Find $\alpha^4+\beta^4$ if $\alpha, \beta$ are the roots of the equation $x^2+x+1=0$.

If $\alpha, \beta$ are the roots of $a x^2+b x+c=0$,then the quadratic equation whose roots are $\sqrt{5} \alpha, \sqrt{5} \beta$ is

If $\alpha, \beta, \gamma, \delta$ are the roots of $x^4 - 100x^3 + 2x^2 + 4x + 10 = 0$,then $\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} + \frac{1}{\delta}$ is equal to:

The cubic equation whose roots are thrice to each of the roots of $x^3+2x^2-4x+1=0$ is

If the roots of the equations $ax^2 + bx + c = 0$ and $px^2 + qx + r = 0$ are $\alpha_1, \alpha_2$ and $\beta_1, \beta_2$ respectively,and the system of linear equations $\alpha_1y + \alpha_2z = 0$ and $\beta_1y + \beta_2z = 0$ has a non-zero solution,then which of the following is true?

If $p$ and $q$ are the roots of the equation $x^2 + pq = (p + 1)x$,then $q=$