A radical is any expression of the form: \(\sqrt[n]{a}\), where \(n\) is a positive integer. This is called the \(n^{th}\) root of \(a\).
As a quick example, consider the expression \(\sqrt[4]{81}\). From exponents, \((3)^4=81\), so \(3\) is a \(4^{th}\) root of \(81\). However, consider \(-3\). Here \((-3)^4=81\), so \(-3\) is another \(4^{th}\) root of \(81\).
Look at \(\sqrt[3]{8}\). Here there is only one root. Since \((2)^3=8\), but \((-2)^3=-8\), not \(8\), only \(2\) is a \(3^{rd}\) root of \(8\). This is sometimes referred to as the cube root of \(8\).
These examples show how roots depend on the index \(n\). The following table summarizes the different cases radicals have:
| Case 1: | If \(n\) is even and \(a\) is positive or \(0\), then \(\sqrt[n]{a}\) is the principal \(n^{th}\) root of \(a\) \(-\sqrt[n]{a}\) is the negative \(n^{th}\) root of \(a\) |
| Case 2: | In \(n\) is even and \(a\) is negative, then \(\sqrt[n]{a}\) is not a real number, it is a complex number. |
| Case 3: | If \(n\) is odd, then \(\sqrt[n]{a}\) is the only \(n^{th}\) root of \(a\) |
Radicals do not always give whole numbers. Consider \(\sqrt[5]{9}\). There is not an integer value for this radical. We can approximate this irrational number using a calculator: \(\sqrt[5]{9}\approx 1.55184557392...\). Using the expression \(\sqrt[5]{9}\) gives an exact form that includes all decimal values.
SPECIAL CASE: The square root is a radical with index of 2: \(\sqrt{a}=\sqrt[2]{a}\), it is often written without the index 2. Both are correct, and even cell phone calculators are now putting the index 2 back on the radical symbol.
A strange thing happens whenever the index is an even number and equal to the exponent under the radical sign, for example something like \(\sqrt[6]{(3)^6}\). Since the following are true:
| \(25=(5)^2\) and \(25=(-5)^2\) |
| \(256=(-4)^4\) and \(256=(4)^4\) |
| \(64=(2)^5\) and \(64=(-2)^5\) |
What is the correct value of \(\sqrt{25}\)? Is it \(5\) or \(-5\)? We need a way to show both roots. The mathematical way to show \(\pm5\) is by absolute value: \(\sqrt{25}=|5|\), because absolute value gives both values. The general formula is: $$\sqrt[n]{(a)^n}=|a|$$ This is more prevalent when dealing with variables instead of numeric values.
Simplify the following radical: \(\sqrt[4]{(-5)^4a^8b^{12}}\).
| Try to write terms under radical to \(4^{th}\) power: | $$\sqrt[\color{red}{4}]{(-5)^\color{red}{4}(a^2)^\color{red}{4}(b^3)^\color{red}{4}}$$ |
| Index matches exponents, apply absolute value: | $$|-5||a^2||b^3|$$ |
| Simplify the absolute value if needed | $$5a^2|b^3|$$ |
NOTE: In the example above the use of exponent rules: \(8=2\cdot4\text{ and }12=3\cdot4\) to rewrite the exponents. Also, the absolute value is not needed on the \(a^2\) since that will always be positive.
Radicals have an alternate form using exponents: \(\sqrt[n]{a}=a^{\frac{1}{n}}\). Notice that the index is the denominator of the fraction.
Let's look at some examples:
| a. \(\sqrt[3]{8}\implies8^{\frac{1}{3}}\implies 2\) |
| b. \(\sqrt{9}\implies9^{\frac{1}{2}}\implies 3\) |
| c. \(-\sqrt[4]{81}\implies -81^{\frac{1}{4}}\implies-3\) |
| d. \(\sqrt[4]{-81}\implies(-81)^{\frac{1}{4}}\implies\) NOT a real number |
| e. \(\sqrt[5]{\frac{1}{32}}\implies\left(\frac{1}{32}\right)^{\frac{1}{5}}\implies\frac{1}{2}\) |
NOTE: Pay attention to problems like c. and d. above, that pesky negative can cause problems!!
Notice that the rational exponents above are of the form: \(\frac{1}{n}\). What about \(a^{\frac{m}{n}}\)? Using rules of exponents, there are two different ways to write it: $$a^{\frac{m}{n}}=\left\{ \begin{array}{ccc} \left(a^{\frac{1}{n}}\right)^m & \implies & \left(\sqrt[n]{a}\right)^m \\ \left(a^m\right)^{\frac{1}{n}} & \implies & \left(\sqrt[n]{a^m}\right) \end{array} \right.$$ These two forms only exist provided \(a^{\frac{1}{n}}\) is a real number, otherwise it is NOT a real number.
Evaluate each exponential below.
| a. \(27^{\frac{2}{3}}\) | b. \(\left(\frac{4}{9}\right)^{\frac{-5}{2}}\) negative exponents form reciprocals |
| \(\implies\left(27^{\frac{1}{3}}\right)^2\) | \(\implies\left(\frac{9}{4}\right)^{\frac{5}{2}}\) |
| \(\implies(3)^2\) | \(\implies\left(\sqrt{\frac{9}{4}}\right)^5\) |
| \(\implies9\) | \(\implies\left(\frac{3}{2}\right)^5\implies\frac{243}{32}\) |