Question

For the equation $\frac{A}{x+3}+\frac{B}{x-3}=\frac{-x+9}{x^2-9}$ to be true for values of $x(x\ne 3)$, the value of AB is ______.

• A. 2
• B. -1
• C. -2
• D. -3

Answer 1

The correct option is C -2

Given equation is

$\frac{A}{x+3}+\frac{B}{x-3}=\frac{-x+9}{x^2-9}$

$⟹\frac{(x-3)A+B(x+3)}{x^2-9}=\frac{-x+9}{x^2-9}$

$⟹A(x-3)+B(x+3)=-x+9$

At $x=-3$

we get $A(-3-3)=-(-3)+9$

$-6A=12⟹A=-2$

At $x=4$

we get $A(4-3)+B(4+3)=-4+9$

$A+7B=5⟹7B=5-(-2)⟹7B=7⟹B=1$

$\therefore AB=(-2)\left(1\right)=-2$