Question 1047252
<font face="Times New Roman" size="+2">


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ y\ =\ -1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ xy\ =\ -1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ -\frac{1}{x}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ 1\ -\ \frac{1}{x}\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ x\ -\ 1\ =\ 0]


Use quadratic formula


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ -\frac{1}{2}\ \pm\ \frac{\sqrt{5}}{2}]


Check:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(-\frac{1}{2}\ +\ \frac{\sqrt{5}}{2}\right)\left(-\frac{1}{2}\ -\ \frac{\sqrt{5}}{2}\right)\ =\ -1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(-\frac{1}{2}\ +\ \frac{\sqrt{5}}{2}\right)\ +\ \left(-\frac{1}{2}\ -\ \frac{\sqrt{5}}{2}\right)\ =\ -1]


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 \ \ \ \ \ \ \ \ \ \  

</font>