Question 331669
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Divide the coefficient on the first degree term by 2, square the result, and add that result to both sides of the equation.


The result on the left will be a perfect square trinomial, factor it.


Take the square root of both sides -- remember both positive and negative square roots.


Move the constant on the left to the right.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 16x\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 16x\ +\ 64\ =\ 64]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ +\ 8)^2\ =\ 64]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ 8\ =\ \pm\sqrt{64}\ =\ \pm8]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ 8\ -\ 8\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ -8\ -\ 8\ =\ -\ 16]



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