10.10 Working with Radicals

Sometimes radical notation is more convenient to use than exponents. In these cases, we usually simplify radical expressions algebraically as much as possible before using a calculator to obtain decimal approximations.

Properties of Radicals

Because $\sqrt[n]{a} = a^{1 / n}$ , we can use the laws of exponents to derive two important properties that are useful in simplifying radicals.

Note

Properties of Radicals

$\sqrt[n]{a b} = \sqrt[n]{a} \sqrt[n]{b}, for a, b \geq 0$
$\sqrt[n]{\frac{a}{b}} = \frac{\sqrt[n]{a}}{\sqrt[n]{b}}, for a \geq 0, b > 0$

As examples, you can verify that

$\sqrt{36} = \sqrt{4} \sqrt{9} and \sqrt[3]{\frac{1}{8}} = \frac{\sqrt[3]{1}}{\sqrt[3]{8}}$

Example 1

Which of the following are true?

Is $\sqrt{36 + 64} = \sqrt{36} + \sqrt{64} ?$
Is $\sqrt[3]{8 (64)} = \sqrt[3]{8} \sqrt[3]{64} ?$
Is $\sqrt{x^{2} + 4} = x + 2 ?$
Is $\sqrt[3]{8 x^{3}} = 2 x ?$

The statements in (b) and (d) are true, and both are examples of the first property of radicals.

Statements (a) and (c) are false. In general, $\sqrt[n]{a + b}$ is not equal to $\sqrt[n]{a} + \sqrt[n]{b}$ , and $\sqrt[n]{a - b}$ is not equal to $\sqrt[n]{a} - \sqrt[n]{b}$ .

Simplifying Radicals

We use Property (1) to simplify radical expressions by factoring the radicand. For example, to simplify $the cube root of 108$ , we look for perfect cubes that divide evenly into $108$ . The easiest way to do this is to try the perfect cubes in order:

$1, 8, 27, 64, 125, and so on$

and so on, until we find one that is a factor. For this example, we find that $108 = 27 \cdot 4$ . Using Property (1), we write

$\sqrt[3]{108} = \sqrt[3]{27} \sqrt[3]{4}$

Simplify the first factor to find

$\sqrt[3]{108} = 3 \sqrt[3]{4}$

This expression is considered simpler than the original radical because the new radicand, $4$ , is smaller than the original, $108$ .

We can also simplify radicals containing variables. If the exponent on the variable is a multiple of the index, we can extract the variable from the radical. For instance,

$\sqrt[3]{12} = x^{12 / 3} = x^{4}$

(You can verify this by noting that $(x^{4})^{3} = x^{12}$ .) If the exponent on the variable is not a multiple of the index, we factor out the highest power that is a multiple. For example,

$\begin{aligned} \sqrt[3]{x^{11}} & = \sqrt[3]{x^{9} \cdot x^{2}} & Apply Property (1). \\ = \sqrt[3]{x^{9}} \cdot \sqrt[3]{x^{2}} & Simplify \sqrt[3]{x^{9}} = x^{9 / 3} . \\ = x^{3} \sqrt[3]{x^{2}} \end{aligned}$

Example 2

Simplify each radical.

$\sqrt{18 x^{5}}$
$\sqrt[3]{24 x^{6} y^{8}}$

The index of the radical is $2$ , so we look for perfect square factors of $18 x to the power 5$ . The factor $9$ is a perfect square, and $x to the power 4$ has an exponent divisible by $2$ . Thus,
$\begin{aligned} \sqrt{18 x^{5}} & = \sqrt{9 x^{4} \cdot 2 x} & Apply Property (1). \\ = \sqrt{9 x^{4}} \sqrt{2 x} & Take square roots. \\ = 3 x^{2} \sqrt{2 x} \end{aligned}$
The index of the radical is $3$ , so we look for perfect cube factors of $24 x^{6} y^{8}$ . The factor $8$ is a perfect cube, and $x to the power 6$ and $y to the power 6$ have exponents divisible by $3$ . Thus,
$\begin{aligned} \sqrt[3]{24 x^{6} y^{8}} & = \sqrt[3]{8 x^{6} y^{6} \cdot 3 y^{2}} & Apply Property (1). \\ = \sqrt[3]{8 x^{6} y^{6}} \sqrt[3]{3 y^{2}} & Take cube roots. \\ = 2 x^{2} y^{2} \sqrt[3]{3 y^{2}} \end{aligned}$

To simplify roots of fractions, we use Property (2), which allows us to write the expression as a quotient of two radicals.

