If $x$ is a solution of the equation, $\sqrt {2x + 1}  - \sqrt {2x - 1}  = 1, \left( {x \ge \frac{1}{2}} \right)$ , then $\sqrt {4{x^2} - 1} $ is equal to 

If $x$ is real and $k = \frac{{{x^2} - x + 1}}{{{x^2} + x + 1}},$ then

The least integral value $\alpha $ of $x$ such that $\frac{{x - 5}}{{{x^2} + 5x - 14}} > 0$ , satisfies

If $\alpha,\beta,\gamma, \delta$ are the roots of $x^4-100x^3+2x^2+4x+10 = 0$ then $\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}+\frac{1}{\delta}$ is equal to :-

In the equation ${x^3} + 3Hx + G = 0$, if $G$ and $H$ are real and ${G^2} + 4{H^3} > 0,$ then the roots are

Solution of the equation $\sqrt {x + 3 - 4\sqrt {x - 1} }  + \sqrt {x + 8 - 6\sqrt {x - 1} }  = 1$ is