If the roots of the equation $q x^{2} + p x + q = 0$ where p, q are real, be complex, then the roots of the equation $x^{2} - 4 q x + p^{2} = 0$ are

Real and unequal

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Real and equal

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Imaginary

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None of these

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Solution

The correct option is A
Real and unequal

The given equations are

$q x^{2}$ + px + q = 0 ....(i)

and~ $x^{2}$ - 4qx + $p^{2}$ = 0 ......(ii)

Roots of (i) are complex, therefore $p^{2} - 4 q^{2}$ < 0

Now discrimninant of (ii) is $16 q^{2} - 4 p^{2} = - 4 (p^{2} - 4 q^{2})$ > 0

Hence, roots are real and unequal

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