The value(s) of $b$ for which the equation, $2\log_\frac{1}{25}(bx + 28) =$ $-\log_5( 12 - 4x - x^2)$ has coincident roots is/are

Using logarithm properties, change the base of the first term to get 2 * log_5(bx + 28) / -2 = -log_5(bx + 28), so the equation becomes log_5((bx + 28)^-1) = -log_5(12 - 4x - x^2). Removing the logarithms gives 1 / (bx + 28) = 12 - 4x - x^2, which rearranges to the quadratic equation bx^2 + (4b + 28)x + 336 - 28 = 0, or bx^2 + (4b + 28)x + 308 = 0. For the original equation to have coincident roots, the discriminant of this quadratic must be zero, meaning (4b + 28)^2 - 4(b)(308) = 0. Expanding and simplifying gives 16b^2 + 224b + 784 - 1232b = 0, which factors to 16b^2 - 1008b + 784 = 0, and dividing by 16 yields b^2 - 63b + 49 = 0. Solving for b gives b = (63 +- sqrt(3969 - 196)) / 2 = (63 +- 61) / 2, which results in b = 4 or b = -12.

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