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Precalculus JumpStart

Section 1.8 Roots and Radicals

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 1.7. This distinction is maintained between even and odd \(n\)-th roots.

Definition 1.41. Even \(n\)-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:
\begin{equation*} \sqrt[n]{A} \gt 0 \quad \text{and} \quad \left(\sqrt[n]{A}\right)^n = A. \end{equation*}
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{1.5} \end{equation}
If \(B \lt 0\text{,}\) then \(\sqrt[n]{B}\) is undefined, while \(\sqrt[n]{0} = 0\text{.}\)

Example 1.42.

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 \(x = \pm\sqrt[4]{16} = \pm 2\text{.}\)

Definition 1.43. Odd \(n\)-th Roots.

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{1.6} \end{equation}

Example 1.44.

For example, \(\sqrt[5]{-\frac{1}{32}} = -\frac{1}{2}\) because
\begin{equation*} \left(-\frac{1}{2}\right)^5 = \left(-\frac{1}{2}\right)\cdot\left(-\frac{1}{2}\right)\cdot\left(-\frac{1}{2}\right)\cdot\left(-\frac{1}{2}\right)\cdot\left(-\frac{1}{2}\right) = -\frac{1}{32}. \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, \(x = \sqrt[5]{1/32} = 1/2\text{.}\)

Example 1.45.

The only real number solution to the equation \(x^5 = -32\) is \(x=\sqrt[5]{-32} = -2\text{.}\)

Warning 1.47. Roots are incompatible with addition.

In general, \(\sqrt[n]{a+b}\) is not \(\sqrt[n]{a}+\sqrt[n]{b}\text{.}\)

Example 1.48.

\begin{equation*} \sqrt[5]{64} = \sqrt[5]{32 \cdot 2} = \sqrt[5]{32} \cdot \sqrt[5]{2} = \sqrt[5]{2^5}\cdot \sqrt[5]{2} = 2\sqrt[5]{2}. \end{equation*}

Example 1.49.

\begin{equation*} \sqrt{\frac{1}{12}} = \frac{\sqrt{1}}{\sqrt{12}} = \frac{1}{\sqrt{4\cdot 3}} = \frac{1}{\sqrt{4}\cdot \sqrt{3}} = \frac{1}{2\sqrt{3}}. \end{equation*}
The final result above is fine, but its customary to rationalize radicals in the denominator. This involves multiply by a form of one
\begin{equation*} \frac{1}{2\sqrt{3}} = \frac{1}{2\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{2 \left(\sqrt{3}\right)^2} = \frac{\sqrt{3}}{2\cdot 3} = \frac{\sqrt{3}}{6}. \end{equation*}

Example 1.50.

Evaluate any sums or differences under radicals first.
\begin{equation*} \sqrt{3^2 +4^2} = \sqrt{9+16} = \sqrt{25} = 5. \end{equation*}

Checkpoint 1.51.

Evaluate the expression without a calculator.
\begin{equation*} \sqrt{\sqrt[4]{2^{8}}} \end{equation*}