Math Core

An answer like $\sqrt[3]{54}$ is correct, but $3\sqrt[3]{2}$ is the same number in a form you can compare, combine and compute with. This lesson extends simplifying to any index, to variables, and to adding, multiplying and dividing radicals.

The product and quotient properties

Roots split over multiplication and division. For any index $n$ (with $a, b \ge 0$ when $n$ is even, and $b \ne 0$ in the quotient):

\sqrt[n]{ab} = \sqrt[n]{a} \cdot \sqrt[n]{b} \qquad\qquad \sqrt[n]{\frac{a}{b}} = \frac{\sqrt[n]{a}}{\sqrt[n]{b}}

Both follow from rational exponents: $(ab)^{1/n} = a^{1/n} b^{1/n}$ .

Simplest radical form

A radical expression with index $n$ is in simplest form when:

The radicand has no factor that is a perfect $n$ th power (other than $1$ ).
The radicand contains no fractions.
No radical appears in a denominator.

To meet rule 1, look for the largest perfect $n$ th power that divides the radicand. For cube roots that means factors like $8, 27, 64, 125$ ; for 4th roots, $16, 81, 256, 625$ .

\sqrt[3]{54} = \sqrt[3]{27 \cdot 2} = 3\sqrt[3]{2} \qquad\qquad \sqrt[4]{48} = \sqrt[4]{16 \cdot 3} = 2\sqrt[4]{3}

Radicals with variables

For variables, split each exponent into a multiple of the index plus a remainder. The multiple comes out; the remainder stays in. In this lesson, assume every variable is positive, so you don't need absolute value bars.

For example, with index $3$ , $x^7 = x^6 \cdot x$ , and $\sqrt[3]{x^6} = x^2$ . So $\sqrt[3]{x^7} = x^2\sqrt[3]{x}$ . A quick way to see it: divide the exponent by the index. $7 \div 3 = 2$ remainder $1$ , so $x^2$ comes out and $x^1$ stays in.

Worked example: Simplifying radicals with variables

Simplify. Assume all variables are positive.

$\sqrt{50x^3y^4}$
$\sqrt[3]{16a^5b^9}$
$\sqrt[4]{\dfrac{32x^6}{81}}$

Solutions.

Split off perfect squares: $50 = 25 \cdot 2$ , $x^3 = x^2 \cdot x$ , and $y^4$ is already a perfect square.

\sqrt{50x^3y^4} = \sqrt{25x^2y^4} \cdot \sqrt{2x} = 5xy^2\sqrt{2x}.

Split off perfect cubes: $16 = 8 \cdot 2$ , $a^5 = a^3 \cdot a^2$ , and $b^9 = \left(b^3\right)^3$ .

\sqrt[3]{16a^5b^9} = \sqrt[3]{8a^3b^9} \cdot \sqrt[3]{2a^2} = 2ab^3\sqrt[3]{2a^2}.

Use the quotient property, then simplify the top: $32 = 16 \cdot 2$ and $x^6 = x^4 \cdot x^2$ .

\sqrt[4]{\frac{32x^6}{81}} = \frac{\sqrt[4]{16x^4} \cdot \sqrt[4]{2x^2}}{\sqrt[4]{81}} = \frac{2x\sqrt[4]{2x^2}}{3}.

Adding and subtracting radicals

You can add or subtract radicals only when they are like radicals: the same index and the same radicand. Then you combine the coefficients, exactly as you combine like terms: $4\sqrt[3]{5} + 2\sqrt[3]{5} = 6\sqrt[3]{5}$ , just as $4t + 2t = 6t$ .

Radicals that don't look alike sometimes become alike after simplifying, so simplify first.

Worked example: Combining like radicals

Simplify $3\sqrt[3]{16} + \sqrt[3]{54} - \sqrt[3]{2}$ .

Simplify each term: $\sqrt[3]{16} = \sqrt[3]{8 \cdot 2} = 2\sqrt[3]{2}$ and $\sqrt[3]{54} = 3\sqrt[3]{2}$ . Now every term is a multiple of $\sqrt[3]{2}$ :

3\left(2\sqrt[3]{2}\right) + 3\sqrt[3]{2} - \sqrt[3]{2} = (6 + 3 - 1)\sqrt[3]{2} = 8\sqrt[3]{2}.

