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The roots of the equation $\left(a^2+b^2\right) t^2-2(a c+b d) t+\left(c^2+d^2\right)=0$ are equal, then
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Verified Answer
The correct answer is:
$\frac{a}{b}=\frac{c}{d}$
Accordingly, $\{2(a c+b d)\}^2=4\left(a^2+b^2\right)\left(c^2+d^2\right)$ (as \(D=0\) or \(B^2=4 A C\) for equal roots)
$\Rightarrow 4 a^2 c^2+4 b^2 d^2+8 a b c d=4 a^2 c^2+4 a^2 d^2$ $+4 b^2 c^2+4 b^2 d^2$
$\Rightarrow 4 a^2 d^2+4 b^2 c^2-8 a b c d=0 \Rightarrow 4(a d-b c)^2=0$
$\Rightarrow \quad a d=b c \Rightarrow \frac{a}{b}=\frac{c}{d}$
$\Rightarrow 4 a^2 c^2+4 b^2 d^2+8 a b c d=4 a^2 c^2+4 a^2 d^2$ $+4 b^2 c^2+4 b^2 d^2$
$\Rightarrow 4 a^2 d^2+4 b^2 c^2-8 a b c d=0 \Rightarrow 4(a d-b c)^2=0$
$\Rightarrow \quad a d=b c \Rightarrow \frac{a}{b}=\frac{c}{d}$
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