like radicals.

Radicals that have the same index and the same radicand are called like radicals. Like radicals can be combined with the use of the distributive property.

Example 6 Adding and Subtracting Radical Expressions

Simplify:

$\sqrt{45} + 7 \sqrt{20}$
$5 \sqrt[3]{80 x} - 3 \sqrt[3]{270 x}$

Solution

We find perfect-square factors for both 45 and 20.

$\begin{array}{rcll} \sqrt{45} + 7 \sqrt{20} & = & \sqrt{9 \cdot 5} + 7 \sqrt{4 \cdot 5} & 45 = 9 \cdot 5; 20 = 4 \cdot 5 \\ = & \sqrt{9} \sqrt{5} + 7 \sqrt{4} \sqrt{5} & Product rule \\ = & 3 \sqrt{5} + 7 \cdot 2 \sqrt{5} & \sqrt{9} = 3; \sqrt{4} = 2 \\ = & 3 \sqrt{5} + 14 \sqrt{5} & Like radicals \\ = & (3 + 14) \sqrt{5} & Distributive property \\ = & 17 \sqrt{5} \end{array}$
This time we find perfect-cube factors for both 80 and 270.

$\begin{array}{rcll} 5 \sqrt[3]{80 x} - 3 \sqrt[3]{270 x} & = & 5 \sqrt[3]{8 \cdot 10 x} - 3 \sqrt[3]{27 \cdot 10 x} & 80 = 8 \cdot 10; 270 = 27 \cdot 10 \\ = & 5 \sqrt[3]{8} \cdot \sqrt[3]{10 x} - 3 \sqrt[3]{27} \cdot \sqrt[3]{10 x} & Product rule \\ = & 5 \cdot 2 \cdot \sqrt[3]{10 x} - 3 \cdot 3 \sqrt[3]{10 x} & \sqrt[3]{8} = 2; \sqrt[3]{27} = 3 \\ = & 10 \sqrt[3]{10 x} - 9 \sqrt[3]{10 x} & Like radicals \\ = & (10 - 9) \sqrt[3]{10 x} & Distributive property \\ = & \sqrt[3]{10 x} \end{array}$

Practice Problem 6

Simplify.
1. $3 \sqrt{12} + 7 \sqrt{3}$
2. $2 \sqrt[3]{125 x} - 3 \sqrt[3]{40 x}$

Warning

In general:

$\sqrt[n]{a^{n} + b^{n}} \neq a + b$

$\sqrt[n]{a + b} \neq \sqrt[n]{a} + \sqrt[n]{b}$

$\sqrt[n]{a} + \sqrt[n]{b}$ cannot be added if $a \neq b .$

$\sqrt[n]{a} + \sqrt[n]{a} = 2 \sqrt[n]{a}$

$\sqrt[n]{a} + \sqrt[m]{a}$ cannot be added if $n \neq m .$

Radicals with Different Indexes

Suppose a is a positive real number and m, n, and r are positive integers. Then the property

$\sqrt[n]{a^{m}} = \sqrt[n r]{a^{m r}}$

is used to multiply radicals with different indexes.

Example 7 Multiplying Radicals with Different Indexes

Write $\sqrt{2} \sqrt[3]{5}$ as a single radical.
Write $\sqrt[4]{x^{3}}$ , $\sqrt[3]{y^{2}}$ , and $\sqrt[6]{z^{5}}$ as radicals with the same lowest index.

Solution

Because $\sqrt{2}$ can also be written as $\sqrt[2]{2}$ we have

$\sqrt{2} = \sqrt[2]{2} = \sqrt[2 \cdot 3]{2^{3}} = \sqrt[6]{2^{3}},$

and similarly,

$\sqrt[3]{5} = \sqrt[2 \cdot 3]{5^{2}} = \sqrt[6]{5^{2}} .$

So

$\sqrt{2} \sqrt[3]{5} = \sqrt[6]{2^{3}} \sqrt[6]{5^{2}} = \sqrt[6]{2^{3} 5^{2}} = \sqrt[6]{200} . \sqrt[n]{a} \cdot \sqrt[n]{b} = \sqrt[n]{a b}$
To express each of the three radicals with the same index of lowest order, we choose the least common multiple of the three indexes. Here, the least common multiple of 4, 3, and 6 is 12. We express each radical with index 12. We have:

$\begin{array}{rcl} \sqrt[4]{x^{3}} & = & \sqrt[4 \cdot 3]{x^{3 \cdot 3}} & = & \sqrt[12]{x^{9}}, \\ \sqrt[3]{y^{2}} & = & \sqrt[3 \cdot 4]{y^{2 \cdot 4}} & = & \sqrt[12]{y^{8}}, \\ \sqrt[6]{z^{5}} & = & \sqrt[6 \cdot 2]{z^{5 \cdot 2}} & = & \sqrt[12]{z^{10}}, \end{array}$

Practice Problem 7

1. Write $\sqrt{3} \sqrt[5]{2}$ as a single radical.
2. Write $\sqrt[6]{x^{5}}$ and $\sqrt[8]{y^{7}}$ with the same lowest index.

