Question 1122815
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In order for a quadratic to have one zero with a multiplicity of two, the discriminant, i.e. *[tex \Large b^2\ -\ 4ac], must equal zero.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 1^2\ -\ 4(k)(100k)\ =\ 0]


Solve for the two possible values of *[tex \LARGE k]
								
								
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 \ \ \ \ \ \ \ \ \ \  
								
{{n}\choose{r}}

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