Question

Find the quadratic formula by completing the square.

Answer 1

ax^2 + bx + c = 0
ax^2 + bx = -c
x^2 + (b/a)x = -c/a
x^2 + (b/a)x + (b/(2a))^2 = (b/(2a))^2 - c/a
(x - b/(2a))^2 = b^2/(4a^2) - 4ac/(4a^2) = (b^2 - 4ac)/(4a^2)
x - b/(2a) = +-sqrt(b^2 - 4ac)/((2a)
x = (-b +-sqrt(b^2 - 4ac) ) / (2a)

Answer 2

http://www.wikihow.com/Derive-the-Quadratic-Formula

Answer 3

We can also multiply instead of divide both sides by a:
a^2x^2 + abx + ac = 0
(ax)^2 + b(ax) = -ac
(ax)^2 + b(ax) +(b/2)^2 = (b/2)^2 - ac
( (ax) + b/2) )^2 = b^2/4 - 4ac/4 = (b^2 - 4ac)/4
ax + b/2 = +-sqrt(b^2 - 4ac) / 2
ax = (-b +-sqrt(b^2 - 4ac) )/2
x = (-b +-sqrt(b^2 - 4ac) ) / (2a)

Answer 4

Nice, I hadn't seen it that way before (multiplying everything by a).

Answer 5

There is yet another derivation not involving completing the square. Want to see it?

Answer 6

Let x = y + h. We are looking for the h that will change the original into a quadratic form missing the middle term. Substituting into the original polynomial,
a(y + h)^2 + b(y + h) + c
= (ay*2 + 2ahy + ah^2) + (by + bh) + c
= ay^2 + (2ah + b)y + (ah^2 + bh + c) = 0. Fot the middle term to be 0,
2ah + b = 0, so h = -b/(2a).
The new polynomial simplifies to ay^2 + a(b^2/(4a^2) ) + b(-b/(2a) + c
= ay^2 + b^2/(4a) - b^2/(2a) + c
= ay^2 - b^2/(4a) + 4ac/(4a)
= ay^2 +(-b^2+ 4ac)/(4a) = 0
ay^2 = (b^2 - 4ac)/(4a)
y^2 = (b^2 - 4ac)/(4a^2)
y = +-sqrt(b^2 - 4ac)/(2a). Getting back to x:
If x = y + h, then y = x - h, where h = -b/(2a).
Then y = x + b/(2a) = +-sqrt(b^2 - 4ac)/(2a)
x = (-b +-sqrt(b^2 - 4ac) ) / (2a)

Answer 7

Wow, that's a little convoluted. But great though, now I know three different ways of doing it! Thanks.