Question 663423
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You guys are working <i><b>way</b></i> too hard.


Try this:  Let *[tex \LARGE u\ =\ m^2\ +\ 5m]


Then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ u^2\ +\ u\ -\ 12\ =\ 0]


Factor


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (u\ +\ 4)(u\ -\ 3)\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ u\ +\ 4\ =\ 0\ \ \Rightarrow\ \ m^2\ +\ 5m\ +\ 4\ =\ 0]


Factor:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (m\ +\ 4)(m\ +\ 1)\ =\ 0]


So *[tex \LARGE m\ =\ -4] or *[tex \LARGE m\ =\ -1]


Then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ u\ -\ 3\ =\ 0\ \ \Rightarrow\ m^2\ +\ 5m\ -\ 3\ =\ 0]


Doesn't factor, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ m\ =\ \frac{-5\ \pm\ \sqrt{25\ -\ 4(1)(-3)}}{2}]


You can simplify to get the last two zeros.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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