Question 361804
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 4x\ =\ 7]


Divide the coefficient on the 1st degree term by 2, square the result, then add that result to both sides.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 4x\ +\ 4\ =\ 11]


Factor the perfect square trinomial in the LHS:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ +\ 2)^2\ =\ 11]


Take the square root of both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ 2\ =\ \pm\ \sqrt{11}]


Add the opposite of the constant term in the LHS to both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ -2\ \pm\ \sqrt{11}]


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