Determine the condition that the equation $z^{2} + (p + i q) z + r + i s = 0$ has real roots only.

s^{2} + p q s + r q^{2} = 0

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s^{2} + p q s + r^{2} = 0

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s^{2} - p q s + r q^{2} = 0

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s^{2} - p q s + r^{2} = 0

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Solution

The correct option is D $s^{2} - p q s + r q^{2} = 0$
Suppose $z = x$ (real) is the solution of the given equation, then
$x^{2} + (p + i q) x + r + i s = 0$
$∴ x^{2} + p x + r = 0$ and $q x + s = 0$
Eliminating x, we get $\frac{s^{2}}{q^{2}} - \frac{p s}{q} + r = 0$
$∴ s^{2} - p q s + r q^{2} = 0$
Ans: C

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