I am not asking the SDopers to derive the quadratic formula! But, I am curious how we were able to derived it. What method from Calculus (I assume) gave us this formula for factoring ALL quadratics? Any math wizards care to explain?
I am especially curious where the “-4ac” term comes from…under the radical. But, if it’s too messy to get into, I’ll just accept it.
The link gives the method to derive it. I remember learning to derive the quadratic formula 'way back in Algebra I in junior high, and we did it by “completing the square”. It’s straightforward algebra, exact and accurate, and no calculus is involved.
Wait a second, I’m missing something here (looking at the derivation found at the link given above). What happened to the linear “x” variable in the problem? When I complete the square, and expand it… I end up with (bx/2a) + (bx/2a) = (2b/2a)x = (b/a)x term. So, where did the linear “x” variable go?
Caution: I haven’t had my coffee yet, so feel free to point out the obvious (gently)… - Jinx
Oh wait…the trick is NOT to expand the left hand side of the equation so we can then take the square root of both sides!
A-ha! Still, it seems like we’re cheating a bit to make this work!
Cheating? Do you mean, it looks like they pulled a trick out of a hat, and it just happens to work? That’s the name of the game, actually. All mathematics is a search for that sort of thing–the right trick. Without really cheating, of course.
Of course, once you have the quadratic formula, you can verify it without any further “tricks”, just some crank-turning algebra. So if your teacher wants you to “prove” the quadratic formula, the way to start is to take a “guess” as to what it might be and check that “guess”, and by “pure luck” just happen to choose the right form as your first guess.
Desmostylus’s link is really just a proof of correctness of the formula. It doesn’t explain at all where it came from.
A proof of correctness for such formulas is easy: Plug in the answer and simplify. In this case take (x-first root)(x-second root), multiply thru and simplify. If you take the sequence of formulas and write them in backwards order, you have an alleged “derivation”. But it is completely unhelpful and shouldn’t be taught. This is one way to tell how bad a Math teachers is. “Presto here it is” vs. “here’s some knowledge about solving problems.”
Here’s how to truly derive a solution:
Note that a two term quadratic is easy to solve:
0 = dz[sup]2[/sup]+e
z = +|- sqrt(-e/d)
It is easy to solve since the parabola is symmetric about the y-axis. A three term quadratic (with non-zero b) isn’t symmetric in y. Let’s do the following:
“Push the parabola sideways” until it’s symmetric around y.
Solve the easy case.
Take back the amount we “pushed” to get our real answer.
How much to push? The “center” (minima or maxima) of a parabola is at -b/(2a). We could get this from Calculus, but that’s too advanced for this problem. Note that the curve crosses the y-axis at x=0, at height y=c. Where else is y=c? Solving
ax[sup]2[/sup]+bx+c=c
is easy since the c’s cancel out. One solution is x=0 of course, the other is x=-b/a. This means the amount to push (where the center is) is half -b/a.
So let’s sub: x = z-b/(2a) in the orig. equation, the z terms cancel, allowing us to easily solve for z as above:
z = +|- sqrt(b[sup]2[/sup]-4ac)/(2a)
Adding back the “push” amount of -b/(2a) gives the quadratic formula.
It is a lot better to learn methods of problem solving that mere facts.
I’m pretty sure the Quadratic Forumula was discovered using Algebra. My college Algebra teacher advised: “If you forget the formula during the exam, don’t panic! Just derive it.” That scared me so much, I memorized the thing!
I’m a little hazy but isn’t the quadratic equation just the natural re-statement of ax^2 + bx + c = 0 where x is isolated? In other words, it’s the same damn equation, just re-jiggered around?
Actually, a version of the formula was derived by the Babylonians who knew how to find two numbers, given their sum and product. Note that you can divide by a to force a = 1. Now if u and v are the two solutions to z^2 + bx + c = 0, then u + v = b and uv = -c, so every quadratic can be reduced to that problem. This was hundreds of years before the Greek geometers. Later on Diophantus derived it too. The real problem was conceptual: dealing with negaitve numbers, not to imagine complex numbers. For that reason, the equations x^2 + bx + c = 0 was consider different from x^2 + bx = c, x^2 + c = bx, and x^2 = bx + c were all considered different and each had its own solution.
In the 15th century, I believe, 3rd and 4th degree analogs of these formulas were found by Italian algebraists. Thus began a search for a 5th degree formula, which ended only in the 1800-30 era with several proofs of impossibility.