How to use the conjugate method to rationalize the denominator
What does it mean to “rationalize the denominator”?
Remember that the phrase “rationalize the denominator” just means “get the square root(s) out of the denominator”.
We already know how to rationalize the denominator if the denominator is just a single square root, and nothing else.
How to rationalize the denominator using the conjugate method
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Rationalizing the denominator, vs. rationalizing the denominator with conjugate method
Example
Rationalize the denominator.
???\frac{6}{\sqrt{13}}???
If we multiply the denominator by ???\sqrt{13}???, we’ll get rid of the square root there. But in order to keep the value of the fraction the same, we have to multiply both the numerator and the denominator by ???\sqrt{13}???.
???\frac{6}{\sqrt{13}}\left(\frac{\sqrt{13}}{\sqrt{13}}\right)???
???\frac{6\sqrt{13}}{13}???
Now that the square root is out of the denominator, we’ve rationalized the denominator.
But how do we rationalize the denominator when it’s not just a single square root? Sometimes we’re going to have a denominator with more than one term, like
???\frac{3}{5-\sqrt{3}}???
In a case like this one, where the denominator is the sum or difference of two terms, one or both of which is a square root, we can use the conjugate method to rationalize the denominator.
The conjugate of a binomial is the same two terms, but with the opposite sign in between. For ???5-\sqrt{3}???, we keep the same two terms, ???5??? and ???\sqrt{3}???, but we change the sign in the middle. Since the sign in this case is negative, we’ll change it to a positive sign.
Original binomial: ???5-\sqrt{3}???
Its conjugate: ???5+\sqrt{3}???
To rationalize the denominator of a fraction where the denominator is a binomial, we’ll multiply both the numerator and denominator by the conjugate.
As we’re doing these problems, let’s also remember these facts:
Fact 1: You can multiply any number by one without changing its value.
???\frac{1}{\sqrt{2} - 5} \cdot 1??? is the same as ???\frac{1}{\sqrt{2}-5}???
Fact 2: You can write ???1??? as any number divided by itself, for example
???\frac{\sqrt{2}+5}{\sqrt{2}+5} = 1???
Example
Simplify the expression.
???\frac{3 - \sqrt{2}}{\sqrt{2}-5}???
We want to use the conjugate method to get the radical out of the denominator. Remember that the conjugate of two terms is just the same two terms with the opposite sign in between the terms. So the conjugate of ???\sqrt{2}-5??? is ???\sqrt{2}+5???. This is the value we need to multiply by both the numerator and denominator.
???\frac{3-\sqrt{2}}{\sqrt{2}-5}???
???\frac{3-\sqrt{2}}{\sqrt{2}-5} \cdot \frac{\sqrt{2}+5}{\sqrt{2}+5}???
Now this becomes a binomial multiplication problem. We need to make sure to multiply our first terms, outer terms, inner terms, and last terms (FOIL).
???\frac{\left(3-\sqrt{2}\right)\left(\sqrt{2}+5\right)}{\left(\sqrt{2}-5\right)\left(\sqrt{2}+5\right)}???
???\frac{3\sqrt{2} + 15 - 2 - 5\sqrt{2}}{2 + 5\sqrt{2} - 5\sqrt{2} - 25}???
Combine like terms.
???\frac{3\sqrt{2} -5\sqrt{2}+ 15 - 2}{2 - 25+ 5\sqrt{2} - 5\sqrt{2}}???
???\frac{-2\sqrt{2} + 13}{-23}???
Multiply by ???(-1)/(-1)??? to remove the negative sign from the denominator.
???\frac{-2\sqrt{2} +13}{-23} \cdot \frac{-1}{-1}???
???\frac{2\sqrt{2} - 13}{23}???
Let’s do another example.
Example
Simplify the expression.
???\frac{\sqrt{5}-\sqrt{7}}{\sqrt{5}+\sqrt{7}}???
We want to use conjugate method to get the radical out of the denominator. Remember that the conjugate of two terms is just the same two terms with the opposite sign in between the terms. So the conjugate of ???\sqrt{5}+\sqrt{7}??? is ???\sqrt{5} - \sqrt{7}???. This is the value we need to multiply by both the numerator and denominator.
???\frac{\sqrt{5} - \sqrt{7}}{\sqrt{5}+\sqrt{7}}???
???\frac{\sqrt{5} - \sqrt{7}}{\sqrt{5}+\sqrt{7}} \cdot \frac{\sqrt{5} - \sqrt{7}}{\sqrt{5} - \sqrt{7}}???
Now this becomes a binomial multiplication problem. We need to make sure to multiply our first terms, outer terms, inner terms, and last terms.
???\frac{\left(\sqrt{5}-\sqrt{7}\right)\left(\sqrt{}5 - \sqrt{7}\right)}{\left(\sqrt{5}+\sqrt{7}\right)\left(\sqrt{5}-\sqrt{7}\right)}???
???\frac{5 - \sqrt{5}\sqrt{7} - \sqrt{5}\sqrt{7} + 7}{5 - \sqrt5\sqrt7+\sqrt5\sqrt7 - 7}???
???\frac{5 - 2\sqrt{5}\sqrt{7}+ 7}{5- 7}???
???\frac{12-2\sqrt{5}\sqrt{7}}{-2}???
Reduce by dividing by ???2???.
???\frac{6-\sqrt{5}\sqrt{7}}{-1}???
???-6+\sqrt{5}\sqrt{7}???
???\sqrt{5}\sqrt{7}-6???
???\sqrt{35}-6???
Let’s try one more with numbers.
Example
Rationalize the denominator.
???\frac{3}{5-\sqrt{3}}???
Since the denominator is a binomial in which one of the terms is a square root, we need to multiply the numerator and denominator by the conjugate of the binomial in order to rationalize the denominator.
???\frac{3}{5-\sqrt{3}}\left(\frac{5+\sqrt{3}}{5+\sqrt{3}}\right)???
???\frac{15+3\sqrt{3}}{25+5\sqrt{3}-5\sqrt{3}-3}???
Using the conjugate method, the two terms in the middle of the denominator will always cancel with each other, leaving only the whole numbers in the denominator.
???\frac{15+3\sqrt{3}}{25-3}???
???\frac{15+3\sqrt{3}}{22}???
Now that the square root is out of the denominator, we’ve rationalized the denominator.
We can even use the conjugate method with variables, and with two square roots in the denominator.
Example
Rationalize the denominator.
???\frac{a+\sqrt{3}}{-3\sqrt{a}+\sqrt{3}}???
First, we’ll flip the order of the denominator so that we lead with a positive term instead of a negative term, just to make things a little simpler.
???\frac{a+\sqrt{3}}{\sqrt{3}-3\sqrt{a}}???
Then we’ll multiply the numerator and denominator by the conjugate of the denominator.
???\frac{a+\sqrt{3}}{\sqrt{3}-3\sqrt{a}}\left(\frac{\sqrt{3}+3\sqrt{a}}{\sqrt{3}+3\sqrt{a}}\right)???
???\frac{a\sqrt{3}+3a\sqrt{a}+3+3\sqrt{3}\sqrt{a}}{3+3\sqrt{3}\sqrt{a}-3\sqrt{3}\sqrt{a}-9a}???
Using the conjugate method, the two terms in the middle of the denominator will always cancel with each other, leaving only whole numbers and the constants in the denominator.
???\frac{a\sqrt{3}+3a\sqrt{a}+3+3\sqrt{3}\sqrt{a}}{3-9a}???
???\frac{a\sqrt{3}+3a\sqrt{a}+3+3\sqrt{3a}}{3-9a}???