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) $\mathrm { A } = \frac { 1 } { 2 } , \mathrm {~B} = 1$
B) $A = 1 , B = \frac { 1 } { 2 }$
C) $\mathrm { A } = - \frac { 1 } { 2 } , \mathrm {~B} = \frac { 1 } { 2 }$
D) $\mathrm { A } = \frac { 1 } { 2 } , \mathrm {~B} = \frac { 1 } { 2 }$

Question

Quizplus · Accepted Answer

To solve for A and B, we can multiply both sides of the equation by the common denominator $(x+2)(x-2)$ to get rid of the fractions. This gives us $x = A(x-2) + B(x+2)$. Expanding and simplifying, we get $x = Ax - 2A + Bx + 2B$. Combining like terms, we have $x = (A + B)x + (-2A + 2B)$. For this to be an identity (true for all x), the coefficients on both sides of the equation must match. Therefore, $A + B = 1$ (since the coefficient of x on the left side is 1) and $-2A + 2B = 0$ (since there is no constant term on the left side). Solving these equations simultaneously gives $A = \frac{1}{2}$ and $B = \frac{1}{2}$.

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 $\frac { x } { ( x + 2 ) ( x - 2 ) } = \frac { A } { ( x + 2 ) } + \frac { B } { ( x - 2 ) }$