The sum of the squares of the roots of $|x-2|^2+|x-2|-2=0$ and the squares of the roots of $x^2-2|x-3|-5=0$ is:

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

Let $S = \left\{ \frac{a^2+b^2+c^2}{ab+bc+ca} : a, b, c \in \mathbb{R}, ab+bc+ca \neq 0 \right\}$ where $\mathbb{R}$ is the set of real numbers. Then,$S$ equals

If $\alpha$ is a root of the equation $x^2-x+1=0$,then $\left(\alpha+\frac{1}{\alpha}\right)^3+\left(\alpha^2+\frac{1}{\alpha^2}\right)^3+\left(\alpha^3+\frac{1}{\alpha^3}\right)^3+\left(\alpha^4+\frac{1}{\alpha^4}\right)^3+\ldots$ to $12$ terms $=$

Let $[r]$ denote the greatest integer not exceeding $r$. The roots of the equation $3 x^2 + 6 x + 5 + \alpha (x^2 + 2 x + 2) = 0$ are complex numbers whenever $\alpha > L$ or $\alpha < M$. If $(L - M)$ is minimum,then the greatest value of $[r]$ such that $L y^2 + M y + r < 0$ for all $y \in R$ is:

For $x \in R$, the least value of $\frac{x^2-6x+5}{x^2+2x+1}$ is