If $a>b>0$ and $a^3 +b^3 +27ab=729$ then the quadratic equation $ax^2+bx-9=0$ has roots $P,Q(P<Q)$. Find the value of $4Q-aP$?

Question

If $a>b>0$ and $a^3 +b^3 +27ab=729$ then the quadratic equation $ax^2+bx-9=0$ has roots $P,Q(P<Q)$. Find the value of $4Q-aP$?

How will I begin with the solution just a hint would be enough.

Answer 1

$$a^3 +b^3 +(-9)^3-3ab(-9)=(a+b-9)(a^2+b^2-ab+9a+9b+81)=0$$ therefore \begin{cases} a+b-9=0\\ \qquad\operatorname{or}\\ a=b=-9 \end{cases} since $a>b>0$ thus $$a+b-9=0$$

Set $f(x)=ax^2+bx-9$. We have $f(1)=a+b-9=0$, thus

$Q=1$ and $P=\frac{-9}{a}$ finally $$4Q-aP=4+9=13$$