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anonymous
 4 years ago
Partial fractions:
x^2 + 18

x(x^2+9)
anonymous
 4 years ago
Partial fractions: x^2 + 18  x(x^2+9)

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\frac{x^2 + 18}{x(x^2 +9)} = \frac{A}{x} + \frac{B}{x^2 + 9} \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\frac{A(x^2+9) + Bx}{(x^2+9)x}\] Compare the coefficient! :D

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Sorry 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 coefficient!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I tried that, but it didn't come out right...?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Hmm try this or better use Wolfram for the multiplication, I hate calculation job \[\frac{x^2 + 18}{x(x^2 +9)} = \frac{A}{x} + \frac{B}{x +3} + \frac{C}{x3}\] 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}{x3} See the alternate form

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0What if you multiplied the right side by the denominator which would give x^2 +18 = A(x^2+9) + (bx+c)x ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Using \[\frac{x^2 + 18}{x(x^2 +9)} = \frac{A}{x} + \frac{Bx+C}{x^2 + 9}\] 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
 4 years ago
Best ResponseYou've already chosen the best response.0not 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
 4 years ago
Best ResponseYou've already chosen the best response.0you get "a" instantly by letting x = 0 t give \[9a=18\] \[x=2\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0right and also \[c=0\] since if a is 2 the left hand side already has 18 as the constant

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and since left hand side has \[2x^2+bx^2\] and right hand side has \[x^2\] you get b = 1 as callum wrote

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you 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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