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Rationalizing a Denominator with a Binomial  Problem 2
Carl Horowitz
Carl Horowitz
University of Michigan
Runs his own tutoring company
Carl taught upperlevel math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!
The concept behind rationalizing the denominator is to get rid of any square roots we have in the denominator of a fraction. The example we have behind me actually has two terms in the denominator and both if which are square roots so in order to rationalize the denominator we have to get rid of both of them.
In order to do that we have to multiply by the conjugate which is the same terms, but the opposite sign. What happens is that when we do that is when we foil it out the end terms get squared and our middle terms cancel out altogether.
So we have to multiply by 1 so we have to multiply by the same thing over the same thing and when we foil out the denominator what happens is we have root 5 times root 5 so that’s just going to be 5. We have root 5 times negative root 2 and root 5 times positive root 2 so what that happens is just cancel out together and we get root 10 minus root 10.
They disappear and whenever we have root 2 times – root 2 which is just going to be minus 2. So what we’ve done is by multiplying by the conjugate got rid of all these square roots from the denominator. We’d also need to distribute the same to the numerator when multiplying roots same power so this is a square root and a square root you can combine them so then we have 3 times 5 which is 15. The square root of 15, root 2 times root 3 which is root 6.
So simplifying the 5 minus 2 what we end up with is root 15 minus root 6 all over 3. Simplifying the denominator by rationalizing it, multiplying by the conjugate.
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Carl Horowitz
B.S. in Mathematics University of Michigan
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Sample Problems (8)
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Rationalizing a Denominator with a Binomial
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Problem 2 2,914 viewsRationalizing the denominator.
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