Question 518603
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Ah ha!  Now that is different.  This is an important problem because it gives you the general solution to the quadratic equation.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ ay^2\ +\ by\ +\ c\ =\ 0]


Add the opposite of the constant term to both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ ay^2\ +\ by\ =\ -c]


Multiply both sides by the reciprocal of the lead coefficient:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y^2\ +\ \frac{b}{a}y\ =\ -\frac{c}{a}]


Divide the first degree term coefficient by 2, square the result, and then add it to both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y^2\ +\ \frac{b}{a}y +\ \frac{b^2}{4a^2}\ =\ \frac{b^2}{4a^2}\ -\ \frac{c}{a}]


Apply the common denominator in the RHS:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y^2\ +\ \frac{b}{a}y +\ \frac{b^2}{4a^2}\ =\ \frac{b^2\ -\ 4ac}{4a^2}]


Factor the perfect square in the LHS:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(y\ +\ \frac{b}{2a}\right)^2\ =\ \frac{b^2\ -\ 4ac}{4a^2}]


Take the square root of both sides:



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


Add the opposite of the LHS constant term:



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


Now you can solve all quadratic equations of the form:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ ay^2\ +\ by\ +\ c\ =\ 0]


just by doing some arithmetic with the coefficients.


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