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 $2,g_1,g_2,\dots \dots ,g_{19},g_{20},p$ are in G.P., then find the value of $g_4{g}_{17}$.

Answer 1

Answer: 12

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

(i)$D\ge 0\Rightarrow 4(p-3{)}^2-36\ge 0$
$\Rightarrow p(p-6)\ge 0$
$\Rightarrow p\le 0\text{ or }p\ge 6$....(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 $6\le p<\frac{27}{4}$
$\therefore$ integral value of ' $p$ ' $=6$
Since $2,g_1,g_2,g_3,\dots \dots \dots ,g_{17},g_{18},g_{19},g_{20},6$ are in G.P.
$\therefore g_4{g}_{17}=2\times 6=12$