Other Ways to Get the Quadratic Formula
Date: 02/19/2010 at 11:51:23 From: Davlon Subject: Quadratic Formula Is there any way to derive the quadratic formula that does not involve completing the square? As a teacher, I like to find many different ways to approach problems, so that my lessons can be more diversified for students. I find it hard to believe there is only one derivation.
Date: 02/21/2010 at 03:41:42 From: Doctor Jacques Subject: Re: Quadratic Formula Hi Davlon, Here are two methods I can think of. The first one is actually completing the square in disguise; the second one is interesting, because it has connections with some deep concepts, but it may use some ideas unfamiliar to the students. In either case, we want to solve the equation: f(x) = ax^2 + bx + c = 0  Method 1 : Substitution ----------------------- We notice that  is easily solved if b = 0 (and ac < 0). We try to transform the equation to that form by using a change of variable. Let us write: x = y + h  where y is the new unknown and h is to be determined to make the equation easier to solve. After substitution in , and re-arranging the terms, the equation (in y) becomes: a*y^2 + (b + 2ah)y + (ah^2 + bh + c) = 0  by choosing h = -b/2a, we can reduce  to the easy form, with the coefficient of y equal to 0. Working out the substitution gives: a*y^2 + (c - b^2/4a) = 0  This equation is easily solved for y, and, using  with h = -b/2a, gives the solutions in x. Method 2 : Symmetry ------------------- Let us write p and q for the unknown roots of the equation. We have the well-known relations: p + q = -b/a pq = c/a  Note that you don't need the quadratic formula to prove these relations. You can prove them by equating coefficients in the identity: f(x) = ax^2 + bx + c = a(x - p)(x - q)  and this last identity can be proved by polynomial division, using the fact that f(p) = f(q) by definition. Now,  gives us the sum p + q. If we could have the difference p - q, the resulting system would be easy to solve for p and q. The problem is that there is no simple (more exactly, rational) expression for p - q. To see why, assume that we have an expression: p - q = g(a,b,c)  where g(a,b,c) is a rational function of the coefficients. The right- hand side of  has a single, well-determined value. However, the choice of p and q for the roots is quite arbitrary -- we don't know which one is p and which one is q. If we interchange p and q, the sign of the expression would have to change, and this is not possible for a rational expression. To get rid of that sign change problem, we consider the expression (p - q)^2. This expression does not change if you swap p and q, and this suggests that we could maybe find a rational expression for (p - q)^2. Let us try it: (p - q)^2 = p^2 + q^2 - 2pq = (p + q)^2 - 4pq = b^2/a^2 - 4c/a = (b^2 - 4ac)/a^2  (We made use of the first relation ). This gives: p - q = (+/-) sqrt(b^2 - 4ac)/a  Notice that swapping p and q changes the sign of the square root; as this expression is defined up to the sign, this is OK. It is now an easy matter to solve the linear system: p + q = -b/a p - q = sqrt(b^2 - 4ac)/a  for p and q (we can choose any sign for the RHS of the second equation, as choosing the other sign will merely interchange p and q). This idea is, in fact, a particular case of a method devised by Lagrange to solve the cubic and quartic equations; this kind of idea is at the heart of Galois theory (an undergraduate-level topic). I believe this proof is accessible to high school students. There are two possible difficulties: * The proof of the relations . * If you have not yet taught complex numbers, an astute student may object that  makes no sense if the original equation has no roots. Please feel free to write back if you require further assistance. - Doctor Jacques, The Math Forum http://mathforum.org/dr.math/
Date: 02/21/2010 at 11:54:39 From: Davlon Subject: Thank you (Quadratic Formula) Thank you very much for the added insight. The second derivation is very interesting. I do try to get my students to think more. Thanks again. Davlon
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