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Algebra 1 · N.RN.A.2

Rationalizing a denominator

A fraction has a square root in its denominator. How can you make the bottom a whole number without changing the fraction’s value?

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Rationalizing a denominator
Simplifying a square root

When you reach for it

Dividing roots sometimes leaves a root below the line, as in \displaystyle \frac{1}{\sqrt{2}}. That is an exact, correct number. Rationalizing the denominator means rewriting such a fraction as an equal fraction whose denominator is a rational number, in practice an integer. The name says what happens: the denominator becomes rational.

If the value does not change, why bother? Three reasons.

Estimating. \displaystyle \sqrt{2} \approx 1.41421. To estimate \displaystyle \frac{1}{\sqrt{2}} directly you would divide 1 by 1.41421, long division by a decimal with five places. Rewritten below, the same number is \displaystyle \frac{\sqrt{2}}{2}, and half of 1.41421 is 0.70711 at a glance.

Comparing and adding. \displaystyle \frac{1}{\sqrt{2}} + \sqrt{2} looks like two unlike things. Written as \displaystyle \frac{\sqrt{2}}{2} + \frac{2\sqrt{2}}{2}, it is two like radicals over a common denominator, and the sum is \displaystyle \frac{3\sqrt{2}}{2}.

One agreed form. \displaystyle \frac{1}{\sqrt{2}}, \displaystyle \frac{\sqrt{2}}{2}, \displaystyle \frac{2}{\sqrt{8}} and \displaystyle \sqrt{\frac{1}{2}} are all the same number. If everyone writes it as \displaystyle \frac{\sqrt{2}}{2}, two answers can be compared by eye. So an expression with square roots counts as simplest form only when no radicand has a perfect-square factor other than 1 (as in Simplifying a square root), no fraction sits under a root, and no root sits in a denominator.

Reach for it whenever a calculation ends with a root below the line: a side of a square from its diagonal, a 45-45-90 triangle, or a quotient of two roots that did not divide evenly.

The procedure

Two facts do all the work.

Multiplying by 1 changes nothing. For any non-zero number m, \displaystyle \frac{m}{m} = 1, so multiplying the top and the bottom of a fraction by the same m gives an equal fraction.

A square root times itself is its radicand. For a \ge 0, \displaystyle \sqrt{a} \cdot \sqrt{a} = a. The product is no longer a root.

One term below the line

When the denominator is a root, or a number times a root, multiply top and bottom by that root:

\frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}

Only the root needs clearing. In \displaystyle \frac{7}{2\sqrt{5}} the 2 is already an integer, so the multiplier is \displaystyle \frac{\sqrt{5}}{\sqrt{5}} and the bottom becomes 2 \cdot 5 = 10.

Two habits keep the numbers small. Simplify the root below the line first, so the multiplier is the smallest root that will do: \displaystyle \sqrt{18} = 3\sqrt{2}, so a fraction with \displaystyle \sqrt{18} below needs only \displaystyle \sqrt{2}. And reduce at the end: once the denominator is an integer, it may share a factor with the numbers outside the roots on top. A fraction under one root, \displaystyle \sqrt{\frac{a}{b}}, is first split into \displaystyle \frac{\sqrt{a}}{\sqrt{b}} by the quotient rule and then treated the same way.

Two terms below the line

Multiplying \displaystyle 3 - \sqrt{2} by \displaystyle \sqrt{2} does not help: the product is \displaystyle 3\sqrt{2} - 2, still with a root. What clears it is the conjugate, the same two terms with the sign between them changed: \displaystyle 3 + \sqrt{2}. By the sum-times-difference pattern of Special products,

(p + q)(p - q) = p^2 - q^2

and the square of a square root has no root left in it:

\left(3 - \sqrt{2}\right)\left(3 + \sqrt{2}\right) = 9 - 2 = 7

The middle terms, \displaystyle +3\sqrt{2} and \displaystyle -3\sqrt{2}, cancel. When both terms are roots the same thing happens:

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

This is the conjugate pair from Operations with radicals, now used on purpose. So for a two-term denominator, multiply top and bottom by its conjugate. Keep the numerator in brackets until the denominator is worked out; a common factor is then easy to see.

Checking with a decimal

Every step is a multiplication by 1 or a rewrite of a root, so the value cannot change. A decimal check catches arithmetic slips: evaluate the original and the answer to four or five places, and they must agree.