Rationalizing Radical Expressions

5 Rationalize denominators or numerators.

Removing radicals in the denominator or the numerator is called rationalizing the denominator or rationalizing the numerator, respectively. The procedure for rationalizing involves multiplying the fraction by 1, written in a special way, so as to obtain a perfect nth power.

Example 8 Rationalizing the Denominator

Rationalize each denominator.

$\frac{\sqrt{5}}{\sqrt{2}}$
$\frac{\sqrt[3]{4}}{\sqrt[3]{9}}$

Solution

$\frac{\sqrt{5}}{\sqrt{2}} = \frac{\sqrt{5}}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{10}}{\sqrt{4}} = \frac{\sqrt{10}}{2}$
$\frac{\sqrt[3]{4}}{\sqrt[3]{9}} = \frac{\sqrt[3]{4}}{\sqrt[3]{9}} \cdot \frac{\sqrt[3]{3}}{\sqrt[3]{3}} = \frac{\sqrt[3]{12}}{\sqrt[3]{27}} = \frac{\sqrt[3]{12}}{3}$

Practice Problem 8

Rationalize each denominator.
1. $\frac{7}{\sqrt{8}}$
2. $\frac{\sqrt[3]{3}}{\sqrt[3]{4}}$

Conjugates

Pairs of expressions of the form $a \sqrt{x} + b \sqrt{y}$ and $a \sqrt{x} - b \sqrt{y}$ are called conjugates. We form the product of these two conjugates.

$\begin{array}{rcll} (a \sqrt{x} + b \sqrt{y}) (a \sqrt{x} - b \sqrt{y}) & = & {(a \sqrt{x})}^{2} - {(b \sqrt{y})}^{2} & Difference of two squares \\ = & a^{2} x - b^{2} y & {(a c)}^{2} = a^{2} c^{2}; {(\sqrt{z})}^{2} = z \end{array}$

We see that the product contains no radicals. So the product of conjugates can be used to rationalize the denominator or the numerator.

Example 9 Rationalizing the Numerator

Rationalize each numerator.

$\frac{\sqrt{5} - \sqrt{2}}{4}$
$\frac{\sqrt{x} + \sqrt{3}}{x^{2} - 9}$

Solution

$\begin{array}{rcll} \frac{\sqrt{5} - \sqrt{2}}{4} & = & \frac{\sqrt{5} - \sqrt{2}}{4} \cdot \frac{\sqrt{5} + \sqrt{2}}{\sqrt{5} + \sqrt{2}} \\ = & \frac{{(\sqrt{5})}^{2} - {(\sqrt{2})}^{2}}{4 (\sqrt{5} + \sqrt{2})} (a - b) (a + b) = a^{2} - b^{2} \\ = & \frac{5 - 2}{4 (\sqrt{5} + \sqrt{2})} = \frac{3}{4 (\sqrt{5} + \sqrt{2})} \end{array}$
$\begin{array}{rcll} \frac{\sqrt{x} + \sqrt{3}}{x^{2} - 9} & = & \frac{\sqrt{x} + \sqrt{3}}{x^{2} - 9} \cdot \frac{\sqrt{x} - \sqrt{3}}{\sqrt{x} - \sqrt{3}} \\ = & \frac{x - 3}{(x^{2} - 9) (\sqrt{x} - \sqrt{3})} (\sqrt{x} + \sqrt{3}) (\sqrt{x} - \sqrt{3}) = x - 3 \\ = & \frac{x - 3}{(x - 3) (x + 3) (\sqrt{x} - \sqrt{3})} x^{2} - 9 = (x - 3) (x + 3) . \\ = & \frac{1}{(x + 3) (\sqrt{x} - \sqrt{3})} Remove common factor . \end{array}$

