The Timpson Lyceum · Intermediate Algebra (2023)

Lecture 16 (Intermediate Algebra) Radicals

Radicals – Lecture 16

Overview

This lecture introduces radicals, starting with square roots and generalizing to cube roots, fourth roots, and $n$ th roots. The central skill is simplifying a radical by factoring out perfect powers, then carrying that idea through every radical operation: combining like radical terms, multiplying and distributing radicals, and dividing them. The lecture closes with rationalizing the denominator — rewriting an expression so that no radical remains underneath a fraction bar. Mastering the "pull out the perfect power" move makes every later technique in the chapter routine.

Key Concepts

Square root. The square root of a number $a$ is the value $x$ that you must square to recover $a$ . Formally, $\sqrt{a} = x$ means $x^2 = a$ . For example $\sqrt{25} = 5$ because $5^2 = 25$ , and $\sqrt{81} = 9$ because $9^2 = 81$ .

Square roots of negatives are undefined (in the reals). A positive times a positive is positive and a negative times a negative is positive, so no real number squared gives a negative result. Thus $\sqrt{-81}$ is undefined over the real numbers; such roots belong to the complex (imaginary) numbers.

The $n$ th root. Extending the idea: $\sqrt[n]{a} = x$ means $x^n = a$ .

Cube root: $\sqrt[3]{8} = 2$ because $2^3 = 8$ .
Fourth root: $\sqrt[4]{16} = 2$ because $2^4 = 16$ .

Odd vs. even roots of negatives. Odd roots of negative numbers exist; even roots do not (over the reals).

$\sqrt[3]{-64} = -4$ , since $(-4)^3 = -64$ .
$\sqrt[4]{-16}$ is undefined.

Product rule for radicals. For non-negative quantities, a radical of a product splits into a product of radicals (and the reverse): $\sqrt[n]{a\,b} = \sqrt[n]{a}\,\sqrt[n]{b}$ . This holds for any index $n$ .

Simplifying radicals (the core skill). To simplify $\sqrt[n]{N}$ , factor out the largest perfect $n$ th power you can find, split the radical with the product rule, and take its root.

Useful perfect squares: $1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144$ .
Useful perfect cubes: $1, 8, 27, 64, 125, 216$ .
Useful perfect fourth powers: $1, 16, 81, 256$ .
If you don't spot the largest perfect power immediately, pull out any perfect power you see (e.g. a $4$ or a $9$ ) and repeat — you'll reach the same fully simplified answer.

Variables under a radical. Pull out one factor of the variable for every $n$ copies; the remainder stays inside. For a square root, $\sqrt{x^4} = x^2$ (since $x^4 = (x^2)^2$ ) and $\sqrt{y^6} = y^3$ . In general, divide the exponent by the index: the quotient comes out, the remainder stays in.

Combining like radical terms. Radicals are added or subtracted only when they are like terms — the same index and the same radicand. Treat $\sqrt{11}$ like a variable: $3\sqrt{11} + 2\sqrt{11} = 5\sqrt{11}$ , but $\sqrt[4]{6}$ and $\sqrt[5]{6}$ cannot be combined, nor can $\sqrt{2}$ and $\sqrt{3}$ . Sometimes terms look unlike until you simplify each one — simplify first, then combine.

Multiplying radicals. With matching indices, multiply the outside coefficients together and the radicands together: $a\sqrt[n]{b}\cdot c\sqrt[n]{d} = ac\,\sqrt[n]{b\,d}$ . Then simplify the result. Distribute across parentheses just as with polynomials, multiplying the radical into each term.

Quotient rule for radicals. Division behaves like multiplication: $\dfrac{\sqrt[n]{a}}{\sqrt[n]{b}} = \sqrt[n]{\dfrac{a}{b}}$ , and coefficients reduce separately.

Rationalizing the denominator. We do not leave radicals in a denominator. Multiply the fraction by a well-chosen form of $1$ that turns the denominator's radicand into a perfect power. For a square root, multiply by the same radical over itself. For higher roots, multiply by whatever makes the radicand a perfect $n$ th power — e.g. to clear $\sqrt[4]{2}$ , multiply by $\dfrac{\sqrt[4]{8}}{\sqrt[4]{8}}$ so the denominator becomes $\sqrt[4]{16} = 2$ .

Worked Example

Simplify $\sqrt{72}$ by pulling out the largest perfect square ( $36 = 6^2$ ):

\begin{aligned} \sqrt{N} &= \sqrt{p^2 \cdot m} = p\sqrt{m} \\ \sqrt{72} &= \sqrt{36 \cdot 2} = \sqrt{36}\,\sqrt{2} = 6\sqrt{2} \end{aligned}

The same answer results from a slower path: $\sqrt{72} = \sqrt{9\cdot 8} = 3\sqrt{8} = 3\sqrt{4\cdot 2} = 3\cdot 2\sqrt{2} = 6\sqrt{2}$ .

As a second example, rationalize the denominator of $\dfrac{6}{\sqrt{5}}$ by multiplying by $\dfrac{\sqrt{5}}{\sqrt{5}}$ :

\begin{aligned} \frac{a}{\sqrt{b}} &= \frac{a}{\sqrt{b}}\cdot\frac{\sqrt{b}}{\sqrt{b}} = \frac{a\sqrt{b}}{b} \\ \frac{6}{\sqrt{5}} &= \frac{6}{\sqrt{5}}\cdot\frac{\sqrt{5}}{\sqrt{5}} = \frac{6\sqrt{5}}{5} \end{aligned}

Summary

$\sqrt[n]{a} = x$ means $x^n = a$ ; square roots of negatives are undefined over the reals, but odd roots (e.g. $\sqrt[3]{-64} = -4$ ) exist while even roots of negatives do not.
The product and quotient rules let you split a radical across a product or quotient: $\sqrt[n]{ab} = \sqrt[n]{a}\,\sqrt[n]{b}$ and $\sqrt[n]{a/b} = \sqrt[n]{a}/\sqrt[n]{b}$ .
Simplify any radical by factoring out the largest perfect $n$ th power; for variables, divide the exponent by the index — quotient out, remainder in.
Add or subtract radicals only when they are like terms (same index and same radicand); simplify each term first, since unlike-looking terms may become like.
Multiply radicals by multiplying coefficients and radicands separately, then simplify; distribute a radical across parentheses term by term.
Rationalize a denominator by multiplying by a form of $1$ that makes the radicand a perfect power, leaving no radical below the fraction bar.