If the equation $(a^{2} + b^{2}) x^{2} - 2 (a c + b d) x + c^{2} + d^{2} = 0$ has equal roots, then

a b = c d

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a d = b c

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a d = \sqrt{b c}

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a b = \sqrt{c d}

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Solution

The correct option is B $a d = b c$
For equation $(a^{2} + b^{2}) x^{2} - 2 (a c + b d) x + (c^{2} + d^{2}) = 0$ to has equal real roots, then the condition to be satisfied is,
$B^{2} - 4 A C = 0$

$\Rightarrow 4 {(a c + b d)}^{2} - 4 (a^{2} + b^{2}) (c^{2} + d^{2}) = 0$

$\Rightarrow 8 a b c d - 4 a^{2} d^{2} - 4 b^{2} c^{2} = 0$

$\Rightarrow 4 {(a d - b c)}^{2} = 0$

$\Rightarrow a d = b c$

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Similar questions

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

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(a^{2} + b^{2}) x^{2} + 2 (a c + b d) x + c^{2} + d^{2} = 0

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