Precalculus, Completeing the square?

Question

Solve x^2-12x+7=0 by completing the square

Solve x^2-5x-11/4=0 by completeing the square

Answer 1

Start by putting the integer coefficient on the other side:

x^2 - 12x = -7

Now take the coefficient on the x-term (-12), divide it by 2 (-6) and square it (36). Add this to both sides:
x^2 - 12x + 36 = 29

Now you can turn the left side into a perfect square:
(x - 6)^2 = 29

Take the square root of both sides:
x - 6 = +/-sqrt(29)

Then solve for x by adding 6 to both sides:
x = 6 +/- sqrt(29)

For the second problem, it is the same method:
x^2 - 5x = 11/4
x^2 -5x + 25/4 = 11/4 + 25/4
(x - 5/2)^2 = -14/4
x - 5/2 = +/- sqrt(-7/2)
x = 5/2 +/- sqrt(7/2) i

The only complications are if the coefficient of x^2 is not 1. If it is something else, then just divide both sides by that at the second step to make it x^2. Also, don't forget to look out for sqrt of negative numbers giving complex answers like #2. I almost missed the sqrt(-7/2) = sqrt(7/2) i. And you might want to simplify that so there is no sqrt in the denominator.

Answer 2

(x-6)^2-29=0 x=+or-(29^1/2)+6=11.385 and 0.615

(x-2.5)^2-9=0 x=5.5 and -0.5