Unit 8 · Lesson 4 of 6

Dividing radicals and rationalizing denominators

About 25 minutes · 5 steps · one idea: no radical left in the denominator

You'll be able to: divide square roots and rewrite fractions like \(\dfrac{6}{3+\sqrt{7}}\) so that no radical is left in the denominator.

  1. 1 Warm-up
  2. 2 The idea
  3. 3 Examples
  4. 4 Practice
  5. 5 Wrap-up

Step 1 of 5Warm-up

Two quick questions from earlier lessons. Answer from memory first. Pulling it out of your head is what makes it stick.

\((5-\sqrt{3})(5+\sqrt{3})\)

A conjugate pair: the middle terms cancel, leaving first squared minus last squared.

\(5^2 - (\sqrt{3})^2 = 25 - 3 = 22\).

Solve \(\dfrac{x}{3} + \dfrac{x}{6} = 5\).

Multiply every term by the LCD, 6: \(2x + x = 30\).

\(3x = 30\), so \(x = 10\).

Step 2 of 5The idea

Dividing square roots works like multiplying them: you can put both under one root. If the numbers divide evenly, divide first.

A radical is fully simplified only when no radical is left in the denominator (see simplified radical). Getting the radical out of the bottom is called rationalizing the denominator.

The trick is to multiply the top and the bottom by the same thing. That is the same as multiplying by 1, so the value doesn't change. Only the look changes.

Divide: \(\dfrac{\sqrt{a}}{\sqrt{b}} = \sqrt{\dfrac{a}{b}}\) (with \(b \gt 0\)).

One square root on the bottom: multiply top and bottom by that root. \(\dfrac{1}{\sqrt{2}} \cdot \dfrac{\sqrt{2}}{\sqrt{2}} = \dfrac{\sqrt{2}}{2}\).

One cube root on the bottom: multiply by what makes a perfect cube inside. \(\dfrac{1}{\sqrt[3]{2}} \cdot \dfrac{\sqrt[3]{4}}{\sqrt[3]{4}} = \dfrac{\sqrt[3]{4}}{\sqrt[3]{8}} = \dfrac{\sqrt[3]{4}}{2}\).

A binomial on the bottom, like \(a + \sqrt{b}\): multiply top and bottom by its conjugate \(a - \sqrt{b}\). The bottom becomes \(a^2 - b\), with no radical.

Why this shows up on the placement test: answer choices are written with rational denominators, so \(\dfrac{1}{\sqrt{2}}\) will appear only as \(\dfrac{\sqrt{2}}{2}\).

Step 3 of 5Worked examples

Example 1: \(\dfrac{\sqrt{72}}{\sqrt{2}}\), then \(\dfrac{6}{\sqrt{3}}\)

  1. Put both under one root and divide: \(\dfrac{\sqrt{72}}{\sqrt{2}} = \sqrt{\dfrac{72}{2}} = \sqrt{36} = 6\).
  2. For \(\dfrac{6}{\sqrt{3}}\), multiply top and bottom by \(\sqrt{3}\): \(\dfrac{6}{\sqrt{3}} \cdot \dfrac{\sqrt{3}}{\sqrt{3}} = \dfrac{6\sqrt{3}}{3}\) (\(\sqrt{3} \cdot \sqrt{3} = 3\)).
  3. Reduce: \(6 \div 3 = 2\), so the answer is \(2\sqrt{3}\). The 3 on the bottom divides the 6 in front, not the 3 under the root.

Example 2: \(\dfrac{6}{3+\sqrt{7}}\)

  1. The conjugate of \(3+\sqrt{7}\) is \(3-\sqrt{7}\). Multiply top and bottom by it: \(\dfrac{6(3-\sqrt{7})}{(3+\sqrt{7})(3-\sqrt{7})}\).
  2. Bottom: \(3^2 - (\sqrt{7})^2 = 9 - 7 = 2\). No radical left.
  3. Now it is \(\dfrac{6(3-\sqrt{7})}{2}\). Reduce \(6 \div 2 = 3\): \(3(3-\sqrt{7})\).
  4. Distribute: \(9 - 3\sqrt{7}\).

Step 4 of 5Practice

Type your answer and press Check. If you miss twice, open the worked solution. Then mark how it went. In every problem, assume all variables are positive. Some problems tell you the shape of the answer and ask for one missing number.

\(\dfrac{\sqrt{75}}{\sqrt{3}}\)

One root: \(\sqrt{\dfrac{75}{3}} = \sqrt{25} = 5\).

Simplify \(\dfrac{\sqrt{48x^{5}}}{\sqrt{3x}}\).

