Question

The condition that the roots of the equation ${{x}^{3}}+3p{{x}^{2}}+3qx+r=0$ are in the Arithmetic Progression is:

• A. $2{{p}^{3}}-3pq+r=0$
• B. $2{{p}^{3}}+3pq+r=0$
• C. $2{{p}^{3}}-3pq-r=0$
• D. ${{p}^{3}}+3pq-r=0$

Answer 1

Answer: (A)

$2{{p}^{3}}-3pq+r=0$

Answer 2

Let $\alpha -\beta ,\,\alpha ,\,\alpha +\beta$
be the roots of the equation
${{x}^{3}}+3p{{x}^{2}}+3qx+r=0$ .
$\therefore$ Sum of the roots
$=-\frac{b}{a}$
$\Rightarrow$ $\alpha -\beta +\alpha +\alpha +\beta =\frac{-3p}{1}$
$\Rightarrow$ $3\alpha =-3p$
$\Rightarrow$ $\alpha =-p$
$\because$ $\alpha$ satisfies the given equation.
Put $x=-p,$
we get
$-{{p}^{3}}+3{{p}^{3}}-3pq+r=0$
$\Rightarrow$ $2{{p}^{3}}-3pq+r=0$