Solve the given two equations and select the correct answer from the given options.
$I. 3x^2 - 4x - 32 = 0$
$II. 2y^2 - 17y + 36 = 0$

  • A
    if $x > y$
  • B
    if $x < y$
  • C
    if $x \ge y$
  • D
    if $x \le y$

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With respect to the roots of $x^{2}-x-2=0$,we can say that

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:

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$I.$ $x^{3} \times 13 = x^{2} \times 247$
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