Rationalizing Denominators

Learning goals

Multiply by a disguised form of one to clear a root
Turn $\frac{a}{\sqrt{b}}$ into $\frac{a\sqrt{b}}{b}$
Use the conjugate on a two-term denominator
Recognize the difference of squares the conjugate creates
Simplify the result, reducing factors and any leftover radical

Multiplying by a disguised form of 1

Every rationalizing step is one idea in disguise, which is to multiply the fraction by $1$ . Any nonzero quantity divided by itself equals $1$ , so $\frac{\sqrt{2}}{\sqrt{2}} = 1$ and $\frac{2 - \sqrt{3}}{2 - \sqrt{3}} = 1$ . Multiplying a number by $1$ leaves its value untouched, which is exactly why the move is allowed. What changes is the form of the fraction, not the number it stands for.

The reason a well-chosen form of $1$ can clear a root comes down to one fact about square roots. For any $b \ge 0$ , a square root times itself gives back the number under it, so $\sqrt{b}\cdot\sqrt{b} = b$ .

Why $\sqrt{b}\cdot\sqrt{b} = b$ #

Recall the product rule for square roots: for nonnegative numbers, $\sqrt{a}\cdot\sqrt{b} = \sqrt{ab}$ , because the square root of a product is the product of the square roots. Take $a = b$ in that rule so the two factors are the same:

\sqrt{b}\cdot\sqrt{b} = \sqrt{b\cdot b} = \sqrt{b^2} = b.

The last step holds because $b$ is nonnegative, so $\sqrt{b^2}$ is $b$ itself. Squaring a square root and multiplying a square root by itself are the same operation, and both undo the root. This is why a lone $\sqrt{b}$ in a denominator disappears the moment you multiply by another $\sqrt{b}$ : the product is the plain number $b$ , with no radical left.

So the whole method is to pick the form of $1$ that turns the denominator into something with no radical. For a single root the choice is easy, and for a two-term denominator it is the conjugate.

Clearing a single square root

When the denominator is a single square root, the form of $1$ to use is that same root over itself. To rationalize $\frac{a}{\sqrt{b}}$ , multiply the numerator and the denominator by $\sqrt{b}$ :

\frac{a}{\sqrt{b}} = \frac{a}{\sqrt{b}}\cdot\frac{\sqrt{b}}{\sqrt{b}} = \frac{a\sqrt{b}}{b}.

The denominator $\sqrt{b}\cdot\sqrt{b}$ became $b$ , a whole number, and the root moved up to the numerator where the convention allows it. Notice that you must multiply the top by $\sqrt{b}$ as well; multiplying only the bottom would change the value.

Worked example 1 Rationalize $\frac{1}{\sqrt{2}}$ and $\frac{3}{\sqrt{5}}$

For $\frac{1}{\sqrt{2}}$ , multiply top and bottom by $\sqrt{2}$ . The bottom becomes $\sqrt{2}\cdot\sqrt{2} = 2$ :

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

As a check, $\frac{1}{\sqrt{2}} \approx 0.707$ and $\frac{\sqrt{2}}{2} \approx 0.707$ , the same number in a tidier form.

For $\frac{3}{\sqrt{5}}$ , multiply top and bottom by $\sqrt{5}$ . The bottom becomes $5$ :

\frac{3}{\sqrt{5}} = \frac{3}{\sqrt{5}}\cdot\frac{\sqrt{5}}{\sqrt{5}} = \frac{3\sqrt{5}}{5}.

The $3$ stays out front and the new root $\sqrt{5}$ rides along in the numerator.

Worked example 2 Rationalize and then simplify

Often the fraction reduces after you rationalize, or the radical simplifies first. Take $\frac{8}{\sqrt{2}}$ . Multiply by $\frac{\sqrt{2}}{\sqrt{2}}$ :

\frac{8}{\sqrt{2}} = \frac{8\sqrt{2}}{2} = 4\sqrt{2}.

The $8$ and the $2$ share a factor of $2$ , so the fraction reduces to $4\sqrt{2}$ .

Now take $\frac{6}{\sqrt{12}}$ . Here it pays to simplify the radical first, since $\sqrt{12} = \sqrt{4\cdot 3} = 2\sqrt{3}$ :

\frac{6}{\sqrt{12}} = \frac{6}{2\sqrt{3}} = \frac{3}{\sqrt{3}} = \frac{3\sqrt{3}}{3} = \sqrt{3}.

After cancelling the $6$ and the $2$ , the fraction $\frac{3}{\sqrt{3}}$ rationalizes to $\frac{3\sqrt{3}}{3}$ , which reduces to $\sqrt{3}$ . Simplifying $\sqrt{12}$ up front kept the numbers small.

