# If $a>b>0$ and $a^3+b^3+27ab=729$ then $ax^2+bx-9=0$ has roots $\alpha,\beta,(\alpha<\beta)$. Find the value of $4\beta -a\alpha$.

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

By looking at the equation I figured out $a+b=9$. Hence one root of of the equation is 1. But I don't know how to proceed further. It would be great if I could get some help with this question.

Hint ( I haven't completely verified) $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$ If this equal zero then $$(a+b+c)=0$$ or $$(a^2+b^2+c^2-ab-bc-ca)=0$$
• Okay how do I use $a^2+b^2+81-ab+9b+9a=0$ to find the other root? – oshhh Mar 19 '17 at 7:52
• In my opinion there is no solution for a^2+b^2+81-ab+9b+9a=0, rather if you solve quadratic equation using a+b=9, we get roots, 1 and -9/a, and we know that a > b >0, so we can say that a is a positive number greater than 0, And also we know $\alpha,\beta,(\alpha<\beta)$, thus, $\alpha = -9/a$ and $\beta = 1$, so just solve $4\beta -a\alpha$ – pritywiz Mar 19 '17 at 8:39