Practice Problem 9

Rationalize each denominator.
1. $\frac{2}{\sqrt{7} - \sqrt{5}}$
2. $\frac{x - 2}{\sqrt{x} + \sqrt{2}}$

Rational Exponents

6 Define and evaluate rational exponents.

Previously, we defined such exponential expressions as $2^{3},$ but what do you think $2^{1 / 3}$ means? If we want to preserve the power rule of exponents, ${(a^{n})}^{m} = a^{n m},$ for rational exponents such as $\frac{1}{3},$ we must have

${(2^{1 / 3})}^{3} = 2^{1 / 3 \cdot 3} = 2^{1} = 2 .$

Because ${(2^{1 / 3})}^{3} = 2,$ it follows that $2^{1 / 3}$ is the cube root of 2, that is, $2^{1 / 3} = \sqrt[3]{2} .$

Such considerations lead to the following definition.

Example 10 Evaluating Expressions with Rational Exponents

Evaluate:

$16^{1 / 2}$
${(- 27)}^{1 / 3}$
${(\frac{1}{16})}^{1 / 4}$
$32^{1 / 5}$
${(- 5)}^{1 / 4}$

Solution

$16^{1 / 2} = \sqrt{16} = 4$
${(- 27)}^{1 / 3} = \sqrt[3]{- 27} = \sqrt[3]{{(- 3)}^{3}} = - 3$
${(\frac{1}{16})}^{1 / 4} = \sqrt[4]{\frac{1}{16}} = \sqrt[4]{{(\frac{1}{2})}^{4}} = \frac{1}{2}$
$32^{1 / 5} = \sqrt[5]{32} = \sqrt[5]{2^{5}} = 2$
Because the base $- 5$ is negative and $n = 4$ is even, ${(- 5)}^{1 / 4}$ is not a real number.

Practice Problem 10

Evaluate.
1. ${(\frac{1}{4})}^{1 / 2}$
2. ${(125)}^{1 / 3}$
3. ${(- 32)}^{1 / 5}$

The box below shows a summary of our work.

The Exponent $\frac{1}{n}$ and Radicals

If a is any real number and n is any integer greater than 1, then

	`n` even	`n` odd
If $a > 0,$	$\sqrt[n]{a} = a^{1 / n}$ is positive.	$\sqrt[n]{a} = a^{1 / n}$ is positive.
If $a < 0,$	$\sqrt[n]{a} = a^{1 / n}$ is not a real number.	$\sqrt[n]{a} = a^{1 / n}$ is negative.
If $a = 0,$	$\sqrt[n]{0} = 0^{1 / n} = 0.$	$\sqrt[n]{0} = 0^{1 / n} = 0.$

We have defined the expression $a^{1 / n} .$ How do we define $a^{m / n},$ where m and n are integers? If we insist that the power-of-a-product rule of exponents holds for rational numbers, then we must have

$a^{m / n} = a^{1 / n \cdot m} = {(a^{1 / n})}^{m} = {(\sqrt[n]{a})}^{m}$

and

$a^{m / n} = a^{m \cdot 1 / n} = {(a^{m})}^{1 / n} = \sqrt[n]{a^{m}} .$

From this, we arrive at the following definition.

Note that for $a^{m / n}$ the numerator m of the exponent $\frac{m}{n}$ is the exponent of the radical expression and the denominator n is the index of the radical in ${(\sqrt[n]{a})}^{m}$ . When n is even and $a < 0,$ the symbol $a^{m / n}$ is not a real number.

When rational exponents with negative denominators such as $\frac{3}{- 5}$ and $\frac{- 2}{- 7}$ are encountered, we use equivalent expressions such as $\frac{- 3}{5}$ and $\frac{2}{7},$ respectively, before applying the definition. Similarly, we reduce the exponent $\frac{m}{n}$ to its lowest terms before applying the definition. Note that ${(- 8)}^{2 / 6} \neq {[{(- 8)}^{1 / 6}]}^{2}$ because ${(- 8)}^{1 / 6} = \sqrt[6]{- 8}$ is not a real number but ${(- 8)}^{2 / 6}$ is a real number. Writing $\frac{2}{6} = \frac{1}{3}$ , we have

${(- 8)}^{2 / 6} = {(- 8)}^{1 / 3} = \sqrt[3]{- 8} = - 2.$

Example 11 Evaluating Expressions Having Rational Exponents

Evaluate:

$8^{2 / 3}$
$- 16^{5 / 2}$
$100^{- 3 / 2}$
${(- 25)}^{7 / 2}$

Solution

$8^{2 / 3} = {(\sqrt[3]{8})}^{2} = 2^{2} = 4$
$- 16^{5 / 2} = - {(\sqrt{16})}^{5} = - 4^{5} = - 1024$
$100^{- 3 / 2} = {(100^{1 / 2})}^{- 3} = {(\sqrt{100})}^{- 3} = \frac{1}{{(\sqrt{100})}^{3}} = \frac{1}{10^{3}} = \frac{1}{1000}$
$\begin{array}{rcll} {(- 25)}^{7 / 2} & = & {[{(- 25)}^{1 / 2}]}^{7} & Not a real number because \\ = & {(\sqrt{- 25})}^{7} & \sqrt{- 25} is not a real number \end{array}$

Practice Problem 11

Evaluate.
1. ${(25)}^{2 / 3}$
2. $- 36^{3 / 2}$
3. $16^{- 5 / 2}$
4. ${(- 36)}^{- 1 / 2}$

All of the properties of exponents we learned for integer exponents hold true for rational exponents. For convenience, we list these properties next.

Example 12 Simplifying Expressions Having Rational Exponents

Simplify. Express the answer with only positive exponents. Assume that x represents a positive real number.

$2 x^{1 / 3} \cdot 5 x^{1 / 4}$
$\frac{21 x^{- 2 / 3}}{7 x^{1 / 5}}$
${(x^{3 / 5})}^{- 1 / 6}$

Solution

$\begin{array}{rcll} 2 x^{1 / 3} \cdot 5 x^{1 / 4} & = & 2 \cdot 5 x^{1 / 3} \cdot x^{1 / 4} & Group factors with the same base . \\ = & 10 x^{1 / 3 + 1 / 4} & Add exponents of factors with the \\ same base . \\ = & 10 x^{4 / 12 + 3 / 12} = 10 x^{7 / 12} \end{array}$
$\begin{array}{rcl} \frac{21 x^{- 2 / 3}}{7 x^{1 / 5}} & = & (\frac{21}{7}) (\frac{x^{- 2 / 3}}{x^{1 / 5}}) Group factors with the same base . \\ = & 3 x^{- 2 / 3 - 1 / 5} = 3 x^{- 10 / 15 - 3 / 15} = 3 x^{- 13 / 15} = 3 \cdot (\frac{1}{x^{13 / 15}}) = \frac{3}{x^{13 / 15}} \end{array}$
${(x^{3 / 5})}^{- 1 / 6} = x^{(3 / 5) (- 1 / 6)} = x^{- 1 / 10} = \frac{1}{x^{1 / 10}}$

Practice Problem 12

Simplify. Express your answer with only positive exponents.
1. $4 x^{1 / 2} \cdot 3 x^{1 / 5}$
2. $\frac{25 x^{- 1 / 4}}{5 x^{1 / 3}}, x > 0$
3. ${(x^{2 / 3})}^{- 1 / 5}, x \neq 0$

Example 13 Simplifying Radicals by Using Rational Exponents

Assume that x represents a positive real number. Simplify.

$\sqrt[8]{x^{2}}$
$\sqrt[4]{9} \sqrt{3}$
$\sqrt{\sqrt[3]{x^{8}}}$

Solution

$\sqrt[8]{x^{2}} = {(x^{2})}^{1 / 8} = x^{2 / 8} = x^{1 / 4} = \sqrt[4]{x}$
$\sqrt[4]{9} \sqrt{3} = 9^{1 / 4} \cdot 3^{1 / 2} = {(3^{2})}^{1 / 4} \cdot 3^{1 / 2} = 3^{2 / 4} \cdot 3^{1 / 2} = 3^{1 / 2} \cdot 3^{1 / 2} = 3^{1 / 2 + 1 / 2} = 3^{1} = 3$
$\sqrt{\sqrt[3]{x^{8}}} = \sqrt{{(x^{8})}^{1 / 3}} = \sqrt{x^{8 / 3}} = {(x^{8 / 3})}^{1 / 2} = x^{8 / 3 \cdot 1 / 2} = x^{4 / 3} = x^{1 + 1 / 3} = x^{1} \cdot x^{1 / 3} = x \sqrt[3]{x}$

Practice Problem 13

Simplify.
1. $\sqrt[6]{x^{4}}$
2. $\sqrt[4]{25} \sqrt{5}$
3. $\sqrt{\sqrt[3]{x^{12}}}$

Example 14 Simplifying an Expression Involving Negative Rational Exponents

Simplify $x {(x + 1)}^{- 1 / 2} + 2 {(x + 1)}^{1 / 2} .$ Express the answer with only positive exponents and then rationalize the denominator. Assume that ${(x + 1)}^{- 1 / 2}$ is defined.