Check your understanding

Rationalize $\frac{5}{\sqrt{3}}$ .

\frac{5\sqrt{3}}{3}

\frac{5}{3}

\frac{\sqrt{3}}{5}

5\sqrt{3}

Clearing a binomial denominator with its conjugate

A denominator with two terms, like $1 + \sqrt{2}$ , cannot be fixed by multiplying by a single root: multiplying $1 + \sqrt{2}$ by $\sqrt{2}$ gives $\sqrt{2} + 2$ , which still has a root. The move that works uses the difference-of-squares pattern you proved earlier in this chapter.

The conjugate of a two-term expression is the same expression with the sign between its terms flipped. The conjugate of $a + \sqrt{b}$ is $a - \sqrt{b}$ , and the conjugate of $a - \sqrt{b}$ is $a + \sqrt{b}$ . Multiplying an expression by its conjugate produces a difference of squares, and the square of a square root is rational, so every radical disappears.

Why $(a + \sqrt{b})(a - \sqrt{b}) = a^2 - b$ #

Multiply the expression by its conjugate exactly as you multiply any difference of squares, treating $\sqrt{b}$ as the second term:

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

The two middle products, $-a\sqrt{b}$ and $+a\sqrt{b}$ , are exact opposites and cancel, which is the same mechanism that made the difference of squares work. The remaining $(\sqrt{b})^2$ is just $b$ , because a square root squared returns the number under it:

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

The result $a^2 - b$ is a plain rational expression with no radical in sight. That is why the conjugate is the right form of $1$ to use: it converts a two-term radical denominator into a rational one in a single stroke.

To rationalize $\frac{1}{a + \sqrt{b}}$ , multiply the numerator and the denominator by the conjugate $a - \sqrt{b}$ :

\frac{1}{a + \sqrt{b}} = \frac{1}{a + \sqrt{b}}\cdot\frac{a - \sqrt{b}}{a - \sqrt{b}} = \frac{a - \sqrt{b}}{a^2 - b}.

The denominator is now rational. As always, the conjugate over itself is just $1$ , so the value is unchanged. The same conjugate device comes back in a later chapter for a different kind of number. In this lesson, though, every root is an ordinary square root of a positive number, so $a^2 - b$ is a plain real number.

Worked example 3 Rationalize $\frac{1}{1 + \sqrt{2}}$

The denominator is $1 + \sqrt{2}$ , so its conjugate is $1 - \sqrt{2}$ . Multiply top and bottom by it:

\frac{1}{1 + \sqrt{2}} = \frac{1}{1 + \sqrt{2}}\cdot\frac{1 - \sqrt{2}}{1 - \sqrt{2}} = \frac{1 - \sqrt{2}}{1^2 - (\sqrt{2})^2}.

The denominator is $1^2 - (\sqrt{2})^2 = 1 - 2 = -1$ :

\frac{1 - \sqrt{2}}{1 - 2} = \frac{1 - \sqrt{2}}{-1} = \sqrt{2} - 1.

Dividing by $-1$ flips both signs, so $1 - \sqrt{2}$ becomes $\sqrt{2} - 1$ . A quick check: $\frac{1}{1 + \sqrt{2}} \approx 0.414$ and $\sqrt{2} - 1 \approx 0.414$ .

Check your understanding

What is the conjugate of $3 - \sqrt{5}$ ?

3 + \sqrt{5}

-3 - \sqrt{5}

3 - \sqrt{5}

\sqrt{5} - 3

Worked example 4 Two more conjugate denominators

Rationalize $\frac{1}{2 - \sqrt{3}}$ . The conjugate is $2 + \sqrt{3}$ , and $(2 - \sqrt{3})(2 + \sqrt{3}) = 2^2 - (\sqrt{3})^2 = 4 - 3 = 1$ :

\frac{1}{2 - \sqrt{3}} = \frac{2 + \sqrt{3}}{4 - 3} = \frac{2 + \sqrt{3}}{1} = 2 + \sqrt{3}.

When the difference of squares comes out to $1$ , the denominator disappears entirely.

Now rationalize $\frac{4}{\sqrt{5} + 1}$ , where the numerator is not $1$ . Multiply top and bottom by the conjugate $\sqrt{5} - 1$ . The denominator is $(\sqrt{5})^2 - 1^2 = 5 - 1 = 4$ , and the numerator is $4(\sqrt{5} - 1)$ :

\frac{4}{\sqrt{5} + 1} = \frac{4(\sqrt{5} - 1)}{5 - 1} = \frac{4(\sqrt{5} - 1)}{4} = \sqrt{5} - 1.