Worked examples

Example 1

One root below the line

Rationalize (a) \displaystyle \frac{5}{\sqrt{6}} \qquad (b) \displaystyle \frac{7}{2\sqrt{5}} \qquad (c) \displaystyle \frac{14}{\sqrt{7}}

Solution

(a) Multiply by \displaystyle \frac{\sqrt{6}}{\sqrt{6}}, which is 1:

\frac{5}{\sqrt{6}} \cdot \frac{\sqrt{6}}{\sqrt{6}} = \frac{5\sqrt{6}}{6}

5 and 6 share no factor, so this is finished.

(b) Only \displaystyle \sqrt{5} is a root, so the multiplier is \displaystyle \frac{\sqrt{5}}{\sqrt{5}}:

\frac{7}{2\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{7\sqrt{5}}{2 \cdot 5} = \frac{7\sqrt{5}}{10}

(c) Multiply by \displaystyle \frac{\sqrt{7}}{\sqrt{7}}, then reduce:

\frac{14}{\sqrt{7}} \cdot \frac{\sqrt{7}}{\sqrt{7}} = \frac{14\sqrt{7}}{7} = 2\sqrt{7}

The 7 below divides the 14 on top, so no fraction is left at all. Another way to see it: \displaystyle 14 = 2 \cdot 7 = 2\sqrt{7} \cdot \sqrt{7}, so one \displaystyle \sqrt{7} cancels directly.

Example 2

Simplify the root first

Rationalize \displaystyle \frac{12}{\sqrt{18}}.

Solution

18 = 9 \cdot 2, so \displaystyle \sqrt{18} = 3\sqrt{2}, and the 3 cancels with the 12:

\frac{12}{\sqrt{18}} = \frac{12}{3\sqrt{2}} = \frac{4}{\sqrt{2}}

Now multiply by \displaystyle \frac{\sqrt{2}}{\sqrt{2}}:

\frac{4}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{4\sqrt{2}}{2} = 2\sqrt{2}

Multiplying by \displaystyle \frac{\sqrt{18}}{\sqrt{18}} straight away also works, with bigger numbers: \displaystyle \frac{12\sqrt{18}}{18} = \frac{12 \cdot 3\sqrt{2}}{18} = \frac{36\sqrt{2}}{18} = 2\sqrt{2}. Same answer, more to simplify.

Check. 12 \div 4.24264 \approx 2.8284, and 2 \times 1.41421 \approx 2.8284.

Example 3

A fraction under one root

Write \displaystyle \sqrt{\frac{7}{20}} in simplest form.

Solution

A fraction under a root is not simplest form. Split it with the quotient rule, simplify the root below, then rationalize:

\begin{aligned} \sqrt{\frac{7}{20}} &= \frac{\sqrt{7}}{\sqrt{20}} = \frac{\sqrt{7}}{2\sqrt{5}} \\ &= \frac{\sqrt{7} \cdot \sqrt{5}}{2\sqrt{5} \cdot \sqrt{5}} = \frac{\sqrt{35}}{10} \end{aligned}

A second route makes the bottom a perfect square while it is still under the root. Since 20 \cdot 5 = 100, multiply the top and bottom of the inner fraction by 5:

\sqrt{\frac{7}{20}} = \sqrt{\frac{35}{100}} = \frac{\sqrt{35}}{10}

35 = 5 \cdot 7 has no square factor, so \displaystyle \sqrt{35} is in simplest form. Check. \displaystyle \sqrt{0.35} \approx 0.59161, and 5.91608 \div 10 \approx 0.59161.

Example 4

An integer and a root below

Rationalize (a) \displaystyle \frac{4}{3 - \sqrt{2}} \qquad (b) \displaystyle \frac{10}{\sqrt{11} + 1}

Solution

(a) The conjugate of \displaystyle 3 - \sqrt{2} is \displaystyle 3 + \sqrt{2}, and the bottom becomes \displaystyle 3^2 - \left(\sqrt{2}\right)^2 = 7:

\begin{aligned} \frac{4}{3 - \sqrt{2}} \cdot \frac{3 + \sqrt{2}}{3 + \sqrt{2}} &= \frac{4\left(3 + \sqrt{2}\right)}{9 - 2} \\ &= \frac{12 + 4\sqrt{2}}{7} \end{aligned}

7 does not divide 4, so nothing cancels. Check. 3 - 1.41421 = 1.58579 and 4 \div 1.58579 \approx 2.5224; also (12 + 5.65685) \div 7 \approx 2.5224.

