Question 251889
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You cannot simply multiply by the denominator because that would square the denominator.  If you square the denominator, you would still have radicals in the denominator.


Let's say that either *[tex \Large a] or *[tex \Large b] (or both) are the square root of an integer.  Then, in general


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(a\,+\,b\right)^2\ =\ a^2\ +\ 2ab\ +\ b^2]


would still be irrational because while the the two 2nd power terms would be rational, the 2ab term in the middle would still be irrational -- and squaring it again would just make things worse.


However, recall the difference of two squares factorization:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (a\,+\,b)(a\,-\,b)\ =\ a^2\,-\,b^2]


So, given a binomial expression in the denominator where at least one of the factors of the binomial is an irrational square root, you need to multiply the entire fraction by 1 in the form of the CONJUGATE of the denominator divided by itself.  Form the conjugate of a binomial by simply changing the sign in the middle, that is:  *[tex \Large a\,-\,b] is the conjugate of *[tex \Large a\,+\,b] (and vice versa, of course).



John
*[tex \LARGE e^{i\pi} + 1 = 0]
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