# Maths help

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#1

Can someone explain how this happens pls (highlighted)
0
1 month ago
#2
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.
2
1 month ago
#3
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
1
#4
(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

This is what I got- can divide by 0 or did I do something wrong?
0
1 month ago
#5
When you multiply through, the denominators cancel. You get something like
4 = A# + B#
Where # are expressions in u.
Last edited by mqb2766; 1 month ago
0
1 month ago
#6
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
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#7
Ahaaaa that made sense thanks everyone! I realised my foundation really is lacking a lot T^T
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1 month ago
#8
(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
1
#9
(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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