If $p$ and $q$ are the roots of the equation $x^{2}+px+q=0$, then:

Two candidates attempt to solve the equation $x^2 + px + q = 0$. One starts with a wrong value of $p$ and finds the roots to be $2$ and $6$,and the other starts with a wrong value of $q$ and finds the roots to be $2$ and $-9$. The roots of the original equation are

Let two numbers have an arithmetic mean of $9$ and a geometric mean of $4$. Then these numbers are the roots of the quadratic equation:

If $\alpha, \beta, \gamma$ are the roots of the equation $x^3 + 2x - 5 = 0$ and the equation $x^3 + bx^2 + cx + d = 0$ has roots $2\alpha + 1, 2\beta + 1, 2\gamma + 1$,then the value of $|b + c + d|$ is (where $b, c, d$ are constants):

If $(\alpha+\sqrt{\beta})$ and $(\alpha-\sqrt{\beta})$ are the roots of the equation $x^{2}+px+q=0$, where $\alpha, \beta, p$ and $q$ are real, then the roots of the equation $(p^{2}-4q)(p^{2}x^{2}+4px)-16q=0$ are

If $\alpha$ and $\beta$ are the roots of $x^2-2x+4=0$,then the value of $\alpha^6+\beta^6$ is