(Original post by theone)
I gave you some rep for just being bothered to type that much!.
I'll quickly go through the other method.
ax^2 + bx +c = a(x^2+bx/a+c/a).
Now we want to complete the square on x^2+bx/a+c/a.
Note that if x^2+bx/a+c/a = (x+m)^2 + n (where m and n are unknown). Since the only x term on the rhs is 2mx, we get b/a = 2m, and so m = b/2a (i.e. we jsut divide the coefficent of x by 2).
Now multiplying out, the constant terms on each side must be equal so n = m^2 + n = c/a. using m = b/2a we get n = c/a - b^2/4a^2.
And our completed square form is a((x+b/2a)^2+(c/a - b^2/4a^2)).
This is a lot harder to grasp with algebra than by just practice and an appreciation of the method.
Look away now if you don't want to see more
Using ax^2 + bx + c = a((x+b/2a)^2+(c/a - b^2/4a^2)).
If ax^2 + bx + c = 0, then a((x+b/2a)^2+(c/a - b^2/4a^2)) = 0
Now a is not 0 so we can say (x+b/2a)^2+(c/a - b^2/4a^2) = 0
So (x+b/2a)^2 = (b^2-4ac)/4a^2.
So x+b/2a = +/- (b^2-4ac)^(1/2)/2a and so x = (-b +/- (b^2-4ac)^(1/2))/2a
Phew - Just thought I'd add that on since I'd done most the hard work anyway....
Time to return the favour Bhaal