Solution

$\begin{array}{rcl} x {(x + 1)}^{- 1 / 2} + 2 {(x + 1)}^{1 / 2} & = & \frac{x}{{(x + 1)}^{1 / 2}} + \frac{2 {(x + 1)}^{1 / 2}}{1} {(x + 1)}^{- 1 / 2} = \frac{1}{{(x + 1)}^{1 / 2}} \\ = & \frac{x}{{(x + 1)}^{1 / 2}} + \frac{2 {(x + 1)}^{1 / 2} {(x + 1)}^{1 / 2}}{{(x + 1)}^{1 / 2}} \frac{a}{1} = \frac{a b}{b} \\ = & \frac{x}{{(x + 1)}^{1 / 2}} + \frac{2 (x + 1)}{{(x + 1)}^{1 / 2}} {(x + 1)}^{1 / 2} {(x + 1)}^{1 / 2} = x + 1 \\ = & \frac{x + 2 (x + 1)}{{(x + 1)}^{1 / 2}} = \frac{3 x + 2}{{(x + 1)}^{1 / 2}} Add fractions and simplify . \end{array}$

To rationalize the denominator ${(x + 1)}^{1 / 2} = \sqrt{x + 1},$ multiply both numerator and denominator by ${(x + 1)}^{1 / 2} .$

$\begin{array}{rcll} \frac{3 x + 2}{{(x + 1)}^{1 / 2}} & = & \frac{(3 x + 2) {(x + 1)}^{1 / 2}}{{(x + 1)}^{1 / 2} {(x + 1)}^{1 / 2}} \\ = & \frac{(3 x + 2) {(x + 1)}^{1 / 2}}{x + 1} & {(x + 1)}^{1 / 2} {(x + 1)}^{1 / 2} = x + 1 \end{array}$

Practice Problem 14

Simplify $x {(x + 3)}^{- 1 / 2} + {(x + 3)}^{1 / 2} .$

Section P.6 Exercises

Concepts and Vocabulary

Any positive number has square roots.
To rationalize the denominator of an expression with denominator $2 \sqrt{5}$ , multiply the and by .
Radicals that have the same index and the same radicand are called .
The radical notation for $7^{1 / 3}$ is .
True or False. For all real x, $\sqrt{x^{2}} = x$ .
True or False. The radicals $- 3 \sqrt{5}$ and $12 \sqrt{5}$ are “like radicals.”
True or False. $\frac{1}{a^{- \frac{m}{n}}} = a^{\frac{m}{n}}$
True or False. $9^{- \frac{1}{2}} = - 3$

Building Skills

In Exercises 9–24, evaluate each root or state that the root is not a real number.

$\sqrt{64}$
$\sqrt{100}$
$\sqrt[3]{64}$
$\sqrt[3]{125}$
$\sqrt[3]{- 27}$
$\sqrt[3]{- 216}$
$\sqrt[3]{- \frac{1}{8}}$
$\sqrt[3]{- \frac{27}{64}}$
$\sqrt{{(- 3)}^{2}}$
$- \sqrt{4 {(- 3)}^{2}}$
$\sqrt[4]{- 16}$
$\sqrt[6]{- 64}$
$- \sqrt[5]{- 1}$
$\sqrt[7]{- 1}$
$\sqrt[5]{{(- 7)}^{5}}$
$- \sqrt[5]{{(- 4)}^{5}}$

In Exercises 25–48, simplify each expression using the product and quotient properties for square roots. Assume that x represents a positive real number.

$\sqrt{32}$
$\sqrt{125}$
$\sqrt{18 x^{2}}$
$\sqrt{27 x^{2}}$
$\sqrt{9 x^{3}}$
$\sqrt{8 x^{3}}$
$\sqrt{6 x} \sqrt{3 x}$
$\sqrt{5 x} \sqrt{10 x}$
$\sqrt{15 x} \sqrt{3 x^{2}}$
$\sqrt{2 x^{2}} \sqrt{18 x}$
$\sqrt[3]{- 8 x^{3}}$
$\sqrt[3]{64 x^{3}}$
$\sqrt[3]{- x^{6}}$
$\sqrt[3]{- 27 x^{6}}$
$\sqrt{\frac{5}{32}}$
$\sqrt{\frac{3}{50}}$
$\sqrt[3]{\frac{8}{x^{3}}}$
$\sqrt[3]{\frac{7}{8 x^{3}}}$
$\sqrt[4]{\frac{2}{x^{5}}}$
$- \sqrt[4]{\frac{5}{16 x^{5}}}$
$\sqrt{\sqrt{x^{8}}}$
$\sqrt{\frac{4 x^{4}}{25 y^{6}}}$
$\sqrt{\frac{9 x^{6}}{y^{4}}}$
$\sqrt[5]{x^{15} y^{5}}$

