using method of partial fractions to decompose this 1/(x-3)(x+3)
is 1/6(x-3)-1/6(x+3) correct or have i gone wrong somewhere
I get
1/(x-3)(x+3) = A/(x-3) + B/(x+3)
1 = (x+3)A + (x-3)B
1 = x(A+B) + 3A -3B
-> A + B = 0 -> A = -B
-> 3A - 3B = 1 -> A = 1/6; B = -1/6
= (1/6)(x-3) - (1/6)/(x+3)
thanks so it is the same thank god for that
To decompose the rational function 1/(x-3)(x+3) using the method of partial fractions, you need to express it as a sum of simpler fractions. The general form of the decomposition for a function of this type is:
1/(x-3)(x+3) = A/(x-3) + B/(x+3)
where A and B are constants that we need to determine.
To find the values of A and B, we can use a common denominator and equate the numerators. Multiplying both sides of the equation by (x-3)(x+3), we have:
1 = A(x+3) + B(x-3)
Expanding the right side and collecting like terms, we get:
1 = (A + B)x + (3A - 3B)
To satisfy this equation for all values of x, the coefficients of x and the constant terms must be equal. Therefore, we have the following system of equations:
A + B = 0
3A - 3B = 1
Solving this system, we find that A = 1/6 and B = -1/6. Thus, the correct decomposition is:
1/(x-3)(x+3) = 1/6(x-3) - 1/6(x+3)
Therefore, you are correct. The expression 1/6(x-3) - 1/6(x+3) is the correct decomposition using the method of partial fractions.