By “roots” I’m referring to things like "square roots" and when I say “radicals”, I mean expressions involving the symbol \(\sqrt{\quad}\text{.}\) Hopefully you saw a clear distinction between square roots and cube roots in Section 3.1. This distinction is maintained between even and odd \(n\)-th roots.
Suppose \(n=2,4,6,\ldots\) is an even positive integer and \(A \gt 0\) is a real number. Then there exists a unique positive real number \(\sqrt[n]{A}\) called the principal \(n\)-th root of \(A\). It satisfies the properties:
When \(n\) is even, there are exactly two real \(n\)-th roots of \(A\). They are \(\sqrt[n]{A}\) and its opposite \(-\sqrt[n]{A}\text{.}\) We say briefly that the \(n\)-th roots of \(A\) are \(\pm \sqrt[n]{A}\text{.}\) To put things another way, the equation \(x^n = A\) has exactly two real number solutions:
\begin{equation}
x^n = A \quad \Rightarrow \quad x = \pm \sqrt[n]{A}.\tag{3.2.1}
\end{equation}
If \(B \lt 0\text{,}\) then \(\sqrt[n]{B}\) is undefined, while \(\sqrt[n]{0} = 0\text{.}\)
For example, \(\sqrt[4]{16} = 2\) because \(2 \gt 0\) and \(2^4 = 2\cdot 2\cdot 2\cdot 2 = 16\text{.}\) In addition, \((-2)^4 = 16\text{.}\) Observe how this is positive due to the even number of factors! The equation \(x^4 = 16\) has exactly two real number solutions. They are \(x = \pm\sqrt[4]{16} = \pm 2\text{.}\)
Suppose \(n=3,5,7,\ldots\) is an odd positive integer and \(A\) is any real number. Then there exists a unique real number \(\sqrt[n]{A}\) called the principal \(n\)-th root of \(A\) satisfying the property:
\begin{equation*}
\left(\sqrt[n]{A}\right)^n = A.
\end{equation*}
When n is odd, there is exactly one real \(n\)-th root of \(A\), namely \(\sqrt[n]{A}\text{.}\) Equivalently, the equation \(x^n = A\) has exactly one real number solution:
\begin{equation}
x^n = A \quad \Rightarrow \quad x = \sqrt[n]{A}.\tag{3.2.2}
\end{equation}
We say that \(1/2\) is the fifth root of \(1/32\text{.}\) Alternatively, the equation \(x^5 = 1/32\) has exactly one real number solution, \(x = \sqrt[5]{1/32} = 1/2\text{.}\)
\(n\)-th roots satisfy various properties. Remember that there is a strong dichotomy between odd and even roots! Suppose \(n = 2,3,4,\ldots\) and \(x\) and \(y\) are real numbers. Assume \(x \geq 0\) and \(y \geq 0\) in the case that \(n\) is even. Then:
If \(n\) is even and \(x\) any real number, including negative, then we must have absolute value to ensure a non-negative result when extracting powers from the radical:
