Solving a Quadratic Equation

January 3, 2010

The standard way to derive “the quadratic formula” is by completing the square, as explained, for example, at cut-the-knot. As most standard things, it is a bit boring, maybe because everybody have seen it so many times and everyone knows it.

Once upon a time I thought about a more entertaining way to derive this well-known formula for the roots of x^2+bx+c=0. Here is what I had come up with.

To find the roots x_1 and x_2 of our equation is the same as to factor its left-hand side like x^2+bx+c=(x-x_1)(x-x_2). By expanding the right-hand side of this identity, we can see that x_1+x_2=-b and x_1 x_2 = c.

Now, it would be nice to have a formula for x_1-x_2 in terms of b and c. Then we could have found the roots. But the trouble is that the equation doesn’t know which one of its roots is x_1 and which one is x_2, b and c don’t change when we interchange x_1 and x_2, while x_1-x_2 changes sign. It looks like we are out of luck.

Still, getting x_1-x_2 up to a sign is good enough, and (x_1-x_2)^2 doesn’t change when we flip x_1 and x_2; finding a formula for that would be good enough. Let us work on this idea.

We already know that b^2=(x_1+x_2)^2=(x_1-x_2)^2+4x_1 x_2, but x_1 x_2=c, whence (x_1-x_2)^2=b^2-4c, and now we can figure out the roots. The quadratic formula will appear, of course.

I have written a mathematica script that shows how the roots of z^n+bz+1=0 depend on b, you may find it amusing.