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If the roots of the equation $x^3+bx^2+cx-1=0$ form an increasing G.P., then $b$ lies in which interval?

$\begin{array}{1 1}(A) \; (-3,\infty)\\(B)\;(-\infty,-3)\\(C)\;(-1,\infty)\\(D)\; (-\infty,-1)\end{array}$

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  • If $\alpha,\:\beta\:\gamma$ are the roots of $ax^3+bx^2+cx+d=0$, then $\alpha+\beta+\gamma=\large-\frac{b}{a},$ and $\alpha\beta\gamma=\large-\frac{d}{a}$
Since the roots of the equation $x^3+bx^2+cx-1=0$ are in G.P.,
let the roots be $\large\frac{a}{r},$ $a,\:ar$
$\Rightarrow\:\large\frac{a}{r}$$+a+ar=-b$ and............ (i)
$\large\frac{a}{r}$$\times a\times ar=1$
$\Rightarrow\:a^3=1\:\Rightarrow\:a=1$ and
$\large\frac{1}{r}$$+1+r=-b$
$\Rightarrow\:r^2+r(1+b)+1=0$
Since $r$ is real the discriminant of this eqn., is $>0$
$\Rightarrow \:(1+b)^2-4> 0$
$\Rightarrow\:1+b>2\:\:or\:\:1+b< -2$
$\Rightarrow\:b>1\:or\:b<-3$
But since G.P. is increasing G.P. $r>1$
$\Rightarrow\:b$ cannot be >1 (from (i))
$\therefore\:b\in (-\infty,-3)$
answered Jul 14, 2013 by rvidyagovindarajan_1
 

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