## Archive for the ‘Algebraic equations’ Category

### 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.