In the quadratic equation $x^{2} + (p + i q) x + 3 i = 0$ , p & q are real. If the sum of the squares of the roots is 8 then,values of $p$ and $q$ are

p = 2, q = 2

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p = - 3, q = 4

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p = 3, q = 1

p = - 3, q = - 1

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p = - 2, q = - 2

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Solution

The correct option is B $p = 3, q = 1$ or $p = - 3, q = - 1$
$x^{2} + (p + i q) x + 3 i = 0$
$α + β = - (p - i q), α β = 3 i$
$α^{2} + β^{2} = (α + β)^{2} - 2 α β = [- (p + i q)]^{2} - 6 i$
$= (p^{2} - q^{2}) + i (2 p q - 6) = 8$
$\Rightarrow p^{2} - q^{2} = 8$ and $p q = 3$
$\Rightarrow p = 3, q = 1$ or $p = - 3, q = - 1$

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