Question 1061654
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ x\ =\ a^2\ +\ a]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ x\ -\ a^2\ -\ a]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{-b\ \pm\ \sqrt{b^2\ -\ 4ac}}{2a}]


Here, *[tex \Large a\ =\ 1], *[tex \Large b\ =\ 1] and *[tex \Large c\ =\ -a^2\ -\ a]


So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{-1\ \pm\ \sqrt{1^2\ -\ 4(1)(-a^2-a)}}{2(1)}]


Simplify to find the two values of *[tex \Large x] in terms of *[tex \Large a]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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