Question 758793
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Your process is correct, and your numerator is correct, although it needs to be simplified, but your denominator is way out of whack.  You have the right idea, you just need to work on your arithmetic.


Here's a hint:  The product of a pair of conjugates is the difference of two squares, that is to say:


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


Your denominator should come out to be *[tex \LARGE 3^2\ -\ \left(\sqrt{7}\right)^2\ =\ 9\ -\ 7\ =\ 2]


Just use the distributive property to simplify the numerator.



John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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