Question

If the roots of the equation $x^3+a{x}^2+bx+c=0$ are in A.P., then $2a^3-9ab$ is equal to

• A. 9c
• B. 18c
• C. 27c
• D. -27c

Answer 1

Answer: (D)

Given equation is
$x^3+a{x}^2+bx+c=0$
Let $(\alpha ,\beta ,\gamma )$ be the roots of the given equation then according to given condition, we have
$2\beta =\alpha +\gamma ...(i)$
$(\because \alpha ,\beta ,\gamma$ are in $AP)$
Now, $\alpha +\beta +\gamma =-a$
$\Rightarrow \beta +2\beta =-a$(Using Eq.(i)]
$\Rightarrow \beta =-\frac{a}{3}$
Since, $\beta$ is a root of given equation.
So, ${\beta }^3+a{\beta }^2+b\beta +c=0$
$\Rightarrow {\left(-\frac{a}{3}\right)}^3+a{\left(\frac{-a}{3}\right)}^2+b\left(\frac{-a}{3}\right)+c=0$
$\Rightarrow -\frac{a^3}{27}+\frac{a^3}{9}-\frac{ab}{3}+c=0$
$\Rightarrow -a^3+3a^3-9ab+27c=0$
$\Rightarrow 2a^3-9ab+27c=0$
$\therefore 2a^3-9ab=-27c$