If the roots of the equation $(a^{2} + b^{2}) x^{2} + 2 x (a c + b d) + c^{2} + d^{2} = 0$ are real, they will be equal.

True

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False

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Solution

The correct option is A True
$(a^{2} + b^{2}) x^{2} + 2 x (a c + b d) + (c^{2} + d^{2}) = 0$
$△ = b^{2} - 4 a c$
$= 4 {(a c + b d)}^{2} - 4 (a^{2} + b^{2}) (c^{2} + d^{2})$
$= 4 (a^{2} c^{2} + b^{2} d^{2} + 2 a c b d) - 4 (a^{2} c^{2} + a^{2} d^{2} + b^{2} c^{2} + b^{2} d^{2})$
$= - 4 ({(a d)}^{2} + {(b c)}^{2} - 2 a b c d)$
$= - 4 {(a d - b c)}^{2} < 0$
$△ < 0$
Roots are not real
If $△ = 0$ then $a d = b c$
Roots should be equal
Hence given statement is true

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