Radicals in Fractions - Rationalizing Denominators w/ Complex Conjugates
You know that when you have a fraction with a square root in the denominator, you have to rationalize the denominator (essentially, multiply the fraction by a fraction, equal to one, that will cancel the radical in the original fraction). Like so:
$\dfrac{2}{\sqrt{3}} = \dfrac{2}{\sqrt{3}}\times\dfrac{\sqrt{3}}{\sqrt{3}}=\dfrac{2\sqrt{3}}{3}$
Sometimes the radical in the denominator of a fraction is part of a binomial. You still have to rationalize it, but you want to do it in a way that will not produce more radicals. That's where special products come in. Do you remember what happens when you multiply binomials that contain the same terms, but one that attaches them with addition and one with subtraction?
$\eqalign{(x+3)(x-3)\\x^2+3x-3x-9\\x^2-9}$
You use foil to multiply the binomials, but because the numbers are the same and one binomial uses addition and the other uses subtraction, the middle terms cancel each other out, leaving just the perfect squares (attached by multiplication).
You can use this rule to rationalize denominators with a radical in a binomial:
$\dfrac{2}{2+\sqrt{3}}\times\dfrac{(2-\sqrt{3})}{(2-\sqrt{3})}=\dfrac{2(2-\sqrt{3})}{(2+\sqrt{3})(2-\sqrt{3})}=\dfrac{4-2\sqrt{3}}{4-3}=\dfrac{4-2\sqrt{3}}{1}=4-2\sqrt{3}$
The denominator won't always disappear completely, but using special products will cancel out the radical and give you numbers you can compute in your denominator.
Practice Problems:
Radicals in Fractions - Rationalizing Denominators w/ Complex Conjugates
Rationalize the denominators in the following fractions:
- $\dfrac{3+\sqrt{17}}{2-3\sqrt{17}}$
- $\dfrac{11+\sqrt{7}}{\sqrt{2}+\sqrt{5}}$
- $\dfrac{4+\sqrt{10}}{\sqrt{2}+\sqrt{5}}$
- $\dfrac{6}{3+4i}$
- $\dfrac{7+5i}{2-2i}$
- $\dfrac{9-9i}{9+9i}$
- $\dfrac{3+\sqrt[3]{4}}{\sqrt[3]{5}}$
- $\dfrac{6-\sqrt{35}}{\sqrt{5}-\sqrt{7}}$
- $\dfrac{5-2i}{3+3i}$
- $\dfrac{2+2i}{3i}$
- $\dfrac{3+\sqrt{17}}{2-3\sqrt{17}}$