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).