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anonymous

  • 4 years ago

Partial fractions: x^2 + 18 -------- x(x^2+9)

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  1. anonymous
    • 4 years ago
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    \[\frac{x^2 + 18}{x(x^2 +9)} = \frac{A}{x} + \frac{B}{x^2 + 9} \]

  2. anonymous
    • 4 years ago
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    \[\frac{A(x^2+9) + Bx}{(x^2+9)x}\] Compare the co-efficient! :-D

  3. anonymous
    • 4 years ago
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    Oh Wait..

  4. anonymous
    • 4 years ago
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    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!

  5. anonymous
    • 4 years ago
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    I tried that, but it didn't come out right...?

  6. anonymous
    • 4 years ago
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    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

  7. anonymous
    • 4 years ago
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    What if you multiplied the right side by the denominator which would give x^2 +18 = A(x^2+9) + (bx+c)x ?

  8. anonymous
    • 4 years ago
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    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?

  9. anonymous
    • 4 years ago
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    not to butt in by \[x^2+9\] does not factor so you will have to grind it out and solve \[a(x^2+9)+(bx+c)x=x^2+18\]

  10. anonymous
    • 4 years ago
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    you get "a" instantly by letting x = 0 t give \[9a=18\] \[x=2\]

  11. anonymous
    • 4 years ago
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    a=2*

  12. anonymous
    • 4 years ago
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    right and also \[c=0\] since if a is 2 the left hand side already has 18 as the constant

  13. anonymous
    • 4 years ago
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    and since left hand side has \[2x^2+bx^2\] and right hand side has \[x^2\] you get b = -1 as callum wrote

  14. anonymous
    • 4 years ago
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    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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