(b) The conjugate of \displaystyle \sqrt{11} + 1 is \displaystyle \sqrt{11} - 1:

\begin{aligned} \frac{10}{\sqrt{11} + 1} \cdot \frac{\sqrt{11} - 1}{\sqrt{11} - 1} &= \frac{10\left(\sqrt{11} - 1\right)}{11 - 1} \\ &= \sqrt{11} - 1 \end{aligned}

The bottom came out as 10, which cancels with the 10 on top. Keeping the numerator in brackets is what made that easy to spot.

Example 5

Two roots below

Rationalize (a) \displaystyle \frac{1}{\sqrt{6} + \sqrt{2}} \qquad (b) \displaystyle \frac{\sqrt{6}}{\sqrt{6} - \sqrt{2}}

Solution

(a) Multiply by the conjugate, \displaystyle \sqrt{6} - \sqrt{2}. The bottom becomes 6 - 2 = 4:

\frac{1}{\sqrt{6} + \sqrt{2}} \cdot \frac{\sqrt{6} - \sqrt{2}}{\sqrt{6} - \sqrt{2}} = \frac{\sqrt{6} - \sqrt{2}}{4}

Check. 1 \div (2.44949 + 1.41421) \approx 0.25882, and (2.44949 - 1.41421) \div 4 \approx 0.25882.

(b) Now the conjugate is \displaystyle \sqrt{6} + \sqrt{2}, and the bottom is again 4. The top needs the distributive property and then simplifying, since \displaystyle \sqrt{6} \cdot \sqrt{2} = \sqrt{12} = 2\sqrt{3}:

\begin{aligned} \frac{\sqrt{6}\left(\sqrt{6} + \sqrt{2}\right)}{4} &= \frac{6 + \sqrt{12}}{4} = \frac{6 + 2\sqrt{3}}{4} \\ &= \frac{3 + \sqrt{3}}{2} \end{aligned}

The last step divides top and bottom by 2. A root on top is allowed; only the denominator has to be free of roots.

Example 6

The side of a square kite

Ailbhe builds a square kite from two crossed spars, each 70 cm long, joined at their middles at right angles. The spars are the diagonals of the square. How long is each side, exactly and to the nearest millimeter?

70\ \text{cm}
70\ \text{cm}
s
Figure 1The kite in Example 6. Each spar is a whole diagonal of the square, 70 cm long.
Solution

A diagonal cuts the square into two 45-45-90 triangles, so the diagonal is \displaystyle \sqrt{2} times the side: \displaystyle s\sqrt{2} = 70. Dividing by \displaystyle \sqrt{2} and rationalizing,

s = \frac{70}{\sqrt{2}} = \frac{70\sqrt{2}}{2} = 35\sqrt{2}

Now the estimate is a multiplication: 35 \times 1.41421 \approx 49.497, so each side is 49.5 cm to the nearest millimeter. From \displaystyle \frac{70}{\sqrt{2}} the same number needs 70 \div 1.41421.

Check a second way. The spars cross at right angles at their middles, so each side is the hypotenuse of a right triangle with legs 35 and 35: \displaystyle s = \sqrt{35^2 + 35^2} = \sqrt{2 \cdot 35^2} = 35\sqrt{2}.

Boundaries

Nothing may be zero. The fraction needs a non-zero denominator to begin with, and \displaystyle \frac{\sqrt{a}}{\sqrt{a}} = 1 needs a > 0. With a letter, \displaystyle \frac{3}{\sqrt{x}} = \frac{3\sqrt{x}}{x} holds only for x > 0.

Real square roots only. Every radicand here is non-negative; a square root of a negative number is not a real number.

The new denominator is rational, not always positive. \displaystyle \left(\sqrt{2} - 3\right)\left(\sqrt{2} + 3\right) = 2 - 9 = -7, so \displaystyle \frac{1}{\sqrt{2} - 3} = -\frac{3 + \sqrt{2}}{7}. Move the minus sign to the front.

One conjugate clears two terms, not three. For \displaystyle 1 + \sqrt{2} + \sqrt{3}, multiplying by \displaystyle 1 + \sqrt{2} - \sqrt{3} gives \displaystyle \left(1 + \sqrt{2}\right)^2 - 3 = 2\sqrt{2}, which still needs one more step.

