Question 1161924
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Your equation, as written, does not have two solutions.  It is a linear equation with one solution, namely *[tex \Large x\ =\ \frac{-k}{\sqrt{2}\,-\,\sqrt{3}}].


That is because √(2x^2) means *[tex \Large \sqrt{2x^2}] which is equal to *[tex \Large (\sqrt{2})x] making your equation:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sqrt{2}x\ -\ \sqrt{3}x\ +\ k\ =\ 0]


So if you meant for your equation to actually be quadratic, then the first term has to be √2(x^2) which means *[tex \Large \sqrt{2}x^2]


<b><i>Repost your question and tell us what you really mean</i></b>
								
								
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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