One root of the quadratic equation $x^{2}-5x+6=0$ is $3$. Find the other root.

If the roots of the equation $({p^2} + {q^2}){x^2} - 2q(p + r)x + ({q^2} + {r^2}) = 0$ are real and equal,then $p, q, r$ will be in

If the roots of the equation $x^2 + x + 1 = 0$ are $\alpha$ and $\beta$,and the roots of the equation $x^2 + px + q = 0$ are $\frac{\alpha}{\beta}$ and $\frac{\beta}{\alpha}$,then $p$ is equal to:

Solve the given two equations and select the correct answer from the given options.
$I.$ $x^{3} \times 13 = x^{2} \times 247$
$II.$ $y^{1 / 3} \times 14 = 294 \div y^{2 / 3}$

If the roots of the equation $x^2 + px + q = 0$ are $\alpha$ and $\beta$,and the roots of the equation $x^2 - xr + s = 0$ are $\alpha^4$ and $\beta^4$,then the roots of the equation $x^2 - 4qx + 2q^2 - r = 0$ will be:

If ${b_1}{b_2} = 2({c_1} + {c_2})$,then at least one of the equations ${x^2} + {b_1}x + {c_1} = 0$ and ${x^2} + {b_2}x + {c_2} = 0$ has