Question

Let ' $p$ ' be an integer for which both roots of the quadratic equation $x^2+2(p-3) x+9=0$ lies in $(-6,1)$. If $2, g_1, g_2, \ldots \ldots, g_{19}, g_{20}, p$ are in G.P., then find the value of $g_4 g_{17}$.

Answer 1

Let $f(x)=x^2+2(p-3) x+9$
Since roots lies in $(-6,1)$, so we should have following conditions.

(i)$D \geq 0 \Rightarrow 4( p -3)^2-36 \geq 0 $
$ \Rightarrow p ( p -6) \geq 0 $
$\Rightarrow p \leq 0 \text { or } p \geq 6$....(1)
(ii) $f (-6)>0 \Rightarrow p <\frac{27}{4}$...(2)
(iii) $f (1)>0 \Rightarrow p >-2$....(3)
(iv) $-6<\frac{\alpha+\beta}{2}<1 \Rightarrow 2< p <9$...(4)
$\therefore $ From(i), (ii), (iii) \&(iv),
We get $6 \leq p <\frac{27}{4}$
$\therefore $ integral value of ' $p$ ' $=6$
Since $2, g _1, g _2, g _3, \ldots \ldots \ldots, g _{17}, g _{18}, g _{19}, g _{20}, 6$ are in G.P.
$\therefore g _4 g _{17}=2 \times 6=12$