There are numbers A and B for which
A/(x-1)+ B/(x+1)=(x+2)/(x^2-1)for every number x≠±1. Find B.
multiply through by x^2-1 = (x-1)(x+1) and you have
A(x+1) + B(x-1) = x+2
Ax+A + Bx-B = x+2
(A+B)x + (A-B) = x + 2
If the two polynomials are the same for all values of x, then the coefficients must match.
Now just solve for B in
A+B = 1
A-B = 2
Now review your section on partial fractions. If you get stuck on a problem, next time show what you have tried, so we can see where you went wrong.
To find the value of B, we can start by multiplying both sides of the equation by the denominator (x^2 - 1) to eliminate the fractions.
A/(x-1)(x^2 - 1) + B/(x+1)(x^2 - 1) = (x+2)
Expanding the denominators, we get:
A(x+1) + B(x-1) = (x+2)
Next, let's simplify and rearrange the equation to isolate B:
Ax + A + Bx - B = x + 2
Combining like terms, we have:
(A + B)x + (A - B) = x + 2
Now, we can equate the coefficients of x on both sides of the equation:
A + B = 1
Since this equation represents the coefficients of x, it holds true for all values of x, except for x = ±1 as stated in the initial problem statement.
From this equation, we can see that A + B = 1. Since we want to find B, we can rearrange the equation to solve for B:
B = 1 - A
However, we still need to find the value of A. To do this, let's consider a specific value of x that satisfies the given condition that x ≠ ±1.
Let's choose x = 2:
A/(2 - 1) + B/(2 + 1) = (2 + 2)/(2^2 - 1)
Simplifying this equation, we have:
A + B/3 = 4/3
Multiplying through by 3 to eliminate the fraction, we get:
A + B = 4
Now, we can substitute B = 1 - A into this equation:
A + (1 - A) = 4
Simplifying, we have:
1 = 4
This equation is inconsistent; there is no value for A that satisfies it.
Therefore, there is no unique value for B that satisfies the given equation.
In conclusion, there is no value of B that satisfies the equation A/(x-1) + B/(x+1) = (x+2)/(x^2-1) for every number x ≠ ±1.