Given a cubic and quadratic share a root, prove $(ac-b^{2})(bd-c^{2})\geq 0$ Here is an interesting problem.  Perhaps someone would be so kind as to give me a shove in the right direction?.
If $ax^{3}+3bx^{2}+3cx+d$ and $ax^{2}+2bx+c$ share a common root, then prove that
$$(ac-b^{2})(bd-c^{2})\geq 0$$
I thought about equating coefficients somehow, but that got messy.
I used the quadratic formula on the given quadratic to find 
$\displaystyle x=\frac{-b\pm\sqrt{b^{2}-ac}}{a}$ are two roots.
So, if the cubic shares on of these, then I should be able to sub this in for x in the cubic.
Upon doing so, I got:
$$\frac{2(ac-b^{2})\sqrt{b^{2}-4ac}}{a^{2}}-\frac{3bc}{a}+\frac{2b^{3}}{a^{2}}+d$$
This is where I got hung up.  This may not even by a good way to go about it. 
I see a part of what is to be proven in the above expression, $ac-b^{2}$
Setting it to 0 does not help much. 
I also thought about dividing them. If they share a root, then it should reduce to a quadratic in the numerator and a linear denominator. But, then what?.
Can anyone give a hint as to the best way to proceed?.
 A: We begin by using a single step of the Euclidean algorithm.  Let $f(x) = ax^3 + 3bx^2 + 3cx + d$ and $g(x) = ax^2 + 2bx + c$.  If $f$ and $g$ share a common root, then the polynomial
$$
h(x) \;=\; f(x) - xg(x) \;=\; bx^2 + 2cx + d
$$
must share that root as well. Now, the roots of $g(x)$ are real when $ac-b^2 \leq 0$, and complex when $ac-b^2 > 0$.  Similarly, the roots of $h(x)$ are real when $bd-c^2 \leq 0$, and complex when $bd-c^2 > 0$.  If these two polynomials share a common root, it follows that $ac-b^2$ and $bd-c^2$ are either both positive or both nonpositive, and therefore $(ac-b^2)(bd-c^2)\geq 0$.
A: Hint: Notice anything coincidental about the derivative of the cubic?
Suppose that $\phi$ is the common root.  Let $f(x)=ax^3+3bx^2+3cx+d$ and let $g(x)=ax^2+2bx+c$.  Notice that $f^'(x)=3g(x)$ so that $f'(\phi)=0$ as well.  Hence $\phi$ is a double root of $f$. Since $f$ is a cubic, and complex roots come in pairs, it follows that $\phi$ is real, and hence all the roots of these polynomials are real.
In particular this implies something a bit stronger, that both $c^2-bd\geq 0$ and $b^2-ac\geq 0$.  (To get $b^2-ac\geq 0$, look at the discriminant.  I leave showing that $c^2-bd\geq 0$ to you.  (I have a solution if you really want.))
Hope that helps,
Edit:  Why do we have $c^2-bd\geq 0$?
Here is the immediate brute force way, there is probably a nicer solution.  The cubic has $\phi$ as a root with multiplicity $2$, and a third root, call it $\gamma$.  Then since $c=a \frac{\phi^2 +\phi\gamma+\phi\gamma}{3}$, $b=-a\frac{\phi+\phi+\gamma}{3}$ and $d=-a\phi^2\gamma$ it follows that $c^2 - bd \geq 0$ is equivalent to $$\phi^2 \gamma \left(\frac{\phi+\phi+\gamma}{3}\right)\leq\left(\frac{\phi^2+\phi \gamma+ \phi\gamma}{3}\right)^2.$$  Dividing by $\phi^2$ and multiplying by $9$, we get $$3\gamma (2\phi + \gamma) \leq (\phi+2\gamma)^2,$$ which is then equivalent to $$0\leq\phi^2 -2\phi\gamma+\gamma^2.$$ This last line clearly holds since it is a square. 
