# solving for A and B

• November 27th 2008, 06:21 PM
imthatgirl
solving for A and B
1/((u^2)-4) = A/(u-2) + B/(u+2)

how do u solve for A and B?

i am told that A and B can be easily solved to be A=.25 B=-.25 but i dont see how to get that. Please help
• November 27th 2008, 06:26 PM
Mathstud28
Quote:

Originally Posted by imthatgirl
1/((u^2)-4) = A/(u-2) + B/(u+2)

how do u solve for A and B?

i am told that A and B can be easily solved to be A=.25 B=-.25 but i dont see how to get that. Please help

Ahh, I'm going to assume that this is partial fractions decomposition? If so you can do this. Multiplying through by $u^2-4$ gives

$1=A(u+2)+B(u-2)\quad{\color{red}\star}$

Now we use this little trick, if we let $u=2$ we get

$1=4A+0B\implies{A=\frac{1}{4}}$

And letting $u=-2$ gives $1=0A-4B\implies{B=\frac{-1}{4}}$

The other way is to compare coefficents. Expanding the starred step gives

$1=\left(A+B\right)u+2A-2B$

Now we know that for this expression to be true that each of the coefficients must match. So $A+B=0$ and $2A-2B=1$. Solving those gives the correct values.
• November 27th 2008, 06:29 PM
imthatgirl
how do u know to let u=2 and -2?
• November 27th 2008, 06:34 PM
Mathstud28
Quote:

Originally Posted by imthatgirl
how do u know to let u=2 and -2?

Because think about it if you have something like

$C=A(x+d)+B(x+e)$

And you want to isolate either $A$ or $B$ you want the other one to dissapear, and how do you do that? You make its coefficient zero, so you basically need to use the value of $x$ that makes either $A$ or $B$ zero. In this case it would be $x=-d$ to Isolate $B$ and $x=-e$ for $A$
• November 27th 2008, 11:44 PM
CaptainBlack
Quote:

Originally Posted by imthatgirl
1/((u^2)-4) = A/(u-2) + B/(u+2)

how do u solve for A and B?

i am told that A and B can be easily solved to be A=.25 B=-.25 but i dont see how to get that. Please help

Consider:

$\frac{A}{u-2}+\frac{B}{u+2}=\frac{A(u+2)+B(u-2)}{(u-2)(u+2)}=\frac{u(A+B)+2(A-B)}{u^2-4}$

Now if:

$\frac{A}{u-2}+\frac{B}{u+2}=\frac{u(A+B)+2(A-B)}{u^2-4}=\frac{1}{u^2-4}$

we must have:

$u(A+B)+2(A-B)=1$

so:

$A+B=0$

and

$A-B=\frac{1}{2}$

Solve these and you will get the given answers.

CB