One root: \(\sqrt{\dfrac{48x^5}{3x}} = \sqrt{16x^4}\) (divide \(48 \div 3 = 16\); subtract exponents, \(5 - 1 = 4\)).

\(\sqrt{16x^4} = 4x^2\), because \((4x^2)^2 = 16x^4\). Answer: \(4x^2\).

Rationalize the denominator of \(\dfrac{10}{\sqrt{5}}\). The simplified answer has the form \(k\sqrt{5}\). What number is \(k\)?

Multiply top and bottom by \(\sqrt{5}\): \(\dfrac{10\sqrt{5}}{5}\).

Reduce \(10 \div 5 = 2\): \(2\sqrt{5}\). Answer: \(k = 2\).

Which is \(\dfrac{6}{\sqrt[3]{2}}\) with a rational denominator?

  1. \(3\sqrt[3]{4}\)
  2. \(3\sqrt[3]{2}\)
  3. \(6\sqrt[3]{4}\)
  4. \(3\sqrt{2}\)

To make a perfect cube on the bottom, you need \(\sqrt[3]{8}\). Since \(2 \cdot 4 = 8\), multiply top and bottom by \(\sqrt[3]{4}\).

\(\dfrac{6\sqrt[3]{4}}{\sqrt[3]{8}} = \dfrac{6\sqrt[3]{4}}{2} = 3\sqrt[3]{4}\). Answer: (a).

(b) multiplies by \(\sqrt[3]{2}\), which only gives \(\sqrt[3]{4}\) on the bottom (still a radical). (c) forgets to divide by 2. (d) treats the cube root like a square root.

Rationalize the denominator of \(\dfrac{4}{3-\sqrt{5}}\). The simplified answer has the form \(a + \sqrt{5}\). What number is \(a\)?

Multiply top and bottom by the conjugate \(3+\sqrt{5}\): \(\dfrac{4(3+\sqrt{5})}{(3-\sqrt{5})(3+\sqrt{5})}\).

Bottom: \(9 - 5 = 4\). So it is \(\dfrac{4(3+\sqrt{5})}{4} = 3 + \sqrt{5}\). Answer: \(a = 3\).

Simplify \(\sqrt{\dfrac{5}{8}}\) with no radical in the denominator. The answer has the form \(\dfrac{\sqrt{10}}{k}\). What number is \(k\)?

Split the root: \(\dfrac{\sqrt{5}}{\sqrt{8}}\). Simplify the bottom: \(\sqrt{8} = 2\sqrt{2}\), so it is \(\dfrac{\sqrt{5}}{2\sqrt{2}}\).

Multiply top and bottom by \(\sqrt{2}\): \(\dfrac{\sqrt{10}}{2 \cdot 2} = \dfrac{\sqrt{10}}{4}\). Answer: \(k = 4\).

Which expression is equal to \(\dfrac{\sqrt{2}}{\sqrt{2}+1}\)?

  1. \(\sqrt{2} - 2\)
  2. \(2 + \sqrt{2}\)
  3. \(2 - 2\sqrt{2}\)
  4. \(2 - \sqrt{2}\)

Multiply top and bottom by the conjugate \(\sqrt{2} - 1\).

Bottom: \((\sqrt{2})^2 - 1^2 = 2 - 1 = 1\). Top: \(\sqrt{2}(\sqrt{2} - 1) = 2 - \sqrt{2}\).

Over a denominator of 1, the answer is \(2 - \sqrt{2}\): (d). Choice (a) has the wrong sign (bottom taken as \(-1\)), (b) flips the sign inside, and (c) gets \(\sqrt{2} \cdot 1\) wrong.

Rationalize the denominator of \(\dfrac{6}{\sqrt{7}+2}\) and simplify. Write the result as \(a\sqrt{7} + b\), then type the pair \((a, b)\).

Multiply top and bottom by the conjugate \(\sqrt{7} - 2\). Bottom: \(7 - 4 = 3\).

\(\dfrac{6(\sqrt{7}-2)}{3} = 2(\sqrt{7} - 2) = 2\sqrt{7} - 4\).

So \(a = 2\) and \(b = -4\). Answer: \((2, -4)\).

Step 5 of 5Wrap-up

You can now divide radicals under one root and clear a radical from a denominator: multiply by the root for one term, and by the conjugate for two terms.

Read or watch next: OpenStax Intermediate Algebra 2e, 8.5 Divide Radical Expressions (free textbook, with more practice at the end of the section).

Also good: Khan Academy: Algebra 1, Simplifying square roots (short videos). Quick rules: Radicals and complex numbers cheat sheet.

Something unclear? Ask your teacher bot a follow-up question, for example: "Why does multiplying by the conjugate get rid of the square root in the bottom?" It knows this course and can explain things another way.