How to Expand a Fraction Using Partial Fractions

aerandir4
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Homework Statement



expand by partial fractions:

Homework Equations



2(s+5)/(1.25*s^2+3s+9)

The Attempt at a Solution



ok I initially used the quadratic formula to get the two roots for the denominator
these being
(s+6/5+12i/5)(s+6/5-12i/5) i.e complex numbers

so now the partial fractions looks like this:

2(s+5)/(s+6/5+12i/5)(s+6/5-12i/5) = A/(s+6/5+12i/5)+B/(s+6/5-12i/5)

solving for B I get 1-19/12i which when multiplied by i/i = 1+19i/12 and A is the conjugate I believe, therefore A=1-19i/12

now the partial fraction looks like this
(1-19i/12)/(s+6/5-12i/5)+(1+19i/12)/(s+6/5+12i/5)

does this look right so far? If so how should I proceed in simplifying the terms?

thanks
 
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Multiply the numerator and denominator of each fraction by the complex conjugate of its denominator.
 
thanks,
have I worked it out right up to the last term?
 
I haven't worked it out, but that suggestion might give you back what you started with.

Do you need to simplify it?
 
Bill Foster said:
I haven't worked it out, but that suggestion might give you back what you started with.

Do you need to simplify it?

I personally prefer to always leave terms in the most simple of ways.
 
aerandir4 said:
ok I initially used the quadratic formula to get the two roots for the denominator
these being
(s+6/5+12i/5)(s+6/5-12i/5) i.e complex numbers

so now the partial fractions looks like this:

2(s+5)/(s+6/5+12i/5)(s+6/5-12i/5) = A/(s+6/5+12i/5)+B/(s+6/5-12i/5)

Careful,

1.25*s^2+3s+9=\frac{5}{4}\left(s+\frac{6}{5}+\frac{12i}{5}\right)\left(s+\frac{6}{5}-\frac{12i}{5}\right)\neq\left(s+\frac{6}{5}+\frac{12i}{5}\right)\left(s+\frac{6}{5}-\frac{12i}{5}\right)[/itex]
 
gabbagabbahey said:
Careful,

1.25*s^2+3s+9=\frac{5}{4}\left(s+\frac{6}{5}+\frac{12i}{5}\right)\left(s+\frac{6}{5}-\frac{12i}{5}\right)\neq\left(s+\frac{6}{5}+\frac{12i}{5}\right)\left(s+\frac{6}{5}-\frac{12i}{5}\right)[/itex]
<br /> <br /> I can&#039;t see how I&#039;ve gone wrong. If you use the quadratic formula straight up with<br /> a=1.25, b=3 and c=9 then <br /> <br /> s1,s2=-3+-sqrt(9-45)/(5/2)<br /> <br /> s1,s2=-3+-sqrt(-36)/(5/2)<br /> <br /> s1,s2=-3+-6i/(5/2)<br /> s1,s2=-6/5+-12i/5<br /> <br /> so s+6/5-12i/5 and s+6/5+12i/5 are the two roots
 
Sure, the roots are s1,s2=-3+-6i/(5/2), but as^2+bs+c=a(s-s1)(s-s2) not just (s-s1)(s-s2)
 
gabbagabbahey said:
Sure, the roots are s1,s2=-3+-6i/(5/2), but as^2+bs+c=a(s-s1)(s-s2) not just (s-s1)(s-s2)

im getting a little confused now :confused:

I don't understand why its as^2+bs+c=a(s-s1)(s-s2).
Is this a special case or something? I have never seen it done like that before

at this stage... s1,s2=-6/5+-12i/5
don't you just take the whole term on the right to the left hand side?
where does the five over four come from here 5/4*(...)?
 
  • #10
It's basic algebra... when you expand (s-s1)(s-s2) using FOIL, you get s^2+bs/a+c/a not as^2+bs+c...you should really know this stuff by now
 
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