Rationalizing the Denominator

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Rationalizing the denominator means rewriting a fraction so there is no radical in the denominator. For one square root, multiply the top and bottom by that root. For a sum like 3 + \sqrt{2}, multiply by its conjugate, 3 - \sqrt{2}. For example, \frac{6}{\sqrt{3}} = \frac{6\sqrt{3}}{3} = 2\sqrt{3}.

Before this lesson: You should know simplifying square roots, multiplying radicals and how to work with fractions.

Why Rationalize the Denominator?

The denominator is the bottom of a fraction. A fraction like \frac{1}{\sqrt{2}} has a radical there. Math teachers and tests expect the answer without it, so \frac{\sqrt{2}}{2} is the simplest form.

The trick is to multiply by a form of 1. Since \frac{\sqrt{3}}{\sqrt{3}} = 1, multiplying by it does not change the value. But it turns \sqrt{3} \cdot \sqrt{3} into 3, a whole number.

The fraction 6 over √3 is multiplied by √3 over √3, which equals 1. The new fraction is 6√3 over 3, because √3 · √3 = 3. It simplifies to 2√3, with no radical in the denominator.
Multiply the top and bottom by √3. The bottom becomes 3, a whole number.

Rules for Rationalizing

\begin{aligned} \frac{a}{\sqrt{b}} &= \frac{a}{\sqrt{b}} \cdot \frac{\sqrt{b}}{\sqrt{b}} = \frac{a\sqrt{b}}{b} \\ \frac{c}{a + \sqrt{b}} &= \frac{c}{a + \sqrt{b}} \cdot \frac{a - \sqrt{b}}{a - \sqrt{b}} = \frac{c(a - \sqrt{b})}{a^2 - b} \end{aligned}

The conjugate of a + \sqrt{b} is a - \sqrt{b}. It has the same terms with the opposite sign in the middle. Their product, a^2 - b, has no radical.

Steps for rationalizing the denominator

  1. Simplify any radicals first.
  2. If the denominator is one square root, multiply the top and bottom by that square root.
  3. If the denominator is a sum or difference, multiply the top and bottom by its conjugate.
  4. Multiply out the top and the bottom.
  5. Simplify. Reduce the fraction if you can.

Worked Examples

Example 1: One square root

Rationalize \frac{5}{\sqrt{2}}.

Step 1. Multiply by \frac{\sqrt{2}}{\sqrt{2}}. It equals 1, so the value doesn’t change.

\frac{5}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}}

Step 2. Multiply the tops and multiply the bottoms (see multiplying and dividing fractions).

\frac{5}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{5\sqrt{2}}{\sqrt{2} \cdot \sqrt{2}}

Step 3. A square root times itself is the number under it: \sqrt{2} \cdot \sqrt{2} = 2.

\frac{5\sqrt{2}}{\sqrt{2} \cdot \sqrt{2}} = \frac{5\sqrt{2}}{2}

Answer: \frac{5\sqrt{2}}{2}. The bottom has no radical now.

Example 2: Reduce after rationalizing

Rationalize \frac{6}{\sqrt{3}}.

Step 1. Multiply by \frac{\sqrt{3}}{\sqrt{3}}.

\frac{6}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{6\sqrt{3}}{3}

Step 2. Divide 6 by 3. The \sqrt{3} stays as it is.

\frac{6\sqrt{3}}{3} = 2\sqrt{3}

Answer: 2\sqrt{3}

Example 3: Simplify the radical first

Rationalize \frac{3}{\sqrt{12}}.

Step 1. Simplify the bottom first: \sqrt{12} = \sqrt{4 \cdot 3} = 2\sqrt{3}.

\frac{3}{\sqrt{12}} = \frac{3}{2\sqrt{3}}

Step 2. Multiply by \frac{\sqrt{3}}{\sqrt{3}}.

\frac{3}{2\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{3\sqrt{3}}{2 \cdot 3}

Step 3. Multiply the bottom.

\frac{3\sqrt{3}}{2 \cdot 3} = \frac{3\sqrt{3}}{6}

Step 4. Divide the top and bottom by 3.

\frac{3\sqrt{3}}{6} = \frac{\sqrt{3}}{2}

Answer: \frac{\sqrt{3}}{2}

Example 4: Use the conjugate

Rationalize \frac{4}{3 + \sqrt{5}}.

Step 1. The bottom has two terms. Multiply by its conjugate over itself. The conjugate of 3 + \sqrt{5} is 3 - \sqrt{5}.

\frac{4}{3 + \sqrt{5}} \cdot \frac{3 - \sqrt{5}}{3 - \sqrt{5}} = \frac{4(3 - \sqrt{5})}{(3 + \sqrt{5})(3 - \sqrt{5})}

Step 2. Multiply the bottom. A sum times a difference is a difference of squares (see special products of binomials).

