Question

If the roots of the equation $x^4+a{x}^3+b{x}^2+cx+d=0$ are in geometric progression, then

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

Answer 1

The correct option is D $c^2=a^{2}d$

Let the roots be $\frac{\alpha }{r^3},\frac{\alpha }{r},\alpha r,\alpha r^3$
Sum of the roots $=-a$
$\Rightarrow \frac{\alpha }{r^3}+\frac{\alpha }{r}+\alpha r+\alpha r^3=-a$
$\Rightarrow \alpha (r+\frac{1}{r}+r^3+\frac{1}{r^3})=-a\dots \left(1\right)$

Product of the roots $=d$
$\Rightarrow {\alpha }^4=d\dots \left(2\right)$

Sum of products of the roots, three at a time $=-c$
$\Rightarrow \frac{{\alpha }^3}{r^3}+{\alpha }^{3}r+{\alpha }^3{r}^3+\frac{{\alpha }^3}{r}=-c$
$\Rightarrow {\alpha }^3(r+\frac{1}{r}+r^3+\frac{1}{r^3})=-c\dots \left(3\right)$

From equation $\left(1\right)$ and $\left(3\right),$
$\frac{1}{{\alpha }^2}=\frac{a}{c}$
$\Rightarrow {\alpha }^4=\frac{c^2}{a^2}$
From equation $\left(2\right),$
$d=\frac{c^2}{a^2}$
$\Rightarrow c^2=a^{2}d$