We can also use Properties (1) and (2) to simplify products and quotients of radicals.

Example 4

Simplify.

$\sqrt[4]{6 x^{2}} \sqrt[4]{8 x^{3}}$
$\frac{\sqrt[3]{16 y^{5}}}{\sqrt[3]{y^{2}}}$

First apply Property (1) to write the product as a single radical, then simplify.
$\begin{aligned} \sqrt[4]{6 x^{2}} \sqrt[4]{8 x^{3}} & = \sqrt[4]{48 x^{5}} & Factor out perfect fourth powers. \\ = \sqrt[4]{16 x^{4}} \sqrt[4]{3 x} & Simplify. \\ = 2 x \sqrt[4]{3 x} \end{aligned}$
Apply Property (2) to write the quotient as a single radical.
$\begin{aligned} \frac{\sqrt[3]{16 y^{5}}}{\sqrt[3]{y^{2}}} & = \sqrt[3]{\frac{16 y^{5}}{y^{2}}} & Reduce. \\ = \sqrt[3]{16 y^{3}} & Simplify: factor out perfect cubes. \\ = \sqrt[3]{8 y^{3}} \sqrt[3]{2} \\ = 2 y \sqrt[3]{2} \end{aligned}$

Sums and Differences of Radicals

You know that sums or differences of like terms can be combined by adding or subtracting their coefficients:

$3 x y + 5 x y = (3 + 5) x y = 8 x y$

Like radicals, that is, radicals of the same index and radicand, can be combined in the same way.

Products of Radicals

According to Property (1), radicals of the same index can be multiplied together.

Thus, for example,

$\sqrt{2} \sqrt{18} \sqrt{36} = 6 and \sqrt[3]{2 x} \sqrt[3]{4 x^{2}} = \sqrt[3]{8 x^{3}} = 2 x$

For products involving binomials, we can apply the distributive law.

Example 6

$\begin{aligned} \sqrt{3} (\sqrt{2 x} + \sqrt{6}) & = \sqrt{3 \cdot 2 x} + \sqrt{3 \cdot 6} \\ = \sqrt{6 x} + \sqrt{18} = \sqrt{6 x} + 3 \sqrt{2} \end{aligned}$
$\begin{aligned} (\sqrt{x} - \sqrt{y}) (\sqrt{x} + \sqrt{y}) & = \sqrt{x^{2}} + \sqrt{x y} - \sqrt{x y} - \sqrt{y^{2}} \\ = x - y \end{aligned}$

Rationalizing the Denominator

It is easier to work with radicals if there are no roots in the denominators of fractions. We can use the fundamental principle of fractions to remove radicals from the denominator. This process is called rationalizing the denominator. For square roots, we multiply the numerator and denominator of the fraction by the radical in the denominator.

Example 7

Rationalize the denominator of each fraction.

$\sqrt{\frac{1}{3}}$
$\frac{\sqrt{2}}{\sqrt{50 x}}$

Apply Property (2) to write the radical as a quotient.
$\begin{aligned} \sqrt{\frac{1}{3}} & = \frac{\sqrt{1}}{\sqrt{3}} \\ = \frac{1}{\sqrt{3}} & Multiply numerator and denominator by \sqrt{3} . \\ = \frac{1 \cdot \sqrt{3}}{\sqrt{3} \cdot \sqrt{3}} \\ = \frac{\sqrt{3}}{3} \end{aligned}$
It is always best to simplify the denominator before rationalizing.
$\begin{aligned} \frac{\sqrt{2}}{\sqrt{50 x}} & = \frac{\sqrt{2}}{5 \sqrt{2 x}} & Multiply numerator and denominator by \sqrt{2 x} . \\ = \frac{\sqrt{2} \cdot \sqrt{2 x}}{5 \sqrt{2 x} \cdot \sqrt{2 x}} & Simplify. \\ = \frac{\sqrt{4 x}}{5 (2 x)} \\ = \frac{2 \sqrt{x}}{10 x} = \frac{\sqrt{x}}{5 x} \end{aligned}$

If the denominator of a fraction is a binomial in which one or both terms is a radical, we can use a special building factor to rationalize it. First, recall that

$(p - q) (p + q) = p^{2} - q^{2}$

where the product consists of perfect squares only. Each of the two factors $p - q$ and $p + q$ is said to be the conjugate of the other.