(3 + \sqrt{5})(3 - \sqrt{5}) = 3^2 - (\sqrt{5})^2 = 9 - 5 = 4

Step 3. Replace the bottom with 4.

\frac{4(3 - \sqrt{5})}{(3 + \sqrt{5})(3 - \sqrt{5})} = \frac{4(3 - \sqrt{5})}{4}

Step 4. Divide the top and bottom by 4.

\frac{4(3 - \sqrt{5})}{4} = 3 - \sqrt{5}

Answer: 3 - \sqrt{5}

Example 5: Word problem

A rectangle has an area of 12 square feet. Its length is \sqrt{6} feet. Find the exact width. Then estimate it, using \sqrt{6} \approx 2.45.

Step 1. Width is area divided by length.

w = \frac{12}{\sqrt{6}}

Step 2. Multiply by \frac{\sqrt{6}}{\sqrt{6}}.

w = \frac{12}{\sqrt{6}} \cdot \frac{\sqrt{6}}{\sqrt{6}} = \frac{12\sqrt{6}}{6}

Step 3. Divide 12 by 6.

w = 2\sqrt{6}

Step 4. Estimate: replace \sqrt{6} with 2.45 and multiply.

2 \cdot 2.45 = 4.9

Answer: The width is 2\sqrt{6} feet, or about 4.9 feet.

Check: Length times width: \sqrt{6} \cdot 2\sqrt{6} = 2 \cdot 6 = 12.

Common Mistakes

MistakeExample of the errorFix
Multiplying only the bottom\frac{5}{\sqrt{2}} = \frac{5}{2}Multiply the top too: \frac{5\sqrt{2}}{2}.
Canceling a number under the radical with one outside\frac{\sqrt{6}}{3} = \sqrt{2}You can’t divide inside the radical by 3. \frac{\sqrt{6}}{3} is already simplified.
Using the same sign instead of the conjugate\frac{1}{2 + \sqrt{3}} \cdot \frac{2 + \sqrt{3}}{2 + \sqrt{3}}The bottom still has a radical. Use 2 - \sqrt{3}: the bottom becomes 4 - 3 = 1.
Squaring the conjugate product wrong(3 + \sqrt{5})(3 - \sqrt{5}) = 9 + 5It is a difference of squares: 9 - 5 = 4.
Forgetting to reduce\frac{6\sqrt{3}}{3}Divide 6 by 3: 2\sqrt{3}.

Square Roots of Fractions

A square root of a fraction splits into a top root and a bottom root. Then rationalize the bottom.

\sqrt{\frac{2}{3}} = \frac{\sqrt{2}}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{6}}{3}
DenominatorMultiply top and bottom byNew denominator
\sqrt{7}\sqrt{7}7
2\sqrt{5}\sqrt{5}10
1 + \sqrt{2}1 - \sqrt{2}1 - 2 = -1
\sqrt{5} - \sqrt{3}\sqrt{5} + \sqrt{3}5 - 3 = 2

The conjugate trick is the same difference of squares pattern: (a + b)(a - b) = a^2 - b^2.

