Deriving the Quadratic FormulaDate: 10/15/98 at 21:03:19 From: Anne Smith Subject: How the quadratic formula was formed from ax^2+bx+c=0 I am wondering how the quadratic equation was derived from ax^2 + bx + c = 0. I have tried numerous ways to figure it out and I am very frustrated. I am working with it in a proof form and I first divide it all by "a" but then I remember that "a" is also on top. Please help me answer this problem. Date: 10/22/98 at 00:21:27 From: Doctor Santu Subject: Re: How the quadratic formula was formed from ax^2+bx+c=0 Anne, this is a classic problem solved by Italian mathematicians of the 12th century (if I remember right), and probably by even earlier mathematicians, using painfully difficult methods, but we're lucky, since we have algebra. First, some background. A. You know that if you multiply (A + B)(A - B) you get A^2 - B^2, right? So if we ever see something of the form A^2 - B^2, we know immediately that it can be factored into (A+B)(A-B). B. What about things like A^2 - 5? Of course 5 isn't obviously of the form (something)^2, but we can still "factor" this in the form [A + sqrt(5)][A - sqrt(5)], where Sqrt(5) is the square root of 5. In particular, if I see something like: (x+r)^2 - (p+q) I can factor it as [(x+r) + SQRT(p+q)][(x+r) - SQRT(p+q)]. C. We need to complete squares. In other words, by inspecting things like: (x+1)^2 = x^2 + 2x + 1 (x+2)^2 = x^2 + 4x + 4 (x+3)^2 = x^2 + 6x + 9 ... (x+12)^2 = x^2 + 24x + 144 and so on, we get to be able to answer the puzzle: "if X^2 + 14X are the first two terms of a perfect square, what should the third term be?" In each line, observe that the last term is the square of half of the middle coefficient. For instance, take 6, halve it (you get 3), and then square that, you get 9. So to get the proper last term for X^2 + 14X, take the middle term, which must be 14X, take the coefficient, 14, halve it, 7, and square it, 49. So the perfect square whose first two terms is X^2 + 14X is X^2 + 14X + 49. That's called completing the square. Now we're ready. Start with: ax^2 + bx + c = 0 Divide through with a. You get: b c x^2 + - x + --- = 0 a a Now complete the square: take half of (b/a) to get b/(2a). Now square that, and you get b^2/(4a^2). We add this, and then we subtract it (which is the same as adding 0 and so doesn't change the equation): 2 2 b b b c x^2 + - x + --- - --- + - = 0 a 2 2 a 4a 4a Now regroup as follows: (x + b/(2a))^2 - (b^2 - 4ac)/(4a^2) = 0 If that's hard to follow, look at the graphic file: which was created using Maple V, which will give you the algebra in almost-normal-looking form. Now you basically have a rather simple equation, something like: (x + something)^2 - (something else) = 0 so you can say: (x + something)^2 = (something else) Then the (x + something) must be either the square root of the something else on the right side, or the negative of the square root of the something else. And that's exactly what you get in the quadratic formula: x + b/2a = (+ or -)SQRT(b^2 - 4ac)/(2a) so: x = -b/(2a) +/- sqrt(b^2 - 4ac)/(2a) For another version of this derivation, see: Derivation of Quadratic Formula http://http://mathforum.org/library/drmath/view/52950.html - Doctor Santu, The Math Forum http://mathforum.org/dr.math/ |
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