Question

If zeros of the polynomial $f\left(x\right)=x^3-3p{x}^2+qx-r$ are in A.P., then

• A. $2p^3=pq+r$
• B. $2p^3=pq-r$
• C. None of these
• D. $p^3=pq-r$

Answer 1

The correct option is B $2p^3=pq-r$

$f\left(x\right)=x^3-3p{x}^2-3p{x}^2+qx-r$
Here $a=1,b=-3p,c=q,d=-r$
Zeros are in AP
Let the zeros be $a-d,a+d$

Then sum of zeros =$\frac{-b}{a}$
$=-\left(\frac{-3p}{1}\right)=3p$
$\Rightarrow \alpha -d+\alpha +\alpha +d=3p$
$\Rightarrow 3\alpha =3p\Rightarrow \alpha =p$
$(\alpha -d)\alpha +\alpha (\alpha +d)+(\alpha +d)(\alpha -d)$
$=\frac{c}{a}=\frac{q}{1}=q$
$\Rightarrow {\alpha }^2-\alpha d+{\alpha }^2+\alpha d+{\alpha }^2-d^2=q$
$\Rightarrow 3{\alpha }^2-d^2=q$
$3p^2-d^2=q\Rightarrow d^2=3p^2-q$

and $(\alpha -d)\times \alpha \times (\alpha +d)=\frac{-c}{a}$
$=\frac{-(-r)}{1}=r$
$\Rightarrow \alpha ({\alpha }^2-d^2)=r$
$p(p^2-3p^2+q)=r\Rightarrow p^3-3p^3+pq=r$
$\Rightarrow -2p^3+pq=r\Rightarrow -2p^3-pq+r$
$\Rightarrow 2p^3=pq-r$