Question

Let the roots of the equation $24 x ^3-14 x ^2+ kx +3=0$ form a geometric sequence of real numbers. If absolute value of $k$ lies between the roots of the equation $x ^2+\alpha^2 x -112=0$, then find the

Answer 1

$24 x ^3-14 x ^2+ kx +3=0$
Let roots be $\frac{ a }{ r }$, a, ar
So, product of roots $\Rightarrow a^3=\frac{-1}{8} \Rightarrow a=\frac{-1}{2}$
Put $a=\frac{-1}{2}$ is root of equation (1), we get $k=-7$
Now, 7 lies between the roots of equation $x ^2+\alpha^2 x -112=0$
$\Rightarrow 49+7 \alpha^2-112<0 \Rightarrow 7 \alpha^2-63<0 \Rightarrow \alpha^2-9<0 $
$\therefore \alpha \in(-3,3)$
The largest integral value of $\alpha$ is 2