Question 184184
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2 + 8x + 13 = 0]


Move the constant term to the right side:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2 + 8x = -13]


Divide the coefficient on the <i>x</i> term by 2 and square the result.  Add this result to both sides.  8 divided by 2 is 4 squared is 16:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2 + 8x + 16 = 3]


You now have a perfect square on the left, factor it:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x+4)^2= 3]


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x+4= \pm sqrt{3}]


Solve for <i>x</i>:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = -4 \pm sqrt{3}]


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