Let $\alpha$ and $\beta$ be the roots of $x^{2}-3x+p=0$ and $\gamma$ and $\delta$ be the roots of $x^{2}-6x+q=0$. If $\alpha, \beta, \gamma, \delta$ form a geometric progression,then the ratio $(2q+p):(2q-p)$ is:

If $\alpha_1, \beta_1, \gamma_1, \delta_1$ are the roots of the equation $a x^4+b x^3+c x^2+d x+e=0$ and $\alpha_2, \beta_2, \gamma_2, \delta_2$ are the roots of the equation $e x^4+d x^3+c x^2+b x+a=0$ such that $0 < \alpha_1 < \beta_1 < \gamma_1 < \delta_1$,$0 < \alpha_2 < \beta_2 < \gamma_2 < \delta_2$,$\alpha_1-\delta_2=2$,$\beta_1-\gamma_2=2$,$\gamma_1-\beta_2=4$,and $\delta_1-\alpha_2=4$,then $a+b+c+d+e=$

$p$ and $q$ are two roots of the equation $x^2+7x+3=0$. If $\frac{3p}{1-2p}$ and $\frac{3q}{1-2q}$ are the roots of $lx^2+mx+n=0$ and the greatest common divisor of $l, m, n$ is $1$,then $l-m+n=$

If the expression $7+6x-3x^2$ attains its extreme value $\beta$ at $x=\alpha$,then the sum of the squares of the roots of the equation $x^2+\alpha x-\beta=0$ is

The sum of the squares of all the roots of the equation $x^2+|2x-3|-4=0$ is:

What is the minimum value of $\frac{1 - x + x^2}{1 + x + x^2}$?