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

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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 = p^2 - 4q^2 Now,consider the 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 rewrite this as $D_2 = 4(4q^2 - p^2)$.Since $p^2  0$,and therefore $D_2 > 0$.Since the discriminant $D_2$ is positive,the roots of the equation $x^2 - 4qx + p^2 = 0$ 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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