There exist real numbers A and B so that 1/(k(k+3)) = A/k + B/(k+3)
for all real numbers k other than 0 and -3. Find the ordered pair (A,B)
+128407 1 A B
________ = _____ + _________
k ( k + 3) k k + 3
We can use the method of partial fractions to solve this
Mutilply through by k ( k +3) and we have
1 = A( k + 3) + Bk simplify
1 = (A + B)k + 3A equate terms and we have this system
A + B = 0
3A = 1 ⇒ A = 1/3
Which means that B = -1/3
So (A, B) = (1/3, -1/3 )
+26367 There exist real numbers A and B so that \(\mathbf{ \dfrac{1}{k(k+3)} } = \mathbf{ \dfrac{A}{k} + \dfrac{B}{k+3} }\)
for all real numbers k other than 0 and -3. Find the ordered pair (A,B)
\(\begin{array}{|lrcll|} \hline &\mathbf{ \dfrac{1}{k(k+3)} } &=& \mathbf{ \dfrac{A}{k} + \dfrac{B}{k+3} } \\\\ &&& \boxed{k \neq 0 \text{ and } k \neq -3 } \\\\ &\dfrac{1}{k(k+3)} &=& \dfrac{A}{k} + \dfrac{B}{k+3} \quad | \quad \cdot k(k+3) \\\\ &\dfrac{k(k+3)}{k(k+3)} &=& \dfrac{Ak(k+3)}{k} + \dfrac{Bk(k+3)}{k+3} \\\\ &\mathbf{ 1 } &=& \mathbf{ A (k+3) + Bk } \\ \hline k=0: & 1 &=& A (0+3) + B\cdot 0 \\ & 1 &=& 3A \\ & \mathbf{A} &=& \mathbf{ \dfrac{1}{3} } \\ \hline k=-3: & 1 &=& A (-3+3) + B\cdot (-3) \\ & 1 &=& A \cdot 0 -3B \\ & 1 &=& -3B \\ & \mathbf{B} &=& \mathbf{ -\dfrac{1}{3} } \\ \hline \end{array} \)
