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. $p^2=pq-1$
• D. None of these

Answer 1

Answer: (A)

$2p^3=pq-r$

Let $\alpha ,\beta$ and $\psi$ be the zeros of the polynomial $f\left(x\right)=x^2-3p{x}^2+qx-r$

$f\left(x\right)=x^2-3p{x}^2+qx-r$

$=(x-\alpha )(x-\beta )(x-\psi )$

$=x^2-(\alpha +\beta +\psi )x^2+(\alpha \beta +\beta \psi +\psi \alpha )x-\alpha \beta \psi$

Equating the coefficients of $x^2$ we have

$-(\alpha +\beta +\psi )=-3p$

$\alpha +\beta +\psi =3p$ ……….$\left(1\right)$

Now, it is given that $\alpha ,\beta$ & $\psi$ are in A.P. Let S be the common difference of terms of the AP.

$\Rightarrow \beta -\alpha =\delta$

$\Rightarrow \alpha =\beta -\delta$ ………….$\left(2\right)$

$\psi -\beta =\delta$

$\Rightarrow \psi =\beta +\delta$ ……………..$\left(3\right)$

Put the values of $\alpha$ and $\psi$ from equation $\left(2\right)$ and $\left(3\right)$ in $\left(1\right)$

$\beta -\delta +\beta +\beta +\delta =3p$

$\Rightarrow 3\beta =3p$

$\Rightarrow \beta =p$

$\Rightarrow$ p is a root of the polynomial $f\left(x\right)$

put $x=p$ in $f\left(x\right)$

$\Rightarrow 0=f\left(p\right)=(p{)}^3-3p(p{)}^2+q\left(p\right)-r$

$p^3-3p^3+qp-r=0$

$-2p^2+qp-r=0$

$\Rightarrow 2p^3=qp-r$.