Math Core

Squaring and cubing are easy to undo when you know square roots and cube roots. But what undoes raising to the 4th power, or the 5th? The answer is the $n$ th root, and knowing when it exists, how many there are, and which one the radical symbol means is the foundation for everything else in this unit.

What an nth root is

You already know that $\sqrt{49} = 7$ because $7^2 = 49$ , and $\sqrt[3]{-64} = -4$ because $(-4)^3 = -64$ . The same idea works for any positive integer power.

Definition

nth root

For an integer $n \ge 2$ , a number $b$ is an $n$ th root of $a$ if

b^n = a.

In the radical $\sqrt[n]{a}$ , the number $n$ is the index and $a$ is the radicand. When no index is written, the index is $2$ .

For example, $3$ is a 4th root of $81$ because $3^4 = 81$ . But $-3$ is also a 4th root of $81$ , since $(-3)^4 = 81$ too. So a number can have more than one $n$ th root, and you need a rule for which one the symbol $\sqrt[n]{\ }$ means.

How many real nth roots are there?

The answer depends on whether the index is even or odd, and on the sign of the radicand. Think about what even and odd powers do to signs:

An even power of any nonzero real number is positive: $(-2)^4 = 16$ and $2^4 = 16$ . So a positive number has two even roots (one positive, one negative), and a negative number has no real even root. Nothing real raised to the 4th power gives $-16$ .
An odd power keeps the sign: $2^5 = 32$ and $(-2)^5 = -32$ . So every real number has exactly one real odd root, with the same sign as the radicand.

	$a > 0$	$a = 0$	$a < 0$
$n$ even	two real roots, $\pm\sqrt[n]{a}$	one root, $0$	no real roots
$n$ odd	one real root, positive	one root, $0$	one real root, negative

The principal nth root

The symbol $\sqrt[n]{a}$ always means one number, the principal $n$ th root:

If $n$ is even and $a \ge 0$ , $\sqrt[n]{a}$ is the nonnegative root. So $\sqrt[4]{81} = 3$ , not $-3$ .
If $n$ is odd, $\sqrt[n]{a}$ is the one real root, whatever the sign of $a$ . So $\sqrt[5]{-32} = -2$ .
If $n$ is even and $a < 0$ , $\sqrt[n]{a}$ is not a real number.

The graphs make the even/odd difference visible. The square root graph starts at the origin and exists only for $x \ge 0$ . The cube root graph passes through the origin and continues into negative $x$ -values, because negative numbers have real cube roots.

y = √x (defined only for x ≥ 0) and y = ∛x (defined for every real x).Open in grapher →

Worked example: Evaluating nth roots

Find each real root, or say that it is not a real number.

$\sqrt[4]{625}$
$\sqrt[5]{-243}$
$\sqrt[6]{-64}$
$-\sqrt[4]{16}$

Solutions.

$5^4 = 625$ , and the principal 4th root is nonnegative, so $\sqrt[4]{625} = 5$ .
$(-3)^5 = -243$ , and an odd root keeps the sign, so $\sqrt[5]{-243} = -3$ .
The index is even and the radicand is negative. No real number raised to the 6th power is negative, so $\sqrt[6]{-64}$ is not a real number.
The negative sign is outside the radical. First $\sqrt[4]{16} = 2$ , then take the opposite: $-\sqrt[4]{16} = -2$ . This is a real number.

Solving equations of the form x^n = k

The radical symbol gives only the principal root, but an equation asks for every number that works. That is where the "how many roots" table matters.

To solve $x^n = k$ :

If $n$ is even and $k > 0$ , there are two solutions: $x = \pm\sqrt[n]{k}$ .
If $n$ is even and $k < 0$ , there are no real solutions.
If $n$ is odd, there is exactly one solution: $x = \sqrt[n]{k}$ .

Before taking a root, isolate the power, just as you would isolate $x^2$ before taking a square root.

Worked example: Solving power equations

Solve each equation over the real numbers.

$x^4 = 81$
$2x^5 + 64 = 0$
$(x - 3)^4 = 16$

Solutions.

Even power, positive right side, so two solutions: $x = \pm\sqrt[4]{81} = \pm 3$ .
Isolate the power: $2x^5 = -64$ , so $x^5 = -32$ . The index is odd, so there is one real solution: $x = \sqrt[5]{-32} = -2$ . Check: $2(-2)^5 + 64 = 2(-32) + 64 = 0$ . ✓
Treat $x - 3$ as a single quantity. Its 4th power is $16$ , so $x - 3 = \pm 2$ .

\begin{aligned} x - 3 &= 2 &\quad&\text{or}& x - 3 &= -2 \\ x &= 5 &&& x &= 1 \end{aligned}

Both work: $(5 - 3)^4 = 2^4 = 16$ and $(1 - 3)^4 = (-2)^4 = 16$ .

Common mistake

Don't mix up the symbol and the equation. $\sqrt[4]{16} = 2$ only, because the symbol means the principal root. But the equation $x^4 = 16$ has two solutions, $x = 2$ and $x = -2$ . Forgetting the $\pm$ when solving an even-power equation is the most common way to lose a solution.

Roots of powers: when you need absolute value

What is $\sqrt{x^2}$ ? It is tempting to say $x$ , but try $x = -5$ : $\sqrt{(-5)^2} = \sqrt{25} = 5$ , not $-5$ . An even root always returns a nonnegative number, so

\sqrt[n]{x^n} = |x| \quad \text{when } n \text{ is even}, \qquad \sqrt[n]{x^n} = x \quad \text{when } n \text{ is odd}.

For example, $\sqrt[4]{(-3)^4} = \sqrt[4]{81} = 3 = |-3|$ , while $\sqrt[3]{(-3)^3} = \sqrt[3]{-27} = -3$ .

In later lessons you'll often be told that variables are positive, and then you can drop the absolute value bars. When no such assumption is given, keep them.

Estimating roots that aren't whole numbers

Most $n$ th roots are irrational. You can still locate them between consecutive integers by listing perfect powers. For example, to estimate $\sqrt[4]{50}$ , note that $2^4 = 16$ and $3^4 = 81$ . Since $50$ is between $16$ and $81$ , $\sqrt[4]{50}$ is between $2$ and $3$ . A calculator gives about $2.66$ , and you can check that $2.66^4 \approx 50.06$ .

Tip

Memorize a few perfect powers to make roots fast: $2^4 = 16$ , $3^4 = 81$ , $4^4 = 256$ , $5^4 = 625$ , $2^5 = 32$ , $3^5 = 243$ , $2^6 = 64$ . Many "nice" roots in Algebra 2 come from this short list.

Practice

Practice 1

Evaluate $\sqrt[4]{256}$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 2

Evaluate $\sqrt[5]{-243}$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 3

Which expression is not a real number?

Practice 4

Solve $x^4 = 16$ . Give all real solutions.

Separate answers with commas, e.g. 2, -5

Practice 5

Solve $3x^3 + 24 = 0$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 6

Between which two consecutive integers is $\sqrt[3]{100}$ ?

Practice 7

Solve $(x + 2)^4 = 81$ . Give all real solutions.

Separate answers with commas, e.g. 2, -5

Practice 8

For every real number $x$ , which expression equals $\sqrt[4]{x^4}$ ?