Finding the point where the tangent is the x-axis

Consider a simple example of a cubic: $px^3+qx^2+rx+s$ with $p,q,r,s\in\mathbb{R}$. I want to find the point $a$ in the following figure, where the cubic has a double root, as a function of the coefficients. Now I know that I can take the discriminant of the cubic and find its root to find $a$. However, I was thinking if I can proceed by considering that $y=0$ is a tangent to the curve and finding a condition on $a$ such that this would hold.

Denoting the cubic as $f$, the equation of the tangent is:

$$y=f(x_0)+f'(x_0)(x-x_0)$$

But I get stuck here because the RHS evaluates to zero (as it should) and I have nothing to work with to get a condition on $a$.

Are there alternate approaches to finding $a$ other than through the discriminant?

I ask this because the actual coefficients of my polynomial are all some functions of say $z$. Calculating the discriminant in Mathematica for the cubic gives an 8th order polynomial in $z$ and the roots of this are extremely messy. However, taking the derivative results in a smaller order polynomial and is more workable.

-
Ix $x=0$ known to be a root as suggested in the diagram? – Mark Bennet Jul 4 '11 at 7:32
no. it was just a schematic. – user7815 Jul 4 '11 at 7:39

You could factorise the polynomial as $p(x-a)^2(x-b)$ and use this to express coefficients in terms of the roots and solve. Or note that $f$ and $f'$ have a common root and use the division algorithm to locate it.