In Exercises 49–62, simplify each expression. Assume that all variables represent nonnegative real numbers.

$2 \sqrt{3} + 5 \sqrt{3}$
$7 \sqrt{2} - \sqrt{2}$
$6 \sqrt{5} - \sqrt{5} + 4 \sqrt{5}$
$5 \sqrt{7} + 3 \sqrt{7} - 2 \sqrt{7}$
$\sqrt{98 x} - \sqrt{32 x}$
$\sqrt{45 x} + \sqrt{20 x}$
$\sqrt[3]{24} - \sqrt[3]{81}$
$\sqrt[3]{54} + \sqrt[3]{16}$
$\sqrt[3]{3 x} - 2 \sqrt[3]{24 x} + \sqrt[3]{375 x}$
$\sqrt[3]{16 x} + 3 \sqrt[3]{54 x} - \sqrt[3]{2 x}$
$\sqrt{2 x^{5}} - 5 \sqrt{32 x} + \sqrt{18 x^{3}}$
$2 \sqrt{50 x^{5}} + 7 \sqrt{2 x^{3}} - 3 \sqrt{72 x}$
$\sqrt{48 x^{5} y} - 4 y \sqrt{3 x^{3} y} + y \sqrt{3 x y^{3}}$
$x \sqrt{8 x y} + 4 \sqrt{2 x y^{3}} - \sqrt{18 x^{5} y}$

In Exercises 63–76, rationalize the denominator of each expression.

$\frac{2}{\sqrt{3}}$
$\frac{10}{\sqrt{5}}$
$\frac{7}{\sqrt{15}}$
$\frac{- 3}{\sqrt{8}}$
$\frac{1}{\sqrt{2} + x}$
$\frac{1}{\sqrt{5} + 2 x}$
$\frac{3}{2 - \sqrt{3}}$
$\frac{- 5}{1 - \sqrt{2}}$
$\frac{1}{\sqrt{3} + \sqrt{2}}$
$\frac{6}{\sqrt{5} - \sqrt{3}}$
$\frac{\sqrt{5} - \sqrt{2}}{\sqrt{5} + \sqrt{2}}$
$\frac{\sqrt{7} + \sqrt{3}}{\sqrt{7} - \sqrt{3}}$
$\frac{\sqrt{x + h} - \sqrt{x}}{\sqrt{x + h} + \sqrt{x}}$
$\frac{a \sqrt{x} + 3}{a \sqrt{x} - 3}$

In Exercises 77–84, rationalize the numerator of each expression.

$\frac{\sqrt{4 + h} - 2}{h}$
$\frac{\sqrt{y + 9} - 3}{y}$
$\frac{2 - \sqrt{4 - x}}{x}$
$\frac{\sqrt{x + 2} - \sqrt{2}}{x}$
$\frac{\sqrt{x} - 2}{x - 4}$
$\frac{\sqrt{x} - \sqrt{5}}{x^{2} - 25}$
$\frac{\sqrt{x^{2} + 4 x} - x}{2}$
$\frac{\sqrt{x^{2} + 2 x + 3} - \sqrt{x^{2} + 3}}{2}$

In Exercises 85–94, evaluate each expression without using a calculator.

$25^{1 / 2}$
$144^{1 / 2}$
${(- 8)}^{1 / 3}$
${(- 27)}^{1 / 3}$
$8^{2 / 3}$
$16^{3 / 2}$
$- 25^{- 3 / 2}$
$- 9^{- 3 / 2}$
${(\frac{9}{25})}^{- 3 / 2}$
${(\frac{1}{27})}^{- 2 / 3}$

In Exercises 95–106, simplify each expression, leaving your answer with only positive exponents. Assume that all variables represent positive numbers.

