In Algebra 1 you met Fractional Exponents: $x^{1/n}$ is another way to write an $n$th root. Here you look at the roots themselves, and at the one question that trips people up: when is a root a real number?
Key idea
$\sqrt[n]{a}$ is the number whose $n$th power is $a$. For odd $n$ there is exactly one real $n$th root of any real number, negatives included: $\sqrt[3]{-8} = -2$. For even $n$, a negative number has no real $n$th root, and a positive number has two ($\pm$), with $\sqrt[n]{a}$ meaning the positive one.
Why it works
Look at the powers. An even power is never negative: $(-3)^4 = 81$ and $3^4 = 81$. So $x^4 = 81$ has two real solutions, $x^4 = -81$ has none, and we let $\sqrt[4]{81}$ name the positive solution. An odd power keeps the sign: $(-2)^3 = -8$. Every real number is the cube (or fifth power) of exactly one real number, so odd roots always exist.
Worked examples
Worked example: Odd and even roots
Evaluate $\sqrt[3]{-125}$, $\sqrt[4]{16}$ and $\sqrt[5]{-32}$.
- $(-5)^3 = -125$, so $\sqrt[3]{-125} = -5$.
- $2^4 = 16$ (and $(-2)^4 = 16$), and the symbol means the positive root: $\sqrt[4]{16} = 2$.
- $(-2)^5 = -32$, so $\sqrt[5]{-32} = -2$.
Answer: $-5$, $2$, $-2$
Worked example: Solving $x^n = a$
Find all real solutions of $x^4 = 81$, $x^3 = -27$ and $x^6 = -64$.
- $x^4 = 81$: $x = \pm\sqrt[4]{81} = \pm 3$.
- $x^3 = -27$: one solution, $x = -3$.
- $x^6 = -64$: even powers are never negative, so there is no real solution.
Answer: $\pm 3$; $-3$; no real solution
Worked example: Rational exponents
Evaluate $27^{2/3}$ and $16^{-3/4}$.
- $27^{2/3} = (\sqrt[3]{27})^2 = 3^2 = 9$.
- $16^{-3/4} = \frac{1}{(\sqrt[4]{16})^3} = \frac{1}{2^3} = \frac{1}{8}$.
Answer: $9$ and $\frac{1}{8}$
Common mistake
Taking an even root of a negative number as if it were real, for example writing $\sqrt[4]{-16} = -2$. Check by raising back: $(-2)^4 = 16$, not $-16$. No real number works, so $\sqrt[4]{-16}$ is not a real number. (Odd roots of negatives are fine.)