Now consider a fraction of the form

$\frac{a}{b + \sqrt{c}}$

If we multiply the numerator and denominator of this fraction by the conjugate of the denominator, we get

$\frac{a (b - \sqrt{c})}{(b + \sqrt{c}) (b - \sqrt{c})} = \frac{a b - a \sqrt{c}}{b^{2} - (\sqrt{c})^{2}} = \frac{a b - a \sqrt{c}}{b^{2} - c}$

The denominator of the fraction no longer contains any radicals—it has been rationalized.

Multiplying numerator and denominator by the conjugate of the denominator also works on fractions of the form

$\frac{a}{\sqrt{b} + c} and \frac{a}{\sqrt{b} + \sqrt{c}}$

We leave the verification of these cases as exercises.

Example 8

Rationalize the denominator: $\frac{x}{\sqrt{2} + \sqrt{x}}$ .

Multiply numerator and denominator by the conjugate of the denominator, $\sqrt{2} - \sqrt{x}$ .

$\frac{x (\sqrt{2} - \sqrt{x})}{(\sqrt{2} + \sqrt{x}) (\sqrt{2} - \sqrt{x})} = \frac{x (\sqrt{2} - \sqrt{x})}{2 - x}$

Simplifying $\sqrt[n]{x^{n}}$

Raising to a power is the inverse operation for extracting roots; that is,

as long as $\sqrt[n]{a}$ is a real number. For example,

${(\sqrt[4]{16})}^{4} = 2^{4} = 16, and {(\sqrt[3]{- 125})}^{3} = (- 5)^{3} = - 125$

Now consider the power and root operations in the opposite order; is it true that $\sqrt[n]{a^{n}} = a$ ? If the index $n$ is an odd number, then the statement is always true. For example,

$\sqrt[3]{2^{3}} = \sqrt[3]{8} = 2 and \sqrt[3]{(- 2)^{3}} = \sqrt[3]{- 8} = - 2$

However, if $n$ is even, we must be careful. Recall that the principal root $\sqrt[n]{x}$ is always positive, so if $a$ is a negative number, it cannot be true that $\sqrt[n]{a^{n}} = a$ . For example, if $a = - 3$ , then

$\sqrt{(- 3)^{2}} = \sqrt{9} = 3$

Instead, we see that, for even roots, $\sqrt[n]{a^{n}} = | a |$ .

We summarize our results in below.

Extraneous Solutions to Radical Equations

It is important to check the solution to a radical equation, because it is possible to introduce false, or extraneous, solutions when we square both sides of the equation. For example, the equation

$\sqrt{x} = - 5$

has no solution, because $\sqrt{x}$ is never a negative number. However, if we try to solve the equation by squaring both sides, we find

$\begin{aligned} (\sqrt{x})^{2} & = (- 5)^{2} \\ x & = 25 \end{aligned}$

You can check that $25$ is not a solution to the original equation, $\sqrt{x} = - 5$ , because $the square root of 25$ does not equal $minus 5$ .

If each side of an equation is raised to an odd power, extraneous solutions will not be introduced. However, if we raise both sides to an even power, we should check each solution in the original equation.

Example 10

Solve the equation $\sqrt{x + 2} + 4 = x$ .

First, isolate the radical expression on one side of the equation. (This will make it easier to square both sides.)

$\begin{aligned} \sqrt{x + 2} & = x - 4 & Square both sides of the equation. \\ {(\sqrt{x + 2})}^{2} & = (x - 4)^{2} \\ x + 2 & = x^{2} - 8 x + 16 & Subtract x + 2 from both sides. \\ x^{2} - 9 x + 14 & = 0 & Factor the left side. \\ x = 2 & or x = 7 \end{aligned}$

Check

Does $\sqrt{2 + 2} + 4 = 2$ ?

$No; 2 is not a solution.$

Does $\sqrt{7 + 2} + 4 = 7$ ?

$Yes; 7 is a solution.$

The apparent solution $2$ is extraneous. The only solution to the original equation is $7$ . We can verify the solution by graphing the equations

$y_{1} = \sqrt{x + 2} and y_{2} = x - 4$

as shown at right. The graphs intersect in only one point, $open parenthesis 7, 3 close parenthesis$ , so there is only one solution, $x equals 7$ .