$x^{1 / 2} \cdot x^{2 / 5}$
$x^{3 / 5} \cdot 5 x^{2 / 3}$
$x^{3 / 5} \cdot x^{- 1 / 2}$
$x^{5 / 3} \cdot x^{- 3 / 4}$
${(8 x^{6})}^{2 / 3}$
${(16 x^{3})}^{1 / 2}$
${(27 x^{6} y^{3})}^{- 2 / 3}$
${(16 x^{4} y^{6})}^{- 3 / 2}$
$\frac{15 x^{3 / 2}}{3 x^{1 / 4}}$
$\frac{20 x^{5 / 2}}{4 x^{2 / 3}}$
${(\frac{x^{- 1 / 4}}{y^{- 2 / 3}})}^{- 12}$
${(\frac{27 x^{- 5 / 2}}{y^{- 3}})}^{- 1 / 3}$

In Exercises 107–116, convert each radical expression to its rational exponent form and then simplify. Assume that all variables represent positive numbers.

$\sqrt[4]{3^{2}}$
$\sqrt[4]{5^{2}}$
$\sqrt[3]{x^{9}}$
$\sqrt[3]{x^{12}}$
$\sqrt[3]{x^{6} y^{9}}$
$\sqrt[3]{8 x^{3} y^{12}}$
$\sqrt[4]{9} \sqrt{3}$
$\sqrt[4]{49} \sqrt{7}$
$\sqrt{\sqrt[3]{x^{10}}}$
$\sqrt{\sqrt[3]{64 x^{6}}}$

In Exercises 117–124, convert the given product to a single radical with lowest index. Assume that all variables represent positive numbers.

$\sqrt[3]{2} \cdot \sqrt[5]{3}$
$\sqrt[3]{3} \cdot \sqrt[4]{2}$
$\sqrt[3]{x^{2}} \cdot \sqrt[4]{x^{3}}$
$\sqrt[5]{x^{2}} \cdot \sqrt[7]{x^{5}}$
$\sqrt[9]{2 m^{2} n} \cdot \sqrt[3]{5 m^{5} n^{2}}$
$\sqrt{3 x y} \cdot \sqrt[3]{9 x^{2} y^{2}}$
$\sqrt[5]{x^{4} y^{3}} \cdot \sqrt[6]{x^{3} y^{5}}$
$\sqrt[5]{3 a^{4} b^{2}} \cdot \sqrt[6]{7 a^{2} b^{6}}$

In Exercises 125–130, factor each expression. Use only positive exponents in your answer.

$\frac{4}{3} x^{1 / 3} (2 x - 3) + 2 x^{4 / 3}$
$\frac{4}{3} x^{1 / 3} (7 x + 1) + 7 x^{4 / 3}$
$4 {(3 x + 1)}^{1 / 3} (2 x - 1) + 2 {(3 x + 1)}^{4 / 3}$
$4 {(3 x - 1)}^{1 / 3} (x + 2) + {(3 x - 1)}^{4 / 3}$
$3 x {(x^{2} + 1)}^{1 / 2} (2 x^{2} - x) + 2 {(x^{2} + 1)}^{3 / 2} (4 x - 1)$
$3 x {(x^{2} + 2)}^{1 / 2} (x^{2} - 2 x) + {(x^{2} + 2)}^{3 / 2} (2 x - 2)$

Applying the Concepts

Sign dimensions. A sign in the shape of an equilateral triangle of area A has sides of length $\sqrt{\frac{4 A}{\sqrt{3}}}$ centimeters. Find the length of a side assuming that the sign has an area of 692 square centimeters. (Use the approximation $\sqrt{3} \approx 1.73 .)$
Return on investment. For an initial investment of P dollars to mature to S dollars after two years when the interest is compounded annually, an annual interest rate of $r = \sqrt{\frac{S}{P}} - 1$ is required. Find r assuming that $P = $ 1, 210, 000$ and $S = $ 1, 411, 344 .$
Terminal speed. The terminal speed V of a steel ball that weighs w grams and is falling in a cylinder of oil is given by the equation $V = \sqrt{\frac{w}{1.5}}$ centimeters per second. Find the terminal speed of a steel ball weighing 6 grams.
Model airplane speed. A spring balance scale attached to a wire that holds a model airplane on a circular path indicates the force F on the wire. The speed of the plane V, in m/sec, is given by $V = \sqrt{(r F) / m},$ where m is the mass of the plane (in kilograms) and r is the length of the wire (in meters). What is the top speed of a 2-kilogram plane on an 18-meter wire if the force on the wire at top speed is 49 newtons?
Electronic game current. The current I (in amperes) in a circuit for an electronic game using W watts and having a resistance of R ohms is given by $I = \sqrt{\frac{W}{R}} .$ Find the current in a game circuit using 1058 watts and having a resistance of 2 ohms.
Radius of the moon. The radius of a sphere having volume V is given by $r = {(\frac{3 V}{4 π})}^{1 / 3} .$ Assuming that we treat the moon as a sphere and are told that its volume is $2.19 \times 10^{19}$ cubic meters, find its radius. Use 3.14 as an approximation for $π .$
Atmospheric pressure. The atmospheric pressure P, measured in pounds per square inch, at an altitude of h miles $(h < 50)$ is given by the equation $P = 14.7 {(0.5)}^{h / 3.25} .$ Compute the atmospheric pressure at an altitude of 16.25 miles.
Force from constrained water. The force F that is exerted on the semicircular end of a trough full of water is approximately $F = 42 r^{3 / 2}$ pounds, where r is the radius of the semicircular end. Find the force assuming that the radius is 3 feet.