Common mistake

Roots do not split over addition. $\sqrt{9 + 16} = \sqrt{25} = 5$ , but $\sqrt{9} + \sqrt{16} = 3 + 4 = 7$ . In the same way, $\sqrt{2} + \sqrt{3}$ is not $\sqrt{5}$ , and $\sqrt{3} + \sqrt[3]{3}$ can't be combined at all (the indices differ). Only like radicals combine.

Multiplying radicals

To multiply radicals with the same index, multiply the radicands, then simplify. For sums, use the distributive property (FOIL) exactly as with polynomials.

\sqrt[3]{4} \cdot \sqrt[3]{10} = \sqrt[3]{40} = \sqrt[3]{8 \cdot 5} = 2\sqrt[3]{5}

\left(2 + \sqrt{3}\right)\left(4 - \sqrt{3}\right) = 8 - 2\sqrt{3} + 4\sqrt{3} - 3 = 5 + 2\sqrt{3}

A special product is especially useful. Expressions of the form $a + \sqrt{b}$ and $a - \sqrt{b}$ are called conjugates, and their product has no radical, because the middle terms cancel:

\left(a + \sqrt{b}\right)\left(a - \sqrt{b}\right) = a^2 - b.

For example, $\left(3 + \sqrt{5}\right)\left(3 - \sqrt{5}\right) = 9 - 5 = 4$ .

Rationalizing the denominator

Rule 3 of simplest form says no radical in a denominator. Removing it is called rationalizing the denominator. You multiply by a clever form of $1$ that turns the denominator's radical into a whole number.

A single square root: multiply top and bottom by that root. $\dfrac{6}{\sqrt{3}} = \dfrac{6\sqrt{3}}{3} = 2\sqrt{3}$ .
A higher root: multiply by whatever completes a perfect $n$ th power. For $\dfrac{5}{\sqrt[3]{2}}$ , you need $\sqrt[3]{2^3}$ , so multiply by $\sqrt[3]{2^2} = \sqrt[3]{4}$ : $\dfrac{5\sqrt[3]{4}}{\sqrt[3]{8}} = \dfrac{5\sqrt[3]{4}}{2}$ .
A binomial with a square root: multiply by the conjugate.

Worked example: Rationalizing with a conjugate

Simplify $\dfrac{4}{3 - \sqrt{5}}$ .

Multiply the numerator and denominator by the conjugate $3 + \sqrt{5}$ :

\frac{4}{3 - \sqrt{5}} \cdot \frac{3 + \sqrt{5}}{3 + \sqrt{5}} = \frac{4\left(3 + \sqrt{5}\right)}{9 - 5} = \frac{4\left(3 + \sqrt{5}\right)}{4} = 3 + \sqrt{5}.

Tip

Check a simplification with a calculator: $\dfrac{4}{3 - \sqrt{5}} \approx \dfrac{4}{0.764} \approx 5.236$ , and $3 + \sqrt{5} \approx 5.236$ . If the decimals match, your algebra is right.

Practice

Practice 1

Write $\sqrt[3]{135}$ in simplest form as $a\sqrt[3]{5}$ . What is $a$ ?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 2

Write $\sqrt[4]{162}$ in simplest form as $a\sqrt[4]{2}$ . What is $a$ ?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 3

Simplify $2\sqrt{75} - \sqrt{27}$ to the form $a\sqrt{3}$ . What is $a$ ?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 4

Assume $x$ and $y$ are positive. Which is the simplest form of $\sqrt{72x^5y^2}$ ?

Practice 5

Which expression equals $\left(3 + \sqrt{2}\right)\left(5 - \sqrt{2}\right)$ ?

Practice 6

Rationalize the denominator: write $\dfrac{10}{\sqrt{5}}$ as $a\sqrt{5}$ . What is $a$ ?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 7

Assume $x$ and $y$ are positive. Which is the simplest form of $\sqrt[3]{24x^4y^6}$ ?

Practice 8

Which expression equals $\dfrac{6}{\sqrt{7} - 2}$ with a rational denominator?