Question 352786
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ 8x\ -\ 29\ \equiv\ (x\ +\ a)^2\ +\ b]


This is a "Complete the Square" problem.


Start by setting the LHS equal to zero.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ 8x\ -\ 29\ =\ 0]


Add the opposite of the constant term to both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ 8x\ =\ 29]


Divide the coefficient on the first degree term by 2, square the result, and then add that result to both sides.  -8 divided by 2 is -4.  -4 squared is 16.  Add 16 to both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ 8x\ +\ 16\ =\ 29\ +\ 16]


The quadradic trinomial in the LHS is now a perfect square.  Factor it:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ -\ 4)^2\ =\ 45]


Add the opposite of the constant term to both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ -\ 4)^2\ -\ 45\ =\ 0]


Since both the original LHS and this new LHS are equal to zero, they must be equivalent, hence:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ 8x\ -\ 29\ \equiv\ (x\ +\ (-4))^2\ +\ (-45)]


But the original LHS was equivalent to *[tex \Large (x\ +\ a)^2\ +\ b], hence:



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ +\ a)^2\ +\ b\ \equiv\ (x\ +\ (-4))^2\ +\ (-45)]


Therefore:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ =\ -4]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ b\ =\ -45]


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