Question

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

Answer 1

Given quadratic equation is $a{x}^2+bx-9=0$
$a^3+b^3+27ab=729\Rightarrow a^3+b^3-729+27ab=0\Rightarrow a^3+b^3+(-9{)}^3=3ab(-9)$
$\Rightarrow a+b-9=0$ or $a=b=-9$
As $a>b>0$, so $a+b-9=0$

Now, $f\left(x\right)=a{x}^2+bx-9$
Putting $x=1,$ we get
$f\left(1\right)=a+b-9=0$
So the quadratic equation has one root as $1$
Other root $=-\frac{9}{a}$
Therefore, $\alpha =-\frac{9}{a},\beta =1$
$\Rightarrow 4\beta -a\alpha =4+9=13$