Cube roots need a different multiplier. \displaystyle \sqrt[3]{2} \cdot \sqrt[3]{2} = \sqrt[3]{4} is still a root; \displaystyle \frac{1}{\sqrt[3]{2}} is cleared by \displaystyle \sqrt[3]{4}, because \displaystyle \sqrt[3]{2} \cdot \sqrt[3]{4} = \sqrt[3]{8} = 2.

Simplest form is an agreement, not a law of arithmetic. \displaystyle \frac{1}{\sqrt{2}} is not wrong, only not in the agreed form. Some later work moves a root into the denominator on purpose.

Where this goes next

Answers from the quadratic formula and lengths in right triangles often arrive as roots inside fractions, and this is how they are put in simplest form. In Algebra 2, dividing complex numbers uses the same conjugate idea to make a denominator real. Much later, in calculus, finding limits by algebra uses the same conjugate trick, sometimes to clear a root from the top of a fraction instead.

Exercises

One root below the line

Rationalize the denominator and simplify.

  1. \displaystyle \frac{3}{\sqrt{11}}
  2. \displaystyle \frac{20}{\sqrt{10}}
  3. \displaystyle \frac{5}{3\sqrt{7}}
  4. \displaystyle \frac{\sqrt{3}}{\sqrt{10}}
  5. \displaystyle \frac{6\sqrt{5}}{\sqrt{15}}

Simplify first

Write in simplest form.

  1. \displaystyle \frac{9}{\sqrt{27}}
  2. \displaystyle \frac{7}{\sqrt{50}}
  3. \displaystyle \sqrt{\frac{2}{45}}
  4. \displaystyle \sqrt{\frac{11}{3}}

Two terms below the line

Rationalize the denominator and simplify.

  1. \displaystyle \frac{2}{5 + \sqrt{3}}
  2. \displaystyle \frac{8}{\sqrt{17} - 3}
  3. \displaystyle \frac{1}{\sqrt{10} - \sqrt{7}}
  4. \displaystyle \frac{\sqrt{3}}{\sqrt{3} + 1}
  5. \displaystyle \frac{\sqrt{7} + 2}{\sqrt{7} - 2}

Checking and using it

  1. Akosua says that \displaystyle \frac{1}{4 - \sqrt{15}} = 4 + \sqrt{15}. Test the claim with decimals, then prove it or correct it.
  2. A rectangular banner has an area of 6 square meters and a length of \displaystyle \left(2 + \sqrt{2}\right) meters. Find its height exactly, in simplest form, and to the nearest centimeter.

