Question

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

Answer 1

$a>b>0$

Simplify the given equation.

$a^3+b^3+27ab=729$

$\Rightarrow a^3+b^3+(-9{)}^3-3ab(-9)=0$

Using:- $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$

$\Rightarrow (a+b-9)(a^2+b^2-ab+9a+9b+81)=0$

therefore

$a+b-9=0$

$a+b=9$

Let, $f\left(x\right)=a{x}^2+bx+c$

$a{x}^2+bx-9=0$ $\to \alpha +\beta =\frac{-b}{a}$

$\Rightarrow \alpha \beta =\frac{-9}{a}(\alpha <\beta )$

$a>b>0$

$f\left(1\right)=a+b-9$

Thus it is clear that $1$ is the root of given quadratic equation.

either $\alpha =1$ or $\beta =1$

if $\beta =1$

$\alpha =\frac{-9}{a}$

We need

$4\beta -a\alpha =4\times 1-a(-\frac{9}{a})$

$=4+9=13$.