If frac{x-4}{x^2-5x-2k} = frac{2}{x-2} - frac | vedclass

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(A) Given the equation: $\frac{x-4}{x^2-5x-2k} = \frac{2}{x-2} - \frac{1}{x+k}$ Simplifying the right side: $\frac{2(x+k) - (x-2)}{(x-2)(x+k)} = \frac

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$-3$

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(A) Given the equation: $\frac{x-4}{x^2-5x-2k} = \frac{2}{x-2} - \frac{1}{x+k}$ Simplifying the right side: $\frac{2(x+k) - (x-2)}{(x-2)(x+k)} = \frac{2x + 2k - x + 2}{(x-2)(x+k)} = \frac{x + 2k + 2}{x^2 + kx - 2x - 2k} = \frac{x + 2k + 2}{x^2 + (k-2)x - 2k}$ Comparing the denominators: $x^2 - 5x - 2k = x^2 + (k-2)x - 2k$ This implies $k-2 = -5$,so $k = -3$ Comparing the numerators: $x-4 = x + 2k + 2$ Substituting $k = -3$: $x-4 = x + 2(-3) + 2 = x - 6 + 2 = x - 4$ Both sides match,therefore $k = -3$.

If $\frac{x-4}{x^2-5x-2k} = \frac{2}{x-2} - \frac{1}{x+k}$,then $k$ is equal to

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