Question 141840
x^2+9x+8=0

To complete the square, we need the digit in front of the x^2 term to be 1. All you have to do to complete the square is take half of the number in front of the x term, then square it and add/subtract it:

{{{x^2+9x+9^2/(2^2)-9^2/(2^2)+8=0}}}

{{{x^2+9x+81/4-81/4+32/4=0}}}

{{{x^2+9x+81/4-49/4=0}}}
{{{x^2+9x+81/4=49/4}}}
Now, the left hand side can be factored:

{{{(x+9/2)^2=49/4}}}
{{{x+9/2=0 +- sqrt(49/4)}}}
{{{x=-9/2 +- sqrt(49/4)}}}
{{{x=-9/2 +- 7/2}}}

Which gives x=-1 or x=-8.


It is important to note that the factoring of the square will always be half of the original term in front of the x. E-mail me at enabla@gmail.com if you still have difficulty or would like it explained differently.