Question

If roots of the equation $x^3+3p{x}^2+3qx+r=0,p,q,r\ne 0$ are in H.P., then which of the following is correct?

• A. $pqr=2q^3+3r^2$
• B. $pqr=2q^3+r^2$
• C. $3pqr=q^3+2r^2$
• D. $3pqr=2q^3+r^2$

Answer 1

The correct option is D $3pqr=2q^3+r^2$

$x^3+3p{x}^2+3qx+r=0$ has roots in H.P.
Putting $x=\frac{1}{y}$
$r{y}^3+3q{y}^2+3py+1=0...\left(1\right)$
So, the roots of equation $\left(1\right)$ are in A.P.

Let the roots be $a-d,a,a+d$
Sum of roots
$a-d+a+a+d=-\frac{3q}{r}\Rightarrow 3a=-\frac{3q}{r}\Rightarrow a=-\frac{q}{r}$

As $a$ is one of the root of the equation $\left(1\right)$, putting $a=-\frac{q}{r}$ in the equation $\left(1\right)$, we get
$r{(-\frac{q}{r})}^3+3q{(-\frac{q}{r})}^2+3p(-\frac{q}{r})+1=0$
$\Rightarrow \frac{-q^3+3q^3-3pqr+r^2}{r^2}=0$
$\Rightarrow 2q^3-3pqr+r^2=0\therefore 3pqr=2q^3+r^2$