We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
+1
128
2
avatar

Find constants $A$ and $B$ such that \(\frac{x + 7}{x^2 - x - 2} = \frac{A}{x - 2} + \frac{B}{x + 1}\) for all $x$ such that \($x\neq -1$\) and \(x\neq 2\). Give your answer as the ordered pair (A,B).

I don't know how to have Only A and B on one side while the rest on the other side

 Apr 20, 2019
 #1
avatar+4322 
+2

Hint: Partial Fraction Decomposition.

 Apr 20, 2019
 #2
avatar+7711 
+2

You don't. That's not how you do it.

You finally get one side as a polynomial of x, with known coefficients, another side as a polynomial of x, with unknown coefficients.

You compare the coefficient of each term, then you can find the unknowns.

 

\(\dfrac{x+7}{x^2-x-2} = \dfrac{A}{x-2}+\dfrac B{x+1}\\ \dfrac{x+7}{x^2-x-2} = \dfrac{A(x+1) + B(x-2)}{x^2-x-2}\\ \text{As the denominators are the same, so the numerators must be the same.}\\ x + 7 = A(x+1) + B(x-2)\\ x + 7 = Ax + A + Bx - 2B\\ x + 7 = (A+B)x + (A-2B)\\ \begin{cases} A + B = 1---(1)\\ A - 2B = 7---(2)\\ \end{cases}\\ (1)-(2):\\3B = -6\\ B = -2\\ A - 2 = 1 \implies A = 3\\ \therefore (A,B) = (3,-2)\)

.
 Apr 20, 2019

13 Online Users

avatar