## Archive for October 19th, 2008

### Using calculus to generate the quadratic formula

October 19, 2008

I get (unreasonable considerable) joy in finding fancy new proofs of elementary results, proofs that might come under the heading of mathematics made difficult or obscure. (I’ve never seen Linderholm’s book, but suspect it would appeal to a twisted part of my psyche.) I don’t quite understand the psychology behind this liking, but there it is nonetheless.

One of my favorite examples follows: solving quadratic equations using integration to complete the square.

Suppose $f(x)=ax^2+bx+c$, and we want to find the solutions to $f(x)=0$.

Note that $f'(x) = 2ax+b$ and $f(0) = c$, and thus it must be that $f(x) = c+ \int_0^x 2at+b \; dt$.

We compute this antiderivative using the change of variables $w=2at+b, \quad dw = 2a\;dt$, which leads to $c + \int_b^{2x+b} \frac1{2a} \; w \; dw$. This last expression is equal to $c + \left( \frac1{4a} (2ax+b)^2 - \frac{b^2}{4a} \right)$.

Thus the roots of $f(x) = 0$ are found by solving the equation $c + \frac{ (2ax+b)^2 - b^2 }{4a} = 0$.

This leads to $(2ax+b)^2 -b^2 = -4ac$, and thus $2ax+b = \pm \sqrt{b^2 - 4ac}$, and so finally we arrive at the roots $x = \frac{ -b \pm \sqrt{b^2 - 4ac}}{2a}$.

Forthcoming: Using Fubini’s Theorem to prove Integration by Parts