anonymous
  • anonymous
Partial fractions: x^2 + 18 -------- x(x^2+9)
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
\[\frac{x^2 + 18}{x(x^2 +9)} = \frac{A}{x} + \frac{B}{x^2 + 9} \]
anonymous
  • anonymous
\[\frac{A(x^2+9) + Bx}{(x^2+9)x}\] Compare the co-efficient! :-D
anonymous
  • anonymous
Oh Wait..

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anonymous
  • anonymous
Sorry I was wrong, this is the right method \[\frac{x^2 + 18}{x(x^2 +9)} = \frac{A}{x} + \frac{Bx+C}{x^2 + 9} \] Now multiply(take lcm) and stuff, then compare co-efficient!
anonymous
  • anonymous
I tried that, but it didn't come out right...?
anonymous
  • anonymous
Hmm try this or better use Wolfram for the multiplication, I hate calculation job \[\frac{x^2 + 18}{x(x^2 +9)} = \frac{A}{x} + \frac{B}{x +3} + \frac{C}{x-3}\] But I think it's same as the previous one, Let me check this on Wolfram. When I say it's same I mean the basic concept or the final answer is same but the constant values may differ. http://www.wolframalpha.com/input/?i=\frac{A}{x}+%2B+\frac{B}{x+%2B3}+%2B+\frac{C}{x-3} See the alternate form
anonymous
  • anonymous
What if you multiplied the right side by the denominator which would give x^2 +18 = A(x^2+9) + (bx+c)x ?
anonymous
  • anonymous
Using \[\frac{x^2 + 18}{x(x^2 +9)} = \frac{A}{x} + \frac{Bx+C}{x^2 + 9}\] I get A=2, B=-1 and C=0 and if you sub these values into the RHS you will get the LHS, so I'm convinced this is right. Are those the values you got order?
anonymous
  • anonymous
not to butt in by \[x^2+9\] does not factor so you will have to grind it out and solve \[a(x^2+9)+(bx+c)x=x^2+18\]
anonymous
  • anonymous
you get "a" instantly by letting x = 0 t give \[9a=18\] \[x=2\]
anonymous
  • anonymous
a=2*
anonymous
  • anonymous
right and also \[c=0\] since if a is 2 the left hand side already has 18 as the constant
anonymous
  • anonymous
and since left hand side has \[2x^2+bx^2\] and right hand side has \[x^2\] you get b = -1 as callum wrote
anonymous
  • anonymous
you also have the option of multiplying out to get \[a(x^2+9)+(bx+c)x=x^2+18\] \[ax^2+9a+bx^2+cx=(a+b)x^2+cx+9a=x^2+18\] and solving \[a+b=1, c =0, 9a=18\]

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