Determine the condition so that the equation $z^{2} + (a + i b) z + (c + i d) = 0$ has
both roots equal.

only

a^{2} - b^{2} = 4 c

is required condition

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only

a b = 2 d

is required condition.

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both

a^{2} - b^{2} = 4 a c

and

a b = 2 d

are required conditions

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none of these

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Solution

The correct option is C both $a^{2} - b^{2} = 4 a c$ and $a b = 2 d$ are required conditions
$z^{2} + (a + i b) z + (c + i d) = 0$
Then for equal roots, we get
$D = 0$
Or
$B^{2} = 4 A C$
Or
$(a + i b)^{2} = 4 (c + i d)$

$a^{2} - b^{2} + i 2 a b = 4 a c + i 4 d$

Comparing real part gives us

$a^{2} - b^{2} = 4 a c$

And by comparing the imaginary part, gives us

$2 a b = 4 d$

Or
$a b = 2 d$ .

Hence the necessary conditions are

$a b = 2 d$ and $a^{2} - b^{2} = 4 a c$

Suggest Corrections

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