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Rationalizing a Denominator with a Binomial - Problem 1
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Rationalizing the denominator when we have a binomial or two terms in the denominator. Whenever we trying to get rid of the square root and we have two terms in the denominator what we have to do is multiply by the conjugate which is the same numbers but there are different sign in between.

So for this particular example we have 2 minus root 5 so in order to get rid of that we need to multiply by 2 plus root 5. Always have to multiply by 1 so we need to do the same thing in the numerator and the denominator and then just we have to follow the denominator.

So denominator becomes 4 that’s 2 times 2. We have 2 root 5 and -2 root 5 those disappear altogether and then we have root 5 times root 5 which is 5 and the negative sign. So 4 minus 5 we end up with negative 1 in the denominator and then we have to distribute the top out. So 3 times 2 is 6, 3 times root 5 is 3 root 5.

So we said that we have a -1 in the denominator I’m actually going to go ahead and get rid of this fraction all together so I’m going to put this -1 distribute it into the numerator leaving us with -6 minus 3 root 5.

Simplifying a sorry rationalizing the denominator of a term with two terms in sorry a binomial in the denominator.

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