Radicals and Rational Exponents, SAT math lesson

Both $5^2$ and $(-5)^2$ equal $25$ , so when you ask which number squares to $25$ there are two candidates. The square root symbol has to pick one, and it always picks the nonnegative one. That single choice explains most of the traps in this topic, from the rules for simplifying radicals to the absolute value that appears when you take the square root of $x^2$ . On the SAT, unless a question says otherwise, every expression is a real number, so these real number rules are the ones that count.

See also Powers, Roots, and Signs in Foundations for Calculus and Power Functions and Root Functions in Precalculus.

Key terms

Principal square root

Definition For a real number $a \ge 0$ , the principal square root $\sqrt{a}$ is the unique real number $b$ with

b \ge 0 \quad \text{and} \quad b^2 = a.

For $a < 0$ there is no real number whose square is $a$ , so $\sqrt{a}$ is not a real number. The equation $x^2 = a$ with $a > 0$ has the two real solutions $x = \sqrt{a}$ and $x = -\sqrt{a}$ .

In plain words The symbol $\sqrt{\phantom{a}}$ always gives the nonnegative answer. If you want both numbers that square to $a$ , write $\pm\sqrt{a}$ yourself.

Example The value of $\sqrt{49}$ is $7$ , not $-7$ , even though $(-7)^2 = 49$ . The equation $x^2 = 49$ has two solutions, $x = 7$ and $x = -7$ . The expression $\sqrt{-49}$ has no real value.

Radical

Definition Let $n \ge 2$ be an integer. The expression $\sqrt[n]{a}$ is called a radical, $n$ is its index and $a$ is its radicand.

If $n$ is even and $a \ge 0$ , then $\sqrt[n]{a}$ is the unique real $b \ge 0$ with $b^n = a$ . If $n$ is even and $a < 0$ , it is not a real number.
If $n$ is odd, then $\sqrt[n]{a}$ is the unique real $b$ with $b^n = a$ , for every real $a$ .

When the radicals are real (all radicands nonnegative for even $n$ ), they follow the product and quotient rules

\begin{aligned} \sqrt[n]{ab} &= \sqrt[n]{a}\,\sqrt[n]{b}, \\[0.4em] \sqrt[n]{\frac{a}{b}} &= \frac{\sqrt[n]{a}}{\sqrt[n]{b}}, \quad b \ne 0. \end{aligned}

There is no such rule for sums, and in general $\sqrt{a + b} \ne \sqrt{a} + \sqrt{b}$ .

In plain words A radical undoes a power. Even roots only accept inputs that are zero or positive and only give outputs that are zero or positive. To simplify a square root, pull out the largest perfect square factor.

Example Since $72 = 36 \cdot 2$ and both factors are positive,

\sqrt{72} = \sqrt{36}\,\sqrt{2} = 6\sqrt{2}.

The sum rule fails, as one quick check shows. The value of $\sqrt{9 + 16}$ is $\sqrt{25} = 5$ , while $\sqrt{9} + \sqrt{16} = 3 + 4 = 7$ .

Cube root

Definition For every real number $a$ , the cube root $\sqrt[3]{a}$ is the unique real number $b$ with $b^3 = a$ . It has the same sign as $a$ , and

\sqrt[3]{x^3} = x \quad \text{for every real } x.

In plain words Every real number has exactly one real cube root, positive or negative. Since a negative number cubed stays negative, cube roots of negative numbers are fine.

Example The cube root of $-27$ is $-3$ , because $(-3)^3 = -27$ . Simplifying works as with square roots,

\sqrt[3]{-54} = \sqrt[3]{-27}\,\sqrt[3]{2} = -3\sqrt[3]{2}.

Absolute value

Definition The absolute value of a real number $x$ is

\lvert x \rvert = \begin{cases} x, & x \ge 0, \\ -x, & x < 0. \end{cases}

For every real number $x$ ,

\sqrt{x^2} = \lvert x \rvert.

The shortcut $\sqrt{x^2} = x$ is true only when $x \ge 0$ . In the same way $\left(\sqrt{x}\right)^2 = x$ holds only for $x \ge 0$ , because $\sqrt{x}$ is not real when $x < 0$ .

In plain words Absolute value is distance from $0$ , so it is never negative. Squaring a number forgets its sign, and the square root brings back the size but not the sign, which is exactly what absolute value records.

Example Take $x = -4$ . Then $x^2 = 16$ and $\sqrt{16} = 4$ , which equals $\lvert -4 \rvert$ and not $-4$ . So a simplification such as

\sqrt{(x - 3)^2} = x - 3

is false at $x = 1$ , where the left side is $\sqrt{4} = 2$ and the right side is $-2$ . The correct statement is $\sqrt{(x - 3)^2} = \lvert x - 3 \rvert$ .

Rational exponent

Definition Let $a > 0$ , let $m$ be an integer and let $n$ be a positive integer. Then

a^{1/n} = \sqrt[n]{a}

and

a^{m/n} = \left(\sqrt[n]{a}\right)^m = \sqrt[n]{a^m}.