Beyond the Basics

Show that $\frac{1}{3 - \sqrt{8}} - \frac{1}{\sqrt{8} - \sqrt{7}} + \frac{1}{\sqrt{7} - \sqrt{6}} - \frac{1}{\sqrt{6} - \sqrt{5}} + \frac{1}{\sqrt{5} - 2} = 5 .$
Show that $\frac{1}{1 + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + \frac{1}{\sqrt{4} + \sqrt{5}} + \frac{1}{\sqrt{5} + \sqrt{6}} + \frac{1}{\sqrt{6} + \sqrt{7}} + \frac{1}{\sqrt{7} + \sqrt{8}} + \frac{1}{\sqrt{8} + \sqrt{9}} = 2 .$
Let $x = 2 + \sqrt{3} .$ Show that
1. $x + \frac{1}{x} = 4 .$
2. $x^{2} + \frac{1}{x^{2}} = 14 .$
Let $x = 3 - 2 \sqrt{2} .$ Show that $x^{2} + \frac{1}{x^{2}} = 34 .$

In Exercises 143–145, arrange the given numbers in increasing order of magnitude without using a calculator.

$\sqrt{5}, \sqrt[3]{11}, \sqrt[6]{123}$
$\sqrt[4]{8}, \sqrt{3}, \sqrt[8]{70}$
$\sqrt{3}, \sqrt[3]{7}, \sqrt[4]{10}$

In Exercises 146–148, simplify each expression.

${(2 \sqrt{3} + 3 \sqrt{2})}^{2}$
$(3 \sqrt{5} + 4 \sqrt{3}) (3 \sqrt{5} - 4 \sqrt{3})$
$(\sqrt{2} + \sqrt{3} - \sqrt{5}) (\sqrt{2} + \sqrt{3} + \sqrt{5})$
Let a, b, x, and y be rational numbers and let $\sqrt{c}$ be an irrational number.
1. Show that if $a + b \sqrt{c} = 0,$ then $a = 0$ or $b = 0 .$
2. Use part (a) to show that if $a + b \sqrt{c} = x + y \sqrt{c},$ then $a = x$ and $b = y .$
Suppose $\sqrt{a + \sqrt{c}} = \sqrt{x} + \sqrt{y},$ show that $\sqrt{a - \sqrt{c}} = | \sqrt{x} - \sqrt{y} | .$
Find x and y if $\sqrt{x} + \sqrt{y} = \sqrt{16 + 2 \sqrt{55}} .$ [Hint: Square both sides to get $x + y = a$ and $x y = c .$ ]
Find $\sqrt{53 - 12 \sqrt{10}} .$ [Hint: $\sqrt{53 - 12 \sqrt{10}} = \sqrt{53 - 2 \sqrt{360}} = \sqrt{x} - \sqrt{y}$ ]

Getting Ready for the Next Section

In Exercises 153–156, perform the indicated operation.

$\frac{7}{3} + \frac{3}{2}$
$\frac{3}{6} - \frac{2}{3}$
$\frac{3}{2} \cdot \frac{12}{7}$
$\frac{3}{5} \div \frac{12}{15}$

In Exercises 157–160 simplify the given expression.

$- 7 x + 2 x$
$x + (- 5 x)$
$(4 x - 5) + (7 x - 3)$
$- 6 x^{2} - (- 2 x^{2})$

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Table of Contents for Section P.6 Rational Exponents and Radicals

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Table of Contents for
Section P.6 Rational Exponents and Radicals