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.


$$a^3 +b^3 +(-9)^3-3ab(-9)=(a+b-9)(a^2+b^2-ab+9a+9b+81)=0$$ 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 $$4Q-aP=4+9=13$$

  • 1
    $\begingroup$ Actually it's not because $a,b>0$, but because $a\neq b$, that $a+b-9=0$. There's no harm in mentioning $a^3+b^3+c^3-3abc=\frac12(a+b+c)((a-b)^2+(b-c)^2+(c-a)^2)$. $\endgroup$ – punctured dusk Jul 21 '16 at 8:21
  • $\begingroup$ Indeed $a\ne b\ne -3 $ $\endgroup$ – Behrouz Maleki Jul 21 '16 at 8:22

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