The roots of the equation $|x^2-x-6|=x+2$ are

Consider the equation $(1+a+b)^2=3(1+a^2+b^2)$ where $a, b$ are real numbers. Then,

If the roots of the equation $x^2 + 2x + p = 0$ are real,then what is the value of $p$?

If $\alpha$ and $\beta$ $(\alpha < \beta)$ are the roots of the equation $(-2+\sqrt{3})(|\sqrt{x}-3|) + (x-6\sqrt{x}) + (9-2\sqrt{3}) = 0$,$x \ge 0$,then $\sqrt{\frac{\beta}{\alpha}} + \sqrt{\alpha\beta}$ is equal to:

The number of solution$(s)$ of the equation $\sqrt{x+1}-\sqrt{x-1}=\sqrt{4x-1}$ is/are

For $x > 2$,the equation $\sqrt{x+2} - \sqrt{x-2} = \sqrt{4x-2}$ has