Equations with More than One Radical

Sometimes it is necessary to square both sides of an equation more than once in order to eliminate all the radicals.

Example 11

Solve $\sqrt{x - 7} + \sqrt{x} = 7$ .

First, isolate the more complicated radical on one side of the equation. (This will make it easier to square both sides.) We will subtract $\sqrt{x}$ from both sides.

$\sqrt{x - 7} = 7 - \sqrt{x}$

Now square each side to remove one radical. Be careful when squaring the binomial $7 - \sqrt{x}$ .

$\begin{aligned} (\sqrt{x - 7})^{2} & = (7 - \sqrt{x})^{2} \\ x - 7 & = 49 - 14 \sqrt{x} + x \end{aligned}$

Collect like terms, and isolate the radical on one side of the equation.

$\begin{aligned} - 56 & = - 14 \sqrt{x} & Divide both sides by - 14. \\ 4 & = \sqrt{x} \end{aligned}$

Now square again to obtain

$\begin{aligned} (4)^{2} & = (\sqrt{x})^{2} \\ 16 & = x \end{aligned}$

Check

Does $\sqrt{16 - 7} + \sqrt{16} = 7$ ?

$Yes. The solution is 16.$

Section Summary

Vocabulary

Look up the definitions of new terms in the Glossary.

Radical
Extraneous
Rationalize
Conjugate
Like radicals
Radicand
Index

SKILLS

Practice each skill in the exercises listed.

Simplify radicals: #1–6
Simplify products and quotients of radcials: #7–10
Combine like radicals: #11–18
Multiply radical expressions: #19–36
Rationalize the denominator: #37–50
Simplify $\sqrt[n]{a^{n}}$ : #51–54
Solve radical equations: #55–80

Exercises A.10

For Problems 1-6, simplify. Assume that all variables represent positive numbers.

$the square root of 18$
$the cube root of 24$
$- \sqrt[4]{64}$

$3 the square root of 2$
$2 the cube root of 3$
$- 2 \sqrt[4]{4} = - 2 \sqrt{2}$

$the square root of 50$
$the cube root of 54$
$- \sqrt[4]{162}$

$the square root of 60, 000$
$the cube root of 900, 000$
$\sqrt[3]{\frac{- 40}{27}}$

$100 the square root of 6$
$10 the cube root of 900$
$\frac{- 2}{3} \sqrt[3]{5}$

$the square root of 800, 000$
$the cube root of 24, 000$
$\sqrt[4]{\frac{80}{625}}$

$\sqrt[3]{x^{10}}$
$the square root of 27 z cubed$
$\sqrt[4]{48 a^{9}}$

$x^{3} \sqrt[3]{x}$
$3 z the square root of 3 z$
$2 a squared the 4th root of 3 a$

$\sqrt[3]{y^{16}}$
$\sqrt{12 t^{5}}$
$\sqrt[3]{81 b^{8}}$

For Problems 7-10, simplify.

$- \sqrt{18 s} \sqrt{2 s^{3}}$
$\sqrt[3]{7 h^{2}} \sqrt[3]{- 49 h}$
$\sqrt{16 - 4 x^{2}}$

$minus 6 s squared$
$minus 7 h$
$2 \sqrt{4 - x^{2}}$

$\sqrt{3 w^{3}} \sqrt{27 w^{3}}$
$- \sqrt[4]{2 m^{3}} \sqrt[4]{8 m}$
$\sqrt{9 Y^{2} + 18}$

$\sqrt[3]{8 A^{3} + A^{6}}$
$\frac{\sqrt{45 x^{3} y^{3}}}{\sqrt{5 y}}$
$\frac{\sqrt[3]{8 b^{7}}}{\sqrt[3]{a^{6} b^{2}}}$

$A \sqrt[3]{8 + A^{3}}$
$3 x y the square root of x$
$\frac{2 b \sqrt[3]{b^{2}}}{a^{2}}$

$\sqrt[3]{b^{9} - 27 b^{3}}$
$\frac{\sqrt{98 x^{2} y^{3}}}{\sqrt{x y}}$
$\frac{\sqrt[3]{16 r^{4}}}{\sqrt[3]{4 t^{3}}}$

For Problems 11-18, simplify and combine like terms.

