Using partial fractions
4/ (u+2) (u-2) = A/(u+2) + B/(u-2)
Multiply both sides by (u+2)(u-2) the substitute in u=2 and u=-2 and you find A and B are both 1
4/ (u+2) (u-2) = A/(u+2) + B/(u-2)
Multiply both sides by (u+2)(u-2) the substitute in u=2 and u=-2 and you find A and B are both 1
Original post by vc94
They've used partial fractions. Assume that 4/(u-2)(u+2) can be written as A/(u-2) + B/(u+2) then solve for A and B.
Original post by Hellllpppp
Using partial fractions
4/ (u+2) (u-2) = A/(u+2) + B/(u-2)
Multiply both sides by (u+2)(u-2) the substitute in u=2 and u=-2 and you find A and B are both 1
4/ (u+2) (u-2) = A/(u+2) + B/(u-2)
Multiply both sides by (u+2)(u-2) the substitute in u=2 and u=-2 and you find A and B are both 1
This is what I got- can divide by 0 or did I do something wrong?
As mqb2766 said 4 = A (u+2) + B (u-2)
So when u=2 4=4A => A=1
And you do the same with u=-2 to get B
So when u=2 4=4A => A=1
And you do the same with u=-2 to get B
Original post by Yazomi
Ahaaaa that made sense thanks everyone! I realised my foundation really is lacking a lot T^T
Partial fractions is something I learnt in year 2 A-level maths so it’s possible that you just haven’t been taught it yet
Original post by Hellllpppp
Partial fractions is something I learnt in year 2 A-level maths so it’s possible that you just haven’t been taught it yet
Me in y13 *sweats anxiously*(^^;;
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