Given that \(\frac{1}{x^2 - 4} = \frac{p}{(x + 2)} + \frac{Q}{(x - 2})\) x \(\neq \pm 2\) Find the value of (P + Q)

Answer Details

We can start by simplifying the right-hand side of the equation using partial fraction decomposition. To do this, we need to find the values of P and Q. We can use a common denominator on the right-hand side of the equation to get: \[\frac{1}{x^2 - 4} = \frac{p(x - 2) + Q(x + 2)}{(x + 2)(x - 2)}\] Next, we can multiply both sides of the equation by the denominator of the left-hand side to get: \[1 = \frac{p(x - 2) + Q(x + 2)}{(x + 2)(x - 2)}(x^2 - 4)\] Simplifying the right-hand side by multiplying out the terms, we get: \[1 = \frac{(p + Q)x^3 - 4p + 4Q}{(x + 2)(x - 2)}\] Since the left-hand side is just the number 1, the numerator on the right-hand side must also be equal to 1. Therefore, we can set up a system of equations to solve for P and Q: \[p + Q = 0\] \[-4p + 4Q = 1\] Solving for P and Q using the system of equations, we get: \[P = -\frac{1}{4}\] \[Q = \frac{1}{4}\] Therefore, P + Q = 0, which is option (D).