Question 289720
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Put the first degree term on the left and the constant on the right:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ 8x\ =\ 5]


Divide the coefficient on the first degree term by 2, square the result, then add that result to both sides of the equation:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ 8x\ + 16\ =\ 5\ +\ 16]


Factor the perfect square on the left and collect terms on the right:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ -\ 4)^2\ =\ 21]


Take the square root of both sides remembering to consider both the positive and negative root.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ -\ 4\ =\ \pm \sqrt{21}]


Add the additive inverse of the constant on the left to both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ 4\ \pm\ \sqrt{21}]


Done.


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