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ArchTangent
 one year ago
My ideals are being ruined in partial fractions pls help.
Particularly the one with the quadratic equation as denominator.
Ex: (7+x) / (x^2 + x + 1)
from what I currently know is that its partial fraction would be at the beginning would look like this:
[(A)(2x+1)+B] / (x^2 + x + 1)
2x +1 coming from the derivative of the quadratic.
But later in our quiz in ECEMath my solution was questioned and professor said it should only be
[(A)(x)+B] / (x^2 + x +1)
x only...
please help me
ArchTangent
 one year ago
My ideals are being ruined in partial fractions pls help. Particularly the one with the quadratic equation as denominator. Ex: (7+x) / (x^2 + x + 1) from what I currently know is that its partial fraction would be at the beginning would look like this: [(A)(2x+1)+B] / (x^2 + x + 1) 2x +1 coming from the derivative of the quadratic. But later in our quiz in ECEMath my solution was questioned and professor said it should only be [(A)(x)+B] / (x^2 + x +1) x only... please help me

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UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2\[\frac{7+x}{x^2+x+1}\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2if I remembered correctly if the denominator factors we would have A B C... only in the numerator.. if the denominator doesn't factor then we are dealing with the Ax+B format

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2(1)^24(1)(1) = 14 = 3 oh great it doesn't factor thanks discriminant formula

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2uh oh ... we don't take derivatives for partial fractions. However towards the end of solving a partial fraction problem we do take antiderivatives.

ArchTangent
 one year ago
Best ResponseYou've already chosen the best response.1says here in my notes (before): if the denominator has distinct quad factor \[ax ^{2}+bx+c\] where \[d = b^2  4ac < 0\] the coresponds one partal frac of the form: \[\frac{ A(2x+b)+B }{ ax^2+bx+c }\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2please don't destroy me, but your professor is right for Ax+B because we have a non factorable denominator. SO for the numerator we have Ax+B "If the denominator of your rational expression has an unfactorable quadratic, then you have to account for the possible "size" of the numerator. If the denominator contains a degreetwo factor, then the numerator might not be just a number; it might be of degree one. So you would deal with a quadratic factor in the denominator by including a linear expression in the numerator." Example \[\frac{x3}{x^3+x}\] I can only factor the x out of the denominator but that \[x^2+1 \] can't be factored \[\frac{x3}{(x)(x^2+1)}\] so your partial fractions will look like this \[\frac{A}{x} +\frac{Bx+C}{x^2+1}\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2so for your problem since there is a nonfactorable in the denominator \[x^2+x+1\] the partial fraction should only be \[\frac{Ax+B}{x^2+x+1}\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2you can't take derivatives.... it doesn't happen for partial fractions. But towards the end of the problem after you solve for A B C... and so on you can take antiderivatives.

ArchTangent
 one year ago
Best ResponseYou've already chosen the best response.1ok its clearing up, this will do for now, cuz our laplace xformations at the moment only consists of \[s^2+1\] which will look like this i hope: \[\frac{ As+B }{ s^2 +1 }\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2yes that is correct... since we can't factor \[s^2+1 \] we have As+B as the numerator... hmmm either this is for Calculus II or Ordinary Differential Equations...because the Laplace Transform is being used.. though I need to refresh my memory.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2I'm curious... are you enrolled in Calculus II or Differential Equations? Laplace transforms aren't used in Calculus II, but they are used in Differential Equations

ArchTangent
 one year ago
Best ResponseYou've already chosen the best response.1wait wait wait, this is me bring paranoid..sory what if its like this: \[\frac{ x+7 }{ 4x^2+x+1 }\] will it still be \[\frac{ Ax+B }{ 4x^2+x+1 }\]

ArchTangent
 one year ago
Best ResponseYou've already chosen the best response.1yes, I am done with Calculus II done with DE. Im now studying Adv Math for Engring. currently solving a Circuit using laplace xforms when this problem shows up and ideals crumble

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2that depends on the discriminant of \[4x^2+x+1\] the discriminant formula is \[b^24ac\] so b = 1, c =1, and a = 4 \[1^24(4)(1)\] \[116 = 15\] not a perfect square so you will have Ax+B on the numerator yes.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2Oh that's good. My university is crazy. They only keep the first part of Differential Equations and the second part is well good luck finding students... especially when that second part is only offered once every 3 semesters. Even the second part of Mathematical Modeling, Numerical Analysis, and Partial Differential Equations is gone . >:/

ArchTangent
 one year ago
Best ResponseYou've already chosen the best response.1ahhh now i get it, if the discriminant is perfect square... that is the time to use \[\frac{ A(2ax+b)+B }{ ax^2+bx+c }\] example is if the discriminant is 4 or 4 and if the discriminant is not perfect square.. ex: 3. that is the time to use \[\frac{ Ax+B }{ quadratic }\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2for the second part when the discriminant is not a perfect square. that is correct but if the discriminant is a perfect square and you can factor, your numerator should only contain letters like A B C...

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2just like the example when I could only factor that x but not the x^2+1

ArchTangent
 one year ago
Best ResponseYou've already chosen the best response.1Enlightenment is upon me. thx ;) your student orgs, should really do about that crazy university of yours if the teachers wont do it by themselves

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2heh they are corrupted ... since the 1970s always has always will. The Applied Mathematics program for the upcoming semester only has 1 course and only 2 out of 30 people enrolled

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2thank goodness I'm in both Secondary Education and Applied Mathematics... I'm taking Discrete Math which is sort of like a refresher to what I have done in Intro to Advanced Mathematics.
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