Deriving the Quadratic Equation Solution: A Closer Look at Completing the Square

[ibc]

Hey
I'm wondering if anyone can give me a proof to the quadratic equation solution which is not the simple add here multiply that proof

(I just recall some cooler proof, which I think requires more "complicated" mathematics than adding and multiplying, though I can't remember what it was)

 

[cristo]

Hey
I'm wondering if anyone can give me a proof to the quadratic equation solution which is not the simple add here multiply that proof

What do you mean by an "add here multiply that" proof?

 

[CompuChip]

Do you want a proof or a derivation?
For a proof, you can just plug
$$x = \frac{-b \pm \sqrt{b^2 - 4 a c}}{2a}$$
into the equation
$$a x^2 + b x + c$$
and show that both give zero.

One derivation (the standard one, I think) is given here, for example. If there is any cooler one, I'd love to see it.

 

[ibc]

Do you want a proof or a derivation?
For a proof, you can just plug
$$x = \frac{-b \pm \sqrt{b^2 - 4 a c}}{2a}$$
into the equation
$$a x^2 + b x + c$$
and show that both give zero.

One derivation (the standard one, I think) is given here, for example. If there is any cooler one, I'd love to see it.

I mean a derivation.
by "add here multiply there" i mean the derivation given above.
and I'd like to see the cooler one too, or maybe I'm wrong and there isn't one, but I strongly recall there is.

 

[Hurkyl]

There are generally lots of ways to prove any mathematical fact. But I would guess you're trying to remember a derivation that completes the square.

 

[ibc]

There are generally lots of ways to prove any mathematical fact. But I would guess you're trying to remember a derivation that completes the square.

nope, that one is practically the same as the one mentioned here (or is it the same? don't know the name of each action =x )

what I'm thinking of is conceptually different one, though if none of you guys heard of it, maybe I'm only dreaming.

 

[Sandee]

ax²+bx+c=0

Taking 'a' common,
a{x²+(b/a)x+(c/a)}=0

Since a is not equal to 0, so
x²+(b/a)x+(c/a)=0

{x+(b/2a)}²+(c/a)-(b/2a)²=0

{x+(b/2a)}²+(c/a)-(b²/4a²)=0

{x+(b/2a)}²=(b²/4a²)-(c/a)

{x+(b/2a)}²=(b²-4ac)/4a²

{x+(b/2a)}=±√{(b²-4ac)/4a²}

x=(b/2a)±√{(b²-4ac)/4a²}

x={-b±√(b²-4ac)}/2a


e.g. x²+3x+4=0

(x+3/2)²+4-(3/2)²=0

It's simple.

 

[HallsofIvy]

ax²+bx+c=0

Taking 'a' common,
a{x²+(b/a)x+(c/a)}=0

Since a is not equal to 0, so
x²+(b/a)x+(c/a)=0

{x+(b/2a)}²+(c/a)-(b/2a)²=0

{x+(b/2a)}²+(c/a)-(b²/4a²)=0

{x+(b/2a)}²=(b²/4a²)-(c/a)

{x+(b/2a)}²=(b²-4ac)/4a²

{x+(b/2a)}=±√{(b²-4ac)/4a²}

x=(b/2a)±√{(b²-4ac)/4a²}

x={-b±√(b²-4ac)}/2a


e.g. x²+3x+4=0

(x+3/2)²+4-(3/2)²=0

It's simple.

Yes, it is. But I imagine that was what ibc was referring to as an "add here multiply there" proof. Apparently he wants one that is more difficult!

 

[symbolipoint]

The most easily understandable derivation for the solution of a quadratic equation is based on "completing the square". This can also be shown graphically. Part of the derivation relies on the fact that you can "complete" the square and then undo the process. The algebraic steps are fairly straight-forward. Many/most intermediate algebra books show a derivation, some with the graphical picture, some without.