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

a b = c d

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a c = d b

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a d + b c = 0

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\frac{a}{b} = \frac{c}{d}

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Solution

The correct option is D $\frac{a}{b} = \frac{c}{d}$
The equation $(a^{2} + b^{2}) t^{2} - 2 (a c + b d) t + (c^{2} + d^{2}) = 0$
Roots are equal $\Rightarrow D = 0$
$\Rightarrow a^{2} d^{2} + b^{2} c^{2} - 2 a b c d = 0$
$\Rightarrow (a d - b c)^{2} = 0$
$\Rightarrow a d = b c$
$\Rightarrow \frac{a}{b} = \frac{c}{d}$
Hence, option 'D' is correct .

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Similar questions

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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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