How to Expand a Fraction Using Partial Fractions

[aerandir4]

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

 

[Bill Foster]

Multiply the numerator and denominator of each fraction by the complex conjugate of its denominator.

 

[aerandir4]

thanks,
have I worked it out right up to the last term?

 

[Bill Foster]

I haven't worked it out, but that suggestion might give you back what you started with.

Do you need to simplify it?

 

[aerandir4]

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.

 

[gabbagabbahey]

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,

[tex]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]

 

[aerandir4]

Careful,

[tex]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]

I can't see how I've gone wrong. If you use the quadratic formula straight up with
a=1.25, b=3 and c=9 then

s1,s2=-3+-sqrt(9-45)/(5/2)

s1,s2=-3+-sqrt(-36)/(5/2)

s1,s2=-3+-6i/(5/2)
s1,s2=-6/5+-12i/5

so s+6/5-12i/5 and s+6/5+12i/5 are the two roots

 

[gabbagabbahey]

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)

 

[aerandir4]

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

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*(...)?

 

[gabbagabbahey]

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