Question

If roots of cubic equation are in G.P. , $a{x}^3+b{x}^2+cx+d$ then :

• A. $c^{3}a=b^{3}d$
• B. $a^{2}c=b^{2}d$
• C. $a{c}^2=b{d}^2$
• D. NOTA

Answer 1

The correct option is A $c^{3}a=b^{3}d$

$\begin{array}{c}Let\\ If\alpha ,\beta ,and\gamma arethethreerootsthenwemusthave\\ a{x}^3+b{x}^2+cx+d=a\left(x-\alpha \right)\left(x-\beta \right)\left(x-\gamma \right)\\ Comparingcoefficientsofvariouspowerofxonbothsidesleadsto\\ \alpha +\beta +\gamma =-\frac{b}{a}----\left(1\right)\\ \alpha \beta +\beta \gamma +\gamma \alpha =\frac{c}{a}----\left(2\right)\\ \alpha \beta \gamma =-\frac{d}{a}---\left(3\right)\\ Onsubstituting\beta =\alpha rand\gamma =\alpha r^{2}weget\\ \alpha \left(1+r+r^2\right)=-\frac{b}{a}---\left(4\right)\\ {\alpha }^2\left(r+r^2+r^3\right)=\frac{c}{a}--\left(5\right)\\ {\alpha }^3{r}^3=-\frac{d}{a}-----\left(6\right)\\ Now,dividing\left(5\right)by\left(4\right)to\alpha r=-\frac{c}{b}andsubstitutingthisin\left(6\right)weget\\ {\left(-\frac{c}{b}\right)}^3=-\frac{d}{a}\\ \Rightarrow -\frac{c^3}{b^3}=-\frac{d}{a}\\ \therefore c^{3}a=b^{3}d.\end{array}$

Hence, the option $A$ is the correct answer.