Question

Let the roots of the equation $24x^3-14x^2+kx+3=0$ 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

Answer: 0002

$24x^3-14x^2+kx+3=0$
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