The following quadratic equation has the equal root of opposite sign. Find the value of k in term of a and b.
$\frac{x^{2} - b x}{a x - c} = \frac{k - 1}{k + 1}$

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Solution

$\frac{x^{2} - b x}{a x - c} = \frac{k - 1}{k + 1} (x^{2} - b x) (k + 1) = (a x - c) (k - 1) k x^{2} - k b x + x^{2} - b x = a k x - k c - a x + c (k + 1) x^{2} + (a - k b - b - a k x + k x - c) = 0 x^{2} + (\frac{a - k b - b - a k}{k + 1}) x + \frac{k c - c}{k + 1} = 0 α = - β α + β = \frac{a - k b - b - a k}{k + 1} α - α = \frac{a - k b - b - a k}{k + 1} 0 = a - b - k (a + b) \frac{b - a}{a + b} = k k = \frac{a - b}{a + b}$

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