Common root between a cubic and a quadratic The equation $x^2-ax+b=0$ & $x^3-px^2+qx=0$, have one common root & the second equation has two equal roots. Prove that $2(q+b) = ap$
($b\neq0 q\neq0$)
I am not able to think of any way to solve this any hint would be appreciated.

Counterexample
Let $a = 1, b = 0, p = 4, q = 4$. Then, $x^2 - ax + b = x(x-1)$ and $x^3 - 4x^2 + 4x = x(x-2)^2$ have one common root, $0$, and the second equation has two equal roots, $2$. However, 
$$
2(q+b) = 2(4+0) = 8\neq 4 = 1\cdot 4 = ap. 
$$
Perhaps you're missing some restrictions?
Edit
Given that $b,q\neq 0$, we can write the polynomials as
$$
x^2-ax+b = (x-r_1)(x-r_2),\qquad x^3-px^2+qx = x(x-r_1)^2. 
$$
Now, try writing $a, b, p, q$ in terms of $r_1$ and $r_2$ and see if you can prove the statement. 
