Question 190280
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{4+sqrt{2}}{4-sqrt{2}}]


You can't just "eliminate" the 4 because 4 is <i><b>not</b></i> a factor of either the numerator or the denominator and you can only eliminate common <i><b>factors</b></i>.  The only thing you can do to this is rationalize the denominator.  That means you get rid of the radical in the denominator making the denominator a rational number.


Just squaring the denominator won't do you any good because you would still end up with a term that has a radical in it. 


What you need to do is multiply the fraction by 1 in the form of the conjugate of the denominator divided by itself.


Remember the difference of two squares factorization?


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


Well,  *[tex \LARGE  a + b] and *[tex \LARGE  a - b] are conjugates.  That means that the conjugate of your denominator is *[tex \LARGE  4 + sqrt{2}]


So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\frac{4+sqrt{2}}{4-sqrt{2}}\right) \left(\frac{4+sqrt{2}}{4+sqrt{2}}\right)]


Multiplying numerator times numerator and denominator times denominator using FOIL:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{16 + 8 sqrt{2} + 2}{16 - 2}\ =\ \frac{18 + 8 sqrt{2}}{14}\ =\ \frac{9 + 4 sqrt{2}}{7}]


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