The Inverse Quadratic

It is often useful to know the roots of a quadratic function. We will first derive an expression for the inverse of a quadratic polynomial $y(x)$ and then find the roots by setting $y=0$. The general form of a quadratic polynomial is

$$y = ax^2 + bx + c$$

We wish to solve for $x$. The most straightforward way to accomplish this is by completing the square. First we divide the equation by the coefficient of the quadratic term, $a$,

$$\frac{y}{a} = x^2 + \frac{b}{a}x + \frac{c}{a}$$

Now move all of the terms involving $x$ to the left side and move everything else to the right side of the equation.

$$x^2 + \frac{b}{a}x = \frac{y}{a} - \frac{c}{a}$$

The left hand side can be made into a perfect square by adding the square of half of the coefficient of the linear term

$$x^2 + \frac{b}{a}x + \left(\frac{b}{2a}\right)^2= \left(\frac{b}{2a}\right)^2 + \frac{y}{a} - \frac{c}{a}$$

then the left can be written as

$$\left(x + \frac{b}{2a}\right)^2 = \left(\frac{b}{2a}\right)^2 + \frac{y}{a} - \frac{c}{a}$$

Now re-write the right side by finding a common denominator,

$$\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 + 4ay - 4ac}{4a^2}$$

$$\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 + 4a(y - c)}{4a^2}$$

taking the square root of both sides,

$$x + \frac{b}{2a} = \frac{\pm\sqrt{b^2 + 4a(y - c)}}{2a}$$

Now isolate $x$ to get the inverse of a quadratic:

$$\boxed{x = \frac{-b\pm\sqrt{b^2 + 4a(y - c)}}{2a}}$$

The special case when $y=0$ yields the quadratic formula:

$$\boxed{x = \frac{-b\pm\sqrt{b^2 - 4ac}}{2a}}$$