The solutions of the quadratic equation ${(3|x| - 3)^2} = |x| + 7$ which belongs to the domain of definition of the function $y = \sqrt {x(x - 3)} $ are given by

Let $a, b, c, d$ be real numbers between $-5$ and $5$ such that  $|a|=\sqrt{4-\sqrt{5-a}},|b|=\sqrt{4+\sqrt{5-b}},|c|=\sqrt{4-\sqrt{5+c}}$ $|d|=\sqrt{4+\sqrt{5+d}}$ Then, the product $a b c d$ is

The product of all real roots of the equation ${x^2} - |x| - \,6 = 0$ is

If $\alpha $ and $\beta $ are the roots of the quadratic equation, $x^2 + x\, sin\,\theta  -2sin\,\theta  = 0$, $\theta  \in \left( {0,\frac{\pi }{2}} \right)$ then $\frac{{{\alpha ^{12}} + {\beta ^{12}}}}{{\left( {{\alpha ^{ - 12}} + {\beta ^{ - 12}}} \right){{\left( {\alpha  - \beta } \right)}^{24}}}}$ is equal to

If the expression $\left( {mx - 1 + \frac{1}{x}} \right)$ is always non-negative, then the minimum value of m must be

Let $p, q$ and $r$  be real numbers $(p \ne q,r \ne 0),$ such that the roots of the equation $\frac{1}{{x + p}} + \frac{1}{{x + q}} = \frac{1}{r}$ are equal in magnitude but opposite in sign, then the sum of squares of these roots is equal to .