The equation $\sqrt {3 {x^2} + x + 5} = x - 3$ , where $x$ is real, has

The set of values of $x$ which satisfy $5x + 2 < 3x + 8$ and $\frac{{x + 2}}{{x - 1}} < 4,$ is

If $|x - 2| + |x - 3| = 7$, then $x =$

Let $S=\left\{\sin ^2 2 \theta:\left(\sin ^4 \theta+\cos ^4 \theta\right) x^2+(\sin 2 \theta) x+\right.$ $\left(\sin ^6 \theta+\cos ^6 \theta\right)=0$ has real roots $\}$. If $\alpha$ and $\beta$ be the smallest and largest elements of the set $S$, respectively, then $3\left((\alpha-2)^2+(\beta-1)^2\right)$ equals....................

The complete solution of the inequation ${x^2} - 4x < 12\,{\rm{ is}}$

If $\alpha, \beta $ and $\gamma$ are the roots of equation ${x^3} - 3{x^2} + x + 5 = 0$ then $y = \sum {\alpha ^2} + \alpha \beta \gamma $ satisfies the equation