The $4$ out front cancels the $4$ in the denominator, leaving $\sqrt{5} - 1$ . Always look for a common factor to cancel at the end.

When both terms are square roots

The conjugate trick works just as well when both terms are square roots. The product $(\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b})$ is again a difference of squares:

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

Both squares collapse to whole numbers, so the denominator becomes the plain difference $a - b$ , with no radical remaining.

Worked example 5 Rationalize $\frac{1}{\sqrt{5} - \sqrt{3}}$ and $\frac{6}{\sqrt{7} - \sqrt{5}}$

For $\frac{1}{\sqrt{5} - \sqrt{3}}$ , the conjugate is $\sqrt{5} + \sqrt{3}$ , and $(\sqrt{5} - \sqrt{3})(\sqrt{5} + \sqrt{3}) = 5 - 3 = 2$ :

\frac{1}{\sqrt{5} - \sqrt{3}} = \frac{\sqrt{5} + \sqrt{3}}{5 - 3} = \frac{\sqrt{5} + \sqrt{3}}{2}.

The denominator $5 - 3 = 2$ has no radical, and the numerator keeps both roots.

For $\frac{6}{\sqrt{7} - \sqrt{5}}$ , multiply by the conjugate $\sqrt{7} + \sqrt{5}$ . The denominator is $7 - 5 = 2$ , and the numerator is $6(\sqrt{7} + \sqrt{5})$ :

\frac{6}{\sqrt{7} - \sqrt{5}} = \frac{6(\sqrt{7} + \sqrt{5})}{7 - 5} = \frac{6(\sqrt{7} + \sqrt{5})}{2} = 3(\sqrt{7} + \sqrt{5}).

The $6$ over $2$ reduces to $3$ , giving $3\sqrt{7} + 3\sqrt{5}$ .

Check your understanding

Rationalize $\frac{1}{\sqrt{6} - \sqrt{5}}$ .

\sqrt{6} + \sqrt{5}

\frac{\sqrt{6} + \sqrt{5}}{11}

\sqrt{6} - \sqrt{5}

\frac{\sqrt{6} + \sqrt{5}}{2}

Common mistakes

Multiplying by the wrong sign. To clear $a + \sqrt{b}$ you multiply by $a - \sqrt{b}$ , not by $a + \sqrt{b}$ again. The same sign gives $(a + \sqrt{b})^2 = a^2 + 2a\sqrt{b} + b$ , which still contains a root.
Forgetting to multiply the numerator. A form of $1$ like $\frac{\sqrt{b}}{\sqrt{b}}$ only equals $1$ when you multiply both the top and the bottom. Changing just the denominator changes the value.
A sign slip in $a^2 - b$ . The conjugate product is $a^2 - b$ , not $a^2 + b$ . For example $(2 - \sqrt{3})(2 + \sqrt{3}) = 4 - 3 = 1$ , not $4 + 3 = 7$ .
Squaring a root incorrectly. A square root squared undoes the root, so $(\sqrt{5})^2 = 5$ , not $25$ .
Adding the squares instead of subtracting. In $(\sqrt{a} - \sqrt{b})(\sqrt{a} + \sqrt{b})$ the middle terms cancel and the result is $a - b$ , so $\frac{1}{\sqrt{6} - \sqrt{5}}$ has denominator $6 - 5 = 1$ , not $6 + 5$ .
Not simplifying at the end. After rationalizing, cancel common factors and simplify the radical: $\frac{8}{\sqrt{2}} = \frac{8\sqrt{2}}{2} = 4\sqrt{2}$ , not left as $\frac{8\sqrt{2}}{2}$ .

Key takeaways

The convention is to leave no square root in a denominator. You rewrite the fraction by multiplying by a form of $1$ , which changes the form but not the value.
For a single root, $\frac{a}{\sqrt{b}} = \frac{a\sqrt{b}}{b}$ , because $\sqrt{b}\cdot\sqrt{b} = b$ .
For a two-term denominator, multiply by the conjugate: $(a + \sqrt{b})(a - \sqrt{b}) = a^2 - b$ , a difference of squares with no radical.
Two-root denominators work the same way: $(\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b}) = a - b$ .
The conjugate flips the sign between the two terms; multiply both the numerator and the denominator by it.
Simplify the result: reduce common factors and simplify any remaining radical.

Rationalizing Denominators

Learning goals

Multiplying by a disguised form of 1

Clearing a single square root

Clearing a binomial denominator with its conjugate

When both terms are square roots

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Free Response Questions (FRQ)