anonymous
  • anonymous
can someone help me with partial fraction ? 4/x(x^2+4)
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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anonymous
  • anonymous
i might
anonymous
  • anonymous
so the function is 4/( x * ( x^2 + 4) )?
anonymous
  • anonymous
yes thats right

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anonymous
  • anonymous
yes thats right
anonymous
  • anonymous
cool, so what's the first step? separate the function based upon the denominator, you have two functions mutliplied by eachother on the bottom: x * (x^2+4)
anonymous
  • anonymous
so you could create a sum of two fractions: one with x as the denonm. and one with (x^2+4) as the denom. can you post those please?
anonymous
  • anonymous
i don't quite get it.. ><
anonymous
  • anonymous
np, ok, the goal is to express the function as the sum of other functions, which are actually "fractions"...where the denominators of these functions that you want to sum are the parts of the denominator of the larger function....clear?
anonymous
  • anonymous
so basically, here you have 4 ?( x * (x^2 + 4) )...the larger function on the denom is (x * (x^2 + 4) )....so can you separate this into two functions?
anonymous
  • anonymous
4 / x + 1/(x^2+4) im not very sure.. i suck at maths
anonymous
  • anonymous
right now, ignore the numerator so you're right, but ignore 4...the setup is now: \[4/(x * (x^2 + 4) ) = A/(x) + B/(x^2+4)\]
anonymous
  • anonymous
does that look familiar, like something in your text?
anonymous
  • anonymous
but shouldn't it be A / x + Bx+c / (x^2+4)?
anonymous
  • anonymous
ok, why do you think that you use "x" twice?
anonymous
  • anonymous
because of the demo has a (x^2)?
anonymous
  • anonymous
sorry, you are right about that equation not being quite right, because there is x^2+4...so you actually need to adjust the numerator of the fraction with (x^2+4)
anonymous
  • anonymous
good catch! the equation becomes: \[4/(x∗(x2+4))=A/(x)+(Bx+C)/(x2+4) \]
anonymous
  • anonymous
yeap.. that about it.. the only thing is i don't know how to solve for Bx+C
anonymous
  • anonymous
ok, now do you think you can make these fractions so that they have a common denominator? hint: multiply by "1"
anonymous
  • anonymous
i did a cross multipy which end up to be 4= A(x^2+4) + X(Bx+c)?
anonymous
  • anonymous
ok, however you are adding these fractions...so you cannot cross multiply...you need to "give" the fractions the same denominator...and easy way would be multiply each fraction by 1, where 1 is actually a fraction with the numerator and denominator equal, in this case you would want to multiply A/x by (x^2+4)/(x^2+4)...see how this multiplying by 1 is OK?
anonymous
  • anonymous
so the fraction on becomes: \[( A * (x^2 + 4))/(x * (x^2 +4))\]
anonymous
  • anonymous
cool?
anonymous
  • anonymous
ohh icic
anonymous
  • anonymous
nice, now can you apply the concept to the other fraction? then post the new equation?
anonymous
  • anonymous
A(x^2+4) / x(x^2+4) + Bx+C / x(x^2+4) am i right?
anonymous
  • anonymous
almost, what about the numerator on the second fraction?
anonymous
  • anonymous
the second fraction demo is x(x^2+4) isnt it same as the 1st fraction 's demo?
anonymous
  • anonymous
yes, that's right, but the numerator is missing something...
anonymous
  • anonymous
\[(A * (x^2 + 4) ) /( x * x^2 +4 )) + ((Bx + C)*(????))/(x * (x^2 + 4))\]
anonymous
  • anonymous
is it 1 ?
anonymous
  • anonymous
why 1?
anonymous
  • anonymous
right now you've only multiplied the second fraction by (1/x)...so you didn't multiply by "1" like you did for the first with (x^2+4)/(X^2+4)
anonymous
  • anonymous
then is it Bx+C * x(x^2+4)?
anonymous
  • anonymous
so you're close, right?
anonymous
  • anonymous
so i got to muplity Bx+C with X(x^2+4)?
anonymous
  • anonymous
not quite, right it is multiplied by 1/x...see how you put x in the denominator? multiply the numerator by x too...so that you have multiplied by "1"..right?
anonymous
  • anonymous
Bx+c * 1/x / x(x^2+4) ?? im confused
anonymous
  • anonymous
no, you were very close with (Bx + c) /x * (x^2 + 4)...how did you get here from (Bx+c)/(x^2 + 4)?
anonymous
  • anonymous
oh cancel out the x from the x(9x^2+4) but then the 1st fraction's demo isnt the same as 2nd one?
anonymous
  • anonymous
typo error x(x^2+4)
anonymous
  • anonymous
er, no canceling now
anonymous
  • anonymous
actually do the opposite....so (Bx+c)/(x^2+4) becomes ((Bx+c)*x)/((x^2+4)*x)...see how you multiplied numerator and denominator by x because you mutliplied the fraction by "1" where "1" = x/x
anonymous
  • anonymous
so, what are the two fractions now, now that they have common denominators of x*(x^2+4)...rewrite the equation
anonymous
  • anonymous
A( x^2+4) / x(x^2+4) + Bx+c * (x) / x (x^2+4)
anonymous
  • anonymous
right on
anonymous
  • anonymous
so the entire equation looks like this: \[4/(x * (x^2 + 4)) = ((A * (x^2 + 4)) + ((Bx + c) *(x)))/(x * (x^2 + 4))\]
anonymous
  • anonymous
yeap..
