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hi_vis360
November, 2024

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Anyone fancy a partial fraction

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90053
Posts: 2565 - Karma: 3,728
wolfram can't handle it...and i keep making mistakes and getting rubbish numbers


(9(x^2)-223x+445)/(((3x-5)^2)(x+10)))

in the form

a/(3x-5)+b/((3x-5)^2)+c/(x+10)

damn computer generated tests....
Oct 6 2013 5:20PM
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cannaguts666
Posts: 504 - Karma: 256
Ugh an 1/8 plus a 1/4 is 3/8.
Oct 6 2013 5:22PM
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Miomo
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69
Oct 6 2013 5:24PM
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90053
Posts: 2565 - Karma: 3,728
Click for larger image
Oct 6 2013 5:26PM
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Mag
Posts: 8155 - Karma: 14,809
You've got one too many )'s on the original message, that could be why wolfram didn't read it right?

IIRC, don't you just factorise and multiply by the original denominator in each case
Oct 6 2013 5:32PM
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imajew
Posts: 324 - Karma: 22
a=-1941/245 b=-199/7 c=-131/49
Oct 6 2013 5:36PM
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90053
Posts: 2565 - Karma: 3,728
yeh i tried again, still exceeds computational power

Oct 6 2013 5:37PM
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90053
Posts: 2565 - Karma: 3,728
you just end up with some ridiculous simultaneous equations
Oct 6 2013 5:38PM
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imajew
Posts: 324 - Karma: 22
but really, up your wolfram game son Click for larger image
Oct 6 2013 5:41PM
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90053
Posts: 2565 - Karma: 3,728
okay i finished the test and the answer was

-6, 7 and 3
Oct 6 2013 5:45PM
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Mag
Posts: 8155 - Karma: 14,809
((9x^2)-233x+445)/(((3x-5)^2)(x+10))=(a/(3x-5))+(b/((3x-5)^2))+(c/(x+10)) <the a,b,c is the bit I'm not sure about.


(9x^2)-233x+445)= (A(3x-5)(x+10))+(B(x+10))+(C((3x-5)^2))

then let x=-10, 5 and then substitute in B and C to get A.


Oct 6 2013 5:46PM
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