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

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Solution

we use this by using $b^{2} - 4 a c = 4 (a c + b d)^{2} - 4 (a^{2} + b^{2}) (c^{2} + d^{2})$
= $4 [(a c)^{2} + 2 a b c d + (b d)^{2}] - 4 ((a c)^{2} + (b c)^{2} + (a c)^{2} + (b d)^{2})$
= $- 4 (b c)^{2} + 8 a b c d - 4 (a d)^{2}$
add $- 2 a b c d$ and $2 a b c d$
we get
$(b c - a d) - 2 a b c d = 0$
$(b c - a d) = 0$
$a d = b c$
$a / b = c / d$
$(a / b) - (c / d) = 0$

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