Question

If the roots of the equation, $x^3+p{x}^2+qx-1=0$ form an increasing G.P where p and q are real, then

• A. $p\in (-\infty ,-3)$
• B. $p\in (-3,\infty )$
• C. $p\in (-\infty ,3)$
• D. $p\in (3,\infty )$

Answer 1

The correct option is A $p\in (-\infty ,-3)$

Since the roots of the equation $x^3+p{x}^2+qx-1=0$ are in G.P.

Let the roots be $\frac{a}{r},a,ar$

$\Rightarrow \frac{a}{r}+a+ar=-p$ and ......$\left(1\right)$

and $\frac{a}{r}\times a\times ar=1$

$\Rightarrow a^3=1\Rightarrow a=1$ and

$\frac{1}{r}+1+r=-p$

$\Rightarrow r^2+r(1+p)+1=0$

Since $r$ is real the discriminant of this equation is $>0$

$\Rightarrow {(1+p)}^2-4>0$

$\Rightarrow 1+p>2$ or $1+p<-2$

$\Rightarrow p>1$ or $p<-3$

But since G.P is increasing G.P, $r>1$

$\Rightarrow p$ cannot be $>1$ from $\left(1\right)$

$\therefore p\in (-\infty ,-3)$