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(A) The given equation is $qx^2 + px + q = 0$. Since the roots are complex,the discriminant $D_1 < 0$.$D_1 = p^2 - 4(q)(q) = p^2 - 4q^2 < 0$,which imp

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Real and unequal

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(A) The given equation is $qx^2 + px + q = 0$. Since the roots are complex,the discriminant $D_1 $D_1 = p^2 - 4(q)(q) = p^2 - 4q^2 Now,consider the second equation $x^2 - 4qx + p^2 = 0$. The discriminant $D_2$ of this equation is given by:$D_2 = (-4q)^2 - 4(1)(p^2) = 16q^2 - 4p^2$.We can factor this as $D_2 = 4(4q^2 - p^2)$.Since $p^2  0$.Therefore,$D_2 > 0$.Since the discriminant of the second equation is positive,its roots are real and unequal.

If the roots of the equation $qx^2 + px + q = 0$,where $p$ and $q$ are real,are complex,then the roots of the equation $x^2 - 4qx + p^2 = 0$ are

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