For the equation, determine the constants A and B that make the equation an identity.
-$$\frac { x } { ( x + 2 ) ( x - 2 ) } = \frac { A } { ( x + 2 ) } + \frac { B } { ( x - 2 ) }$$
A) $\mathrm { A } = - \frac { 1 } { 2 } , \mathrm {~B} = \frac { 1 } { 2 }$
B) $A = \frac { 1 } { 2 } , B = 1$
C) $\mathrm { A } = \frac { 1 } { 2 } , \mathrm {~B} = \frac { 1 } { 2 }$
D) $\mathrm { A } = 1 , \mathrm {~B} = \frac { 1 } { 2 }$

Question

Quizplus · Accepted Answer

The equation is an identity if and only if the numerators on both sides are equal. So, we have:$$x = A(x - 2) + B(x + 2)$$$$x = Ax - 2A + Bx + 2B$$$$x = (A + B)x + (2B - 2A)$$Equating the coefficients of x and the constant terms, we get:$$A + B = 1$$$$2B - 2A = 0$$Solving for A and B, we get:$$A = \frac{1}{2}$$$$B = \frac{1}{2}$$

For the Equation, Determine the Constants a and B That x(x+2)(x−2)=A(x+2)+B(x−2)\frac { x } { ( x + 2 ) ( x - 2 ) } = \frac { A } { ( x + 2 ) } + \frac { B } { ( x - 2 ) }(x+2)(x−2)x​=(x+2)A​+(x−2)B​

For the Equation, Determine the Constants a and B That $\frac { x } { ( x + 2 ) ( x - 2 ) } = \frac { A } { ( x + 2 ) } + \frac { B } { ( x - 2 ) }$