$3 \sqrt{7} + 2 \sqrt{7}$

$5 the square root of 7$

$5 \sqrt{2} - 3 \sqrt{2}$

$4 \sqrt{3} - \sqrt{27}$

$the square root of 3$

$\sqrt{75} + 2 \sqrt{3}$

$\sqrt{50 x} + \sqrt{32 x}$

$9 the square root of 2 x$

$\sqrt{8 y} - \sqrt{18 y}$

$3 \sqrt[3]{16} - \sqrt[3]{2} - 2 \sqrt[3]{54}$

$- \sqrt[3]{2}$

$\sqrt[3]{81} + 2 \sqrt[3]{24} - 3 \sqrt[3]{3}$

For Problems 19-32, multiply.

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

$6 - 2 \sqrt{5}$

$5 (2 - \sqrt{7})$

$\sqrt{2} (\sqrt{6} + \sqrt{10})$

$2 \sqrt{3} + 2 \sqrt{5}$

$\sqrt{3} (\sqrt{12} - \sqrt{15})$

$\sqrt[3]{2} (\sqrt[3]{20} - 2 \sqrt[3]{12})$

$2 \sqrt[3]{5} - 4 \sqrt[3]{3}$

$\sqrt[3]{3} (2 \sqrt[3]{18} + \sqrt[3]{36})$

$(\sqrt{x} - 3) (\sqrt{x} + 3)$

$x minus 9$

$(2 + \sqrt{x}) (2 - \sqrt{x})$

$(\sqrt{2} - \sqrt{3}) (\sqrt{2} + 2 \sqrt{3})$

$- 4 + \sqrt{6}$

$(\sqrt{3} - \sqrt{5}) (2 \sqrt{3} + \sqrt{5})$

$(\sqrt{5} - \sqrt{2})^{2}$

$7 - 2 \sqrt{10}$

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

$(\sqrt{a} - 2 \sqrt{b})^{2}$

$a - 4 \sqrt{a b} + 4 b$

$(\sqrt{2 a} - 2 \sqrt{b}) (\sqrt{2 a} + 2 \sqrt{b})$

For Problems 33-36, verify by substitution that the number is a solution of the quadratic equation.

$x^{2} - 2 x - 2 = 0, 1 + \sqrt{3}$

$(1 + \sqrt{3})^{2} - 2 (1 + \sqrt{3}) - 2 = 0$

$x^{2} + 4 x - 1 = 0, - 2 + \sqrt{5}$

$x^{2} + 6 x - 9 = 0, - 3 + 3 \sqrt{2}$

$(- 3 + 3 \sqrt{2})^{2} + 6 (- 3 + 3 \sqrt{2}) - 9 = 0$

$4 x^{2} - 20 x + 22 = 0, \frac{5 - \sqrt{3}}{2}$

For Problems 37-50, rationalize the denominator.

$\frac{6}{\sqrt{3}}$

$2 the square root of 3$

$\frac{10}{\sqrt{5}}$

$\sqrt{\frac{7 x}{18}}$

$\frac{\sqrt{14 x}}{6}$

$\sqrt{\frac{27 x}{20}}$

$\sqrt{\frac{2 a}{b}}$

$\frac{\sqrt{2 a b}}{b}$

$\sqrt{\frac{5 p}{q}}$

$\frac{2 \sqrt{3}}{\sqrt{2 k}}$

$\frac{\sqrt{6 k}}{k}$

$\frac{6 \sqrt{2}}{\sqrt{3 v}}$

$\frac{4}{1 + \sqrt{3}}$

$- 2 (1 - \sqrt{3})$

$\frac{3}{7 - \sqrt{2}}$

$\frac{x}{x - \sqrt{3}}$

$\frac{x (x + \sqrt{3})}{x^{2} - 3}$

$\frac{y}{\sqrt{5} - y}$

$\frac{\sqrt{6} - 3}{2 - \sqrt{6}}$

$\frac{\sqrt{6}}{2}$

$\frac{\sqrt{x} + \sqrt{y}}{\sqrt{x} - \sqrt{y}}$

Use your calculator to graph each function, and explain the result.

$y = \sqrt{x^{2}}$
$y = \sqrt[3]{x^{3}}$

$y = \sqrt{x^{2}} = | x |$
$y = \sqrt[3]{x^{3}} = x$

Use your calculator to graph each function, and explain the result.

