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Thread starter 14 years ago
#1
Find the value of A, B & C.

x²/(x-2)(x-6) = A + B/(x-2) + C/(x-6)

x² = A(x-2)(x-6) + B(x-6) + C(x-2)

When x=2, B=-1

When x=6, C=9

How do I find A?
0
14 years ago
#2
you have to expand the RHS, and set the coefficients equal.
0
14 years ago
#3
when x = 1
using x²/(x-2)(x-6) = A + B/(x-2) + C/(x-6)
1/(-1)(-5) = a - 1/(-1) + 9/(-5)
a = 1

Alt method
when x = 1
using x² = A(x-2)(x-6) + B(x-6) + C(x-2)
1 = 5a +5 -9
5a = 5
a=1
0
Thread starter 14 years ago
#4
Hmm, I get A=1.

Is that right?

Manps, How did you get A=3?
0
14 years ago
#5
sorry i spotted the mistake and then changed it
0
Thread starter 14 years ago
#6
Oh, alrighty.
0
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