Question 757768
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \  -2x^2\ +\ kx\ +\ 6\ =\ 0]


Use the quadratic formula:


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{-k\ \pm\ \sqrt{k^2\ -\ 4(-2)(6)}}{2(-2)}\ = \frac{-k\ \pm\sqrt{k^2\ +\ 48}}{-4}]


Which you should recognize as a conjugate pair the product of which is the difference of two squares:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{k^2}{16}\ -\ \frac{k^2\ +\ 48}{16}\ =\ \frac{-48}{16}\ =\ -3]


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