Because $a > 0$ , the value depends only on the fraction $\frac{m}{n}$ , not on how it is written. For positive bases $a$ and $b$ the exponent rules from integer exponents still hold for all rational exponents $r$ and $s$ ,

\begin{gathered} a^r a^s = a^{r + s}, \qquad \left(a^r\right)^s = a^{rs}, \\[0.4em] (ab)^r = a^r b^r, \qquad a^{-r} = \frac{1}{a^r}. \end{gathered}

The boundary cases need care. For the base $0$ , the power $0^{m/n}$ equals $0$ whenever $\frac{m}{n} > 0$ . A negative power of $0$ is undefined, and $0^0$ is usually left undefined in algebra. For a negative base, $a^{1/n}$ with odd $n$ is often read as the real root $\sqrt[n]{a}$ , so $(-8)^{1/3} = -2$ , and with even $n$ it is not a real number. Even where negative bases give real values, the rules above are no longer guaranteed. Take $a = -8$ , $r = 2$ and $s = \frac{1}{6}$ . The rule $\left(a^r\right)^s = a^{rs}$ would claim that $\left((-8)^2\right)^{1/6}$ equals $(-8)^{1/3}$ , but

\begin{gathered} \left((-8)^2\right)^{1/6} = 64^{1/6} = 2, \\[0.4em] (-8)^{1/3} = -2. \end{gathered}

In plain words The bottom of the fraction is the root and the top is the power. Take the root first when that keeps the numbers small. Use the exponent rules freely when the base is positive.

Example Here is a short table of values.

Expression	Root first	Value
$8^{2/3}$	$\left(\sqrt[3]{8}\right)^2 = 2^2$	$4$
$27^{4/3}$	$\left(\sqrt[3]{27}\right)^4 = 3^4$	$81$
$25^{-3/2}$	$1 / \left(\sqrt{25}\right)^3 = 1/5^3$	$\frac{1}{125}$

With a variable base $x > 0$ the rules let you rewrite a radical as one power,

\sqrt{x} \cdot \sqrt[3]{x} = x^{1/2 + 1/3} = x^{5/6}.

Solving for a positive quantity

Definition If $u^2 = c$ with $c > 0$ , then $u = \sqrt{c}$ or $u = -\sqrt{c}$ . When the situation also requires $u > 0$ , for example because $u$ is a length, the only admissible value is $u = \sqrt{c}$ . More generally, if $u > 0$ and $u^{m/n} = c$ with $c > 0$ and $m \ne 0$ , then raising both sides to the power $\frac{n}{m}$ gives the unique positive solution

u = c^{n/m}.

In plain words Taking a square root to solve an equation gives two candidates. The context decides which one to keep, and a radius or a length must be positive.

Example A cylinder has volume $V = \pi r^2 h$ , where $V > 0$ and $h > 0$ . Dividing by $\pi h$ , which is positive, gives $r^2 = \frac{V}{\pi h}$ , and since a radius is positive,

r = \sqrt{\frac{V}{\pi h}}.

For $V = 72\pi$ cubic centimeters and $h = 2$ centimeters this gives $r = \sqrt{36} = 6$ centimeters. The other candidate, $-6$ , is thrown out because a radius cannot be negative.

Check yourself

What is the value of $16^{3/4}$ ? (A) $\frac{1}{8}$ (B) $8$ (C) $12$ (D) $64$

Show the answer

The answer is (B). Taking the fourth root first gives $\sqrt[4]{16} = 2$ , and then $2^3 = 8$ . Choice (A) is $16^{-3/4}$ , which treats the exponent as negative. Choice (C) multiplies $16$ by $\frac{3}{4}$ , which is not what an exponent does. Choice (D) is $16^{3/2}$ , with the root and the power swapped.

Which statement is true for every real number $x$ ? (A) $\sqrt{x^2} = x$ (B) $\sqrt{x^2} = \lvert x \rvert$ (C) $\sqrt[3]{x^3} = \lvert x \rvert$ (D) $\left(\sqrt{x}\right)^2 = x$

Show the answer

The answer is (B). Choice (A) fails at $x = -2$ , where the left side is $2$ and the right side is $-2$ . Choice (C) fails at $x = -2$ too, since the cube root of $-8$ is $-2$ , not $2$ . Choice (D) looks safe, but for $x = -2$ the expression $\sqrt{-2}$ is not a real number, so the statement does not hold for every real $x$ .

A cylinder has volume $V = \pi r^2 h$ . If $V = 200\pi$ cubic centimeters and $h = 8$ centimeters, what is the radius $r$ , in centimeters?

Show the answer

The answer is $5$ . Solving for the positive radius gives

r = \sqrt{\frac{V}{\pi h}} = \sqrt{\frac{200\pi}{8\pi}} = \sqrt{25} = 5.

The equation $r^2 = 25$ also has the solution $-5$ , but a radius must be positive.

Radicals and Rational Exponents

Theory, part 41

Practice, part 42

Key terms

Principal square root

Radical

Cube root

Absolute value

Rational exponent

Solving for a positive quantity

Check yourself

How this lesson connects

Builds on

Used later in