Question

lf the roots of $x^3+3p{x}^2+3qx+r=0$ are in G.P, then

• A. $pr=q^3$
• B. $p^{2}r=q^3$
• C. $p^{3}r=q^3$
• D. $p{r}^3=q^3$

Answer 1

The correct option is C $p^{3}r=q^3$

As roots are in G.P the let $\alpha ,\beta ,\gamma$ are the roots of $x^3+3p{x}^2+3qx+r=0$
Such that $\alpha \gamma ={\beta }^2$
$S_3=\alpha \beta \gamma =-r\Rightarrow {\beta }^3=-r\Rightarrow \beta ={(-r)}^{\frac{1}{3}}$
Substituting this in equation, we get
${\left({(-r)}^{\frac{1}{3}}\right)}^3+3p{\left({(-r)}^{\frac{1}{3}}\right)}^2+3q{(-r)}^{\frac{1}{3}}+r=0\Rightarrow -r+3p{(-r)}^{\frac{2}{3}}+3q\left({(-r)}^{\frac{1}{3}}\right)+r=0\Rightarrow p^{3}r=q^3$