Question 442303
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ ax^2\ +\ bx\ +\ c\ =\ 0]


Step 1:  Move the constant term to the RHS


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ ax^2\ +\ bx\ =\ -c]


Step 2:  Divide by the lead coefficient


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


Step 3:  Divide the 1st degree coefficient by 2, square the result, and add that result to both sides of the equation.  *[tex \Large \frac{b}{a}] is the 1st degree coefficient.  Divide by 2: *[tex \Large \frac{b}{2a}].  Square:  *[tex \Large \frac{b^2}{4a^2}] 


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


Step 4:  Factor the perfect square trinomial in the LHS:


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


Step 5:  Apply the LCD in the RHS


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


Step 6:  Take the square root of both sides.  Remember to consider both positive and negative roots.


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


Step 7:  Add the opposite of the constant term in the RHS to both sides:


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


Of course, when you do it with a problem that has specific numbers rather than general representations of the coefficients, you have to do the arithmetic and simplify the answer as much as possible.


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