Question

If the roots of $x^3+a{x}^2+bx-27=0$ are in G.P. with common ratio r, where a, b $ϵ$ R and $a+b+6=0$, then

• A. $a=5$
• B. $r=-1$
• C. $b=-3$
• D. $r+b=-4$

Answer 1

The correct option is B $r=-1$

Let the roots of the equation be $\frac{A}{r},A,Ar$
Now, $\frac{A}{r}\times A\times Ar=A^3=27\Rightarrow A=\sqrt[3]{327}=3$
Hence, 3 is a root of the equation.
Hence,
$27+9a+3b-27=0$
$\Rightarrow 3a+b=0$
Also, $a+b+6=0$
On solving we get,
$a=3$ , $b=-9$
Now, the sum of the roots $=-a$
$\Rightarrow \frac{A}{r}+A+Ar=-a$
$\Rightarrow \frac{3}{r}+3+3r=-3$
$r+\frac{1}{r}=-2$
$\Rightarrow r=-1$
Hence, option B is only correct.