Let $(a-3)x^2+12x+(a+6)>0, \forall x \in R$ and $a \in (\ell, \infty)$. If $\alpha$ is the least positive integral value of $a$,then the roots of $(\alpha-3)x^2+12x+(\ell+2)=0$ are

If $\alpha, \beta, \gamma, \delta$ are the roots of the equation $x^4+x^2+1=0$,then $\frac{\alpha^3+\beta^3+\gamma^3+\delta^3}{\alpha^6+\beta^6+\gamma^6+\delta^6}=$

If $x = \frac{1}{2} \left( \sqrt{7} + \frac{1}{\sqrt{7}} \right)$,then the value of $\frac{\sqrt{x^2 - 1}}{x - \sqrt{x^2 - 1}}$ is equal to

Consider the equation $x^2 + \alpha x + \beta = 0$ having roots $\alpha, \beta$ such that $\alpha \neq \beta$. Also consider the inequality $||y - \beta| - \alpha| < \alpha$,then:

If $a$ and $b$ are arbitrary positive real numbers, then the least possible value of $\frac{6a}{5b} + \frac{10b}{3a}$ is

For $0 < c < b < a$,let $(a+b-2c)x^2 + (b+c-2a)x + (c+a-2b) = 0$ and $\alpha \neq 1$ be one of its roots. Then,among the two statements:
$(I)$ If $\alpha \in (-1, 0)$,then $b$ cannot be the geometric mean of $a$ and $c$.
$(II)$ If $\alpha \in (0, 1)$,then $b$ may be the geometric mean of $a$ and $c$.