Find constants A and B such that

\frac{2x + 11 - x + 13}{x^2 - 4} = \frac{A}{x - 2} + \frac{B}{x + 2}

for all x such that x\neq -1 and x\neq 2.

kelhaku Dec 31, 2023

#1**0 **

To find the constants A and B, we'll employ a technique called partial fraction decomposition. Here's how it works:

Factor the Denominator:

We can factor the denominator x^2 - 4 as (x - 2)(x + 2).

Set Up Partial Fractions:

We write the original fraction as a sum of two fractions with the denominators x - 2 and x + 2: \frac{x + 24}{(x - 2)(x + 2)} = \frac{A}{x - 2} + \frac{B}{x + 2}

Clear Denominators:

Multiply both sides by the common denominator (x - 2)(x + 2): x + 24 = A(x + 2) + B(x - 2)

Solve for A and B:

To find A:

Substitute x = 2 into the equation to eliminate B: 2 + 24 = A(2 + 2) + B(2 - 2) 26 = 4A A = 13/2

To find B:

Substitute x = -2 into the equation to eliminate A: -2 + 24 = A(-2 + 2) + B(-2 - 2) 22 = -4B B = -11/2

Therefore, the constants A and B that satisfy the given equation are A = 13/2 and B = -11/2.

BuiIderBoi Dec 31, 2023