Practice Problems

  1. Rationalize \frac{7}{\sqrt{7}}.
  2. Rationalize \frac{3}{\sqrt{5}}.
  3. Rationalize \frac{10}{\sqrt{2}}.
  4. Simplify \frac{\sqrt{2}}{\sqrt{6}}.
  5. Rationalize \frac{5}{\sqrt{20}}.
  6. Simplify \sqrt{\frac{5}{8}}.
  7. Rationalize \frac{2}{1 + \sqrt{3}}.
  8. Rationalize \frac{6}{4 - \sqrt{10}}.
  9. Rationalize \frac{1}{\sqrt{5} - \sqrt{3}}.
  10. A rectangle has an area of 15 square meters and a length of \sqrt{5} meters. Find the exact width. Then estimate it, using \sqrt{5} \approx 2.24.
Show answers
  1. Multiply by \frac{\sqrt{7}}{\sqrt{7}}: \frac{7}{\sqrt{7}} \cdot \frac{\sqrt{7}}{\sqrt{7}} = \frac{7\sqrt{7}}{7}
    Divide 7 by 7: \frac{7\sqrt{7}}{7} = \sqrt{7}
  2. Multiply by \frac{\sqrt{5}}{\sqrt{5}}: \frac{3}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{3\sqrt{5}}{5}
    3 and 5 have no common factor, so this is the answer
  3. Multiply by \frac{\sqrt{2}}{\sqrt{2}}: \frac{10}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{10\sqrt{2}}{2}
    Divide 10 by 2: \frac{10\sqrt{2}}{2} = 5\sqrt{2}
  4. Multiply by \frac{\sqrt{6}}{\sqrt{6}}: \frac{\sqrt{2}}{\sqrt{6}} \cdot \frac{\sqrt{6}}{\sqrt{6}} = \frac{\sqrt{12}}{6}
    Simplify the top: \sqrt{12} = 2\sqrt{3}: \frac{\sqrt{12}}{6} = \frac{2\sqrt{3}}{6}
    Divide the top and bottom by 2: \frac{2\sqrt{3}}{6} = \frac{\sqrt{3}}{3}
  5. Simplify the bottom: \sqrt{20} = 2\sqrt{5}: \frac{5}{\sqrt{20}} = \frac{5}{2\sqrt{5}}
    Multiply by \frac{\sqrt{5}}{\sqrt{5}}: \frac{5}{2\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{5\sqrt{5}}{10}
    Divide the top and bottom by 5: \frac{5\sqrt{5}}{10} = \frac{\sqrt{5}}{2}
  6. Take the square root of the top and bottom: \sqrt{\frac{5}{8}} = \frac{\sqrt{5}}{\sqrt{8}}
    Simplify the bottom: \sqrt{8} = 2\sqrt{2}: \frac{\sqrt{5}}{\sqrt{8}} = \frac{\sqrt{5}}{2\sqrt{2}}
    Multiply by \frac{\sqrt{2}}{\sqrt{2}}: \frac{\sqrt{5}}{2\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{10}}{2 \cdot 2}
    Multiply the bottom: \frac{\sqrt{10}}{2 \cdot 2} = \frac{\sqrt{10}}{4}
  7. Multiply by the conjugate over itself, \frac{1 - \sqrt{3}}{1 - \sqrt{3}}. The bottom is 1^2 - (\sqrt{3})^2 = 1 - 3: \frac{2}{1 + \sqrt{3}} \cdot \frac{1 - \sqrt{3}}{1 - \sqrt{3}} = \frac{2(1 - \sqrt{3})}{1 - 3}
    Subtract on the bottom: \frac{2(1 - \sqrt{3})}{1 - 3} = \frac{2(1 - \sqrt{3})}{-2}
    Divide 2 by -2: \frac{2(1 - \sqrt{3})}{-2} = -(1 - \sqrt{3})
    Distribute the negative sign: -(1 - \sqrt{3}) = \sqrt{3} - 1
  8. Multiply by the conjugate over itself, \frac{4 + \sqrt{10}}{4 + \sqrt{10}}. The bottom is 4^2 - (\sqrt{10})^2 = 16 - 10: \frac{6}{4 - \sqrt{10}} \cdot \frac{4 + \sqrt{10}}{4 + \sqrt{10}} = \frac{6(4 + \sqrt{10})}{16 - 10}
    Subtract on the bottom: \frac{6(4 + \sqrt{10})}{16 - 10} = \frac{6(4 + \sqrt{10})}{6}
    Divide the top and bottom by 6: \frac{6(4 + \sqrt{10})}{6} = 4 + \sqrt{10}
  9. Multiply by the conjugate over itself, \frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} + \sqrt{3}}. The bottom is (\sqrt{5})^2 - (\sqrt{3})^2 = 5 - 3: \frac{1}{\sqrt{5} - \sqrt{3}} \cdot \frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} + \sqrt{3}} = \frac{\sqrt{5} + \sqrt{3}}{5 - 3}
    Subtract on the bottom: \frac{\sqrt{5} + \sqrt{3}}{5 - 3} = \frac{\sqrt{5} + \sqrt{3}}{2}
  10. Width is area divided by length: w = \frac{15}{\sqrt{5}}
    Multiply by \frac{\sqrt{5}}{\sqrt{5}}: \frac{15}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{15\sqrt{5}}{5}
    Divide 15 by 5: \frac{15\sqrt{5}}{5} = 3\sqrt{5}
    Estimate: replace \sqrt{5} with 2.24 and multiply: 3 \cdot 2.24 = 6.72
    The width is 3\sqrt{5} meters, or about 6.72 meters.

FAQ

Why do we rationalize the denominator?

It puts answers in one standard form, so they are easy to compare and check. It also makes a value easier to estimate. \frac{\sqrt{2}}{2} is clearly about half of 1.41.

What is a conjugate?

The conjugate of a + \sqrt{b} is a - \sqrt{b}. Multiplying them gives a^2 - b, which has no radical.

Can you leave a radical in the denominator?

The value is the same, but most teachers and tests want it rationalized. Our style is to always rationalize final answers.

How do you rationalize a denominator with two terms?

Multiply the top and bottom by the conjugate of the denominator. For \frac{2}{1 + \sqrt{3}}, multiply by \frac{1 - \sqrt{3}}{1 - \sqrt{3}}.

Summary

  • Rationalizing removes radicals from the denominator.
  • For one square root, multiply by \frac{\sqrt{b}}{\sqrt{b}}. That is multiplying by 1.
  • For a sum or difference, multiply by the conjugate.
  • Simplify radicals first, and reduce the fraction at the end.
  • Never cancel a number inside a radical with one outside it.

Previous lesson: Multiplying Radicals

Next lesson: Graphing Quadratic Functions

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