Reasoning

  1. Multiplying the top and bottom of \displaystyle \frac{5}{4 - \sqrt{3}} by \displaystyle \sqrt{3} does not rationalize the denominator. Explain why the root does not go away and why multiplying by the conjugate \displaystyle 4 + \sqrt{3} does clear it, then finish the job.
  2. Show in two different ways that \displaystyle \frac{1}{\sqrt{2}} and \displaystyle \frac{\sqrt{2}}{2} are the same number: once by rationalizing, and once by squaring both.
Answers
  1. \displaystyle \frac{3\sqrt{11}}{11}. Multiply by \displaystyle \frac{\sqrt{11}}{\sqrt{11}}.
  2. \displaystyle 2\sqrt{10}. \displaystyle \frac{20\sqrt{10}}{10}, and 10 divides 20.
  3. \displaystyle \frac{5\sqrt{7}}{21}. Multiply by \displaystyle \frac{\sqrt{7}}{\sqrt{7}}; the bottom is 3 \cdot 7 = 21.
  4. \displaystyle \frac{\sqrt{30}}{10}. Multiply by \displaystyle \frac{\sqrt{10}}{\sqrt{10}}; the top is \displaystyle \sqrt{3} \cdot \sqrt{10} = \sqrt{30}.
  5. \displaystyle 2\sqrt{3}. First \displaystyle \frac{\sqrt{5}}{\sqrt{15}} = \frac{1}{\sqrt{3}}, so the fraction is \displaystyle \frac{6}{\sqrt{3}} = \frac{6\sqrt{3}}{3}.
  6. \displaystyle \sqrt{3}. \displaystyle \sqrt{27} = 3\sqrt{3}, so \displaystyle \frac{9}{3\sqrt{3}} = \frac{3}{\sqrt{3}} = \frac{3\sqrt{3}}{3}.
  7. \displaystyle \frac{7\sqrt{2}}{10}. \displaystyle \sqrt{50} = 5\sqrt{2}; multiply \displaystyle \frac{7}{5\sqrt{2}} by \displaystyle \frac{\sqrt{2}}{\sqrt{2}}.
  8. \displaystyle \frac{\sqrt{10}}{15}. \displaystyle \frac{\sqrt{2}}{\sqrt{45}} = \frac{\sqrt{2}}{3\sqrt{5}}; multiply by \displaystyle \frac{\sqrt{5}}{\sqrt{5}} to get \displaystyle \frac{\sqrt{10}}{3 \cdot 5}.
  9. \displaystyle \frac{\sqrt{33}}{3}. \displaystyle \frac{\sqrt{11}}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}}. Or \displaystyle \sqrt{\frac{11}{3}} = \sqrt{\frac{33}{9}}.
  10. \displaystyle \frac{5 - \sqrt{3}}{11}. The bottom is 25 - 3 = 22, so the fraction is \displaystyle \frac{2\left(5 - \sqrt{3}\right)}{22}, and 2 cancels.
  11. \displaystyle \sqrt{17} + 3. Multiply by \displaystyle \sqrt{17} + 3; the bottom is 17 - 9 = 8, so the fraction is \displaystyle \frac{8\left(\sqrt{17} + 3\right)}{8} and the 8 cancels completely.
  12. \displaystyle \frac{\sqrt{10} + \sqrt{7}}{3}. Multiply by \displaystyle \sqrt{10} + \sqrt{7}; the bottom is 10 - 7 = 3.
  13. \displaystyle \frac{3 - \sqrt{3}}{2}. Multiply by \displaystyle \sqrt{3} - 1; the top is \displaystyle 3 - \sqrt{3} and the bottom 3 - 1 = 2.
  14. \displaystyle \frac{11 + 4\sqrt{7}}{3}. Multiply by \displaystyle \sqrt{7} + 2. The top is \displaystyle \left(\sqrt{7} + 2\right)^2 = 7 + 4\sqrt{7} + 4 and the bottom is 7 - 4 = 3.
  15. The claim is true. Decimals: 4 - 3.87298 = 0.12702 and 1 \div 0.12702 \approx 7.873, while 4 + 3.87298 \approx 7.873. Proof: multiplying by the conjugate \displaystyle 4 + \sqrt{15} gives a bottom of 16 - 15 = 1, so the fraction equals \displaystyle 4 + \sqrt{15} exactly.
  16. \displaystyle 6 - 3\sqrt{2} meters, about 1.76 m. The height is the area divided by the length, \displaystyle \frac{6}{2 + \sqrt{2}}. Multiply by the conjugate \displaystyle 2 - \sqrt{2}: the bottom is 4 - 2 = 2, so the height is \displaystyle \frac{6\left(2 - \sqrt{2}\right)}{2} = 3\left(2 - \sqrt{2}\right). Check: 6 - 4.24264 = 1.75736.
  17. \displaystyle \frac{20 + 5\sqrt{3}}{13}. Multiplying by \displaystyle \sqrt{3} gives a bottom of \displaystyle \left(4 - \sqrt{3}\right)\sqrt{3} = 4\sqrt{3} - 3: the 4 picks up the root, so a root is still there. Multiplying by \displaystyle 4 + \sqrt{3} makes the bottom a sum times a difference, \displaystyle \left(4 - \sqrt{3}\right)\left(4 + \sqrt{3}\right) = 4^2 - \left(\sqrt{3}\right)^2 = 13. The two middle terms, \displaystyle +4\sqrt{3} and \displaystyle -4\sqrt{3}, cancel, and the only root left is squared. The top is \displaystyle 5\left(4 + \sqrt{3}\right) = 20 + 5\sqrt{3}, and 13 shares no factor with 20 and 5.
  18. Rationalizing: \displaystyle \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}, a multiplication by 1. Squaring: \displaystyle \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} and \displaystyle \left(\frac{\sqrt{2}}{2}\right)^2 = \frac{2}{4} = \frac{1}{2}. Both numbers are positive and have the same square. Two different positive numbers cannot have the same square (the larger one has the larger square), so they are equal.
kg1128 · updated 2026-10-10 · CC-BY-SA-4.0