Question 392768
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If


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ bx\ +\ c\ =\ 0]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ b^2\ -\ 4c\ >\ 0]


(which must be true since you were given two different real number roots for the original equation)


Then there are two numbers *[tex \Large \alpha] and *[tex \Large \beta] such that


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ -\ \alpha)(x\ -\ \beta)\ =\ 0]


You are given *[tex \Large \alpha\ =\ -4] and *[tex \Large \beta\ =\ 6] so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(x\ -\ (-4)\right)\left(x\ -\ 6\right)\ =\ 0]


Just multiply (using FOIL):


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(x\ +\ 4\right)\left(x\ -\ 6\right)]


and the resulting coefficients on the 1st degree and constant terms will be your values for *[tex \Large b] and *[tex \Large c].


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