Determine the condition so that the equation $z^{2} + (a + i b) z + (c + i d) = 0$ has
Only one root real

d^{2} - a d b + b^{2} c = 0

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d^{2} + a d b + b^{2} c = 0

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d^{2} - a d b - b^{2} c = 0

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d^{2} + a d b - b^{2} c = 0

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Solution

The correct option is A $d^{2} - a d b + b^{2} c = 0$
Let $z = k$ ( real ) be a root,

then $k^{2} + (a + i b) k + (c + i d) = 0$

$∴ k^{2} + a k + c = 0$ and $b k + d = 0$

Eliminating k we have the condition as $\frac{d^{2}}{b^{2}} - \frac{a d}{b} + c = 0$ or $d^{2} - a d b + b^{2} c = 0$
Ans: A

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