For the equation $|x^2| + |x| - 6 = 0$,the roots are

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:

If $f(x)$ is a quadratic expression such that $f(1) + f(2) = 0$,and $-1$ is a root of $f(x) = 0$,then the other root of $f(x) = 0$ is

If $\alpha$ and $\beta$ are the roots of $x^2+3(a+3)x-9a=0$ such that the roots are equal for different values of $a$ (where $\alpha > \beta$ is not applicable as roots are equal,but let $\alpha$ be the root for $a=-9$ and $\beta$ be the root for $a=-1$),then the minimum value of the expression $x^2+\alpha x-\beta$ is:

If the roots of the equation $ax^2 + x + b = 0$ are real,then the roots of the equation $x^2 - 4\sqrt{ab}x + 1 = 0$ will be

Find the product of all real roots of the equation $|x|^{6/5} - 26|x|^{3/5} - 27 = 0$.