$y = (x^{4})^{1 / 4}$
$y = (x^{5})^{1 / 5}$

For Problems 53-54, do not assume that variables represent positive numbers. Use absolute value bars as necessary to simplify the radicals.

$the square root of 4 x squared$
$\sqrt{(x - 5)^{2}}$
$\sqrt{x^{2} - 6 x + 9}$

$2 | x |$
$| x minus 5 |$
$| x minus 3 |$

$\sqrt{9 x^{2} y^{4}}$
$\sqrt{(2 x - 1)^{2}}$
$\sqrt{9 x^{2} - 6 x + 1}$

For Problems 55-78, solve

$\sqrt{x} - 5 = 3$

$64$

$\sqrt{x} - 4 = 1$

$\sqrt{y + 6} = 2$

$minus 2$

$\sqrt{y - 3} = 5$

$4 \sqrt{z} - 8 = - 2$

$\frac{9}{4}$

$- 3 \sqrt{z} + 14 = 8$

$5 + 2 \sqrt{6 - 2 w} = 13$

$minus 5$

$8 - 3 \sqrt{9 + 2 w} = - 7$

$3 z + 4 = \sqrt{3 z + 10}$

$\frac{- 1}{3}$

$2 x - 3 = \sqrt{7 x - 3}$

$2 x + 1 = \sqrt{10 x + 5}$

$\frac{- 1}{2}, 2$

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

$\sqrt{y + 4} = y - 8$

$12$

$4 \sqrt{x - 4} = x$

$\sqrt{2 y - 1} = \sqrt{3 y - 6}$

$5$

$\sqrt{4 y + 1} = \sqrt{6 y - 3}$

$\sqrt{x - 3} \sqrt{x} = 2$

$4$

$\sqrt{x} \sqrt{x - 5} = 6$

$\sqrt{y + 4} = \sqrt{y + 20} - 2$

$5$

$4 \sqrt{y} + \sqrt{1 + 16 y} = 5$

$\sqrt{x} + \sqrt{2} = \sqrt{x + 2}$

$0$

$\sqrt{4 x + 17} = 4 - \sqrt{x + 1}$

$\sqrt{5 + x} + \sqrt{x} = 5$

$4$

$\sqrt{y + 7} + \sqrt{y + 4} = 3$

Explain why the following first step for solving the radical equation is incorrect:

$\begin{aligned} \sqrt{x - 5} + \sqrt{2 x - 1} & = 8 \\ (x - 5) + (2 x - 1) & = 64 \end{aligned}$

We cannot square each term separately; we must square each side of the equation.

Explain why the following first step for solving the radical equation is incorrect:

$\begin{aligned} \sqrt{x + 2} + 1 & = \sqrt{2 x - 3} \\ (x + 2) + 1 & = 2 x - 3 \end{aligned}$

For Problems 81-84, write the complex fraction as a simple fraction in lowest terms, and rationalize the denominator.

$\frac{\frac{2}{\sqrt{7}}}{1 - \frac{\sqrt{3}}{\sqrt{7}}}$

$\frac{\sqrt{7} + \sqrt{3}}{2}$

$\frac{\frac{1}{4}}{\frac{\sqrt{5}}{2 \sqrt{2}} - \frac{\sqrt{3}}{2 \sqrt{2}}}$

$\frac{\frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}}}{1 - \frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{2}}}$

$\frac{6 \sqrt{3} + 7 \sqrt{2}}{5}$

$\frac{\frac{1}{\sqrt{3}} - \frac{\sqrt{5}}{3}}{1 + \frac{1}{\sqrt{3}} \cdot \frac{\sqrt{5}}{3}}$

Modeling, Functions, and Graphs by Katherine Yoshiwara (yoshiwarabooks.org), GNU Free Documentation License 1.2 or later. Adapted for the XYZ HTML edition with the authors' permission (recorded 2026-07-04). License: GFDL-1.2-or-later.

10.10 Working with Radicals

Properties of Radicals

Simplifying Radicals

Sums and Differences of Radicals

Products of Radicals

Rationalizing the Denominator

Simplifying x n n

Extraneous Solutions to Radical Equations

Equations with More than One Radical

Section Summary

Vocabulary

SKILLS

Exercises A.10

Simplifying $\sqrt[n]{x^{n}}$