The solution of the equation $2x^2 + 3x - 9 \le 0$ is given by

If the expression $\left( mx - 1 + \frac{1}{x} \right)$ is non-negative for all positive real numbers $x$,then what must be the minimum value of $m$?

If $x^2-5x-14 > 0$ implies $x$ lies outside $[\alpha, \beta]$,then find the value of $\frac{\alpha}{\beta}$.

If $\alpha_1, \alpha_2$ and $\alpha_3$ are the roots of $x^3+3x+2=0$,then $\alpha_1^5+\alpha_2^5+\alpha_3^5=$

Statement $(I)$: The set of solutions of $|x|^2 - 4|x| + 3 < 0$ is the interval $(-3, 3)$.
Statement $(II)$: If $x < 3$ or $x > 5$,then $x^2 - 8x + 15 > 0$.
Which of the above statements is (are) true?

The set of solutions satisfying both $x^2+5x+6 \geq 0$ and $x^2+3x-4 < 0$ is