Question 326332
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Let's presume you are faced with the equation:


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


Where *[tex \Large \alpha], *[tex \Large \beta], and *[tex \Large \gamma] are constants.


Step one is to apply FOIL to the two binomials in the LHS.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ (\alpha\ +\ \beta)x\ +\ \alpha\beta\ =\ \gamma]


Then add *[tex \Large -\gamma] to both sides.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ (\alpha\ +\ \beta)x\ +\ \alpha\beta\ -\  \gamma\ =\ 0]


Then combine the constants so that *[tex \Large b\ =\ -(\alpha\ +\ \beta)] and *[tex \Large c\ =\ \alpha\beta\ -\ \gamma] to write:


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


Then, if you can find numbers *[tex \Large p] and *[tex \Large q] such that *[tex \Large p\ +\ q\ =\ b] and *[tex \Large pq\ =\ c], then you can factor your equation:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ +\ p)(x\ +\ q)\ =\ 0]


which has roots *[tex \Large -p] and *[tex \Large -q]


barring that you can use the quadratic formula to write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{-b\ \pm\ \sqrt{b^2\ -\ 4(1)c}}{2(1)}]


to find your two roots.



John
*[tex \LARGE e^{i\pi} + 1 = 0]
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