anonymous
  • anonymous
nice, now you can ignore denominators for the time being...so we actually want to look at the equality between the numerators
anonymous
  • anonymous
\[4 = (A *(x^2 + 4)) + (Bx+c)*x \]
anonymous
  • anonymous
this eqn isnt it same as what just now i had cross multiply them together?
anonymous
  • anonymous
i'm not sure what you mean by cross multiply...
anonymous
  • anonymous
as stated just now A/x + Bx+c/ x^2+4 i cross mutiply them.. meaning. A * (x^2+4) and x* (bx+c)
anonymous
  • anonymous
in the original equation you're saying that you multiplied everything by the denominator?
anonymous
  • anonymous
yea.
anonymous
  • anonymous
cross multiply refers to when you have two fractions: like a/b = c/d , cross multiplying would yield a*d = c*b...
anonymous
  • anonymous
that is what cross-multiplying technically is...it is best to the procedure as above so as to not move too hastily and miss a step...
anonymous
  • anonymous
ohh..
anonymous
  • anonymous
so anyway, we are close to solving
anonymous
  • anonymous
4=(A∗(x2+4))+(Bx+c)∗x
anonymous
  • anonymous
\[4 = (A * (x^2 + 4)) +(Bx + c)*(x)\] expand this
anonymous
  • anonymous
4 = Ax^2 + 4A + Bx^2 + cx?
anonymous
  • anonymous
\[4 = (A * (x^2 + 4)) + (Bx + C) * x = Ax^2 + 4A + Bx^2 + Cx\]
anonymous
  • anonymous
so, now, let's consider this, both equations are assumed to be equal for all values of x...right, because it's an equation
anonymous
  • anonymous
so you see how only one term does not have x in it?
anonymous
  • anonymous
the 4A
anonymous
  • anonymous
right on! so now we need to figure out what each letter should equal
anonymous
  • anonymous
therefore A = 1
anonymous
  • anonymous
dude, nice!
anonymous
  • anonymous
yeah, the simplest case was to consider that x = 0, now we have that A =1...so let's find C
anonymous
  • anonymous
there are no x terms on the left side, so, what do you think C equals?
anonymous
  • anonymous
but as from just now eqn 4 = A(x^2+4) + x(Bx+c) can i sub X = 0 for Bx+c(x) so that 4=A(0^2+4)? which turn out to be 4 = 4A?
anonymous
  • anonymous
Bx^2+Cx...that is the idea
anonymous
  • anonymous
Ax^2+4A+Bx^2+Cx, ok we should group these with like terms --> (A+B)*x^2 +(C)*x + 4A
anonymous
  • anonymous
(A+B)*x^2 +(C)*x + 4A = 4
anonymous
  • anonymous
you figured out that A = 1 by looking at the case when x=0
anonymous
  • anonymous
yea
anonymous
  • anonymous
and that where i struck finding b and c
anonymous
  • anonymous
now, on the right side here there are no terms with x^2 nor x...so what do you think C would be equal to?
anonymous
  • anonymous
C = 0?
anonymous
  • anonymous
right on!
anonymous
  • anonymous
now, you know A=1, C=0, now...to find B....so we look at where we have B, only ing (A+B)*x^2
anonymous
  • anonymous
which (1 + B) *x^2 because A=1
anonymous
  • anonymous
again, there are no x^2 terms on the other side..so B must equal what?
anonymous
  • anonymous
3?
anonymous
  • anonymous
anonymous
  • anonymous
why 3?
anonymous
  • anonymous
the idea is that on the other side the term is 4 + 0*x^2, so we have (1+B)*x^2 = (0)*x^2
anonymous
  • anonymous
ohh.. its -1
anonymous
  • anonymous
yes, so what are A, B and C
anonymous
  • anonymous
A= 1 B= -1 C= 0
anonymous
  • anonymous
one you have them all, substitute them back in the fractions
anonymous
  • anonymous
oicic
anonymous
  • anonymous
where they were separted
anonymous
  • anonymous
in here: 4/(x*(x^2+4))=A/(x)+B/(x^2+4)
anonymous
  • anonymous
ok got it.. thx a lot man~ sry to wate such a long time.. hope u don't mind
anonymous
  • anonymous
no problem, now do you think you can solve another partial fraction problem?
anonymous
  • anonymous
not really.. cos' theres a lot of type of partial fraction..
anonymous
  • anonymous
well the premise is the same....separate the denominator into a product of functions...then create the letter fractions with these smaller functions...solve for the letters...
anonymous
  • anonymous
just use this conversation as notes....try another and good luck!
anonymous
  • anonymous
ok thanks a lot
anonymous
  • anonymous
np

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