Factoring Problems: Solving for Inverse-Laplace Transformation

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Homework Help Overview

The discussion revolves around factoring a given expression to facilitate the inverse Laplace transformation. The specific expression under consideration is \(\frac{0.25s}{s^2+0.25s+0.25}\), which involves completing the square in the denominator.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants are attempting to complete the square for the denominator and are discussing whether the resulting expression can be factored into a form suitable for inverse Laplace transformation. There are questions about the correctness of the completed square and whether certain terms can be expressed as squares.

Discussion Status

There are differing views on the interpretation of the denominator's structure. Some participants suggest that the denominator is already a square, while others express confusion and seek clarification on the algebra involved. Guidance has been offered regarding the interpretation of the term \(\frac{15}{64}\) as a square, but no consensus has been reached.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the types of assistance they can provide. There is an emphasis on ensuring that the expression is factored correctly to meet the requirements of the inverse Laplace transformation.

EvLer
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How can i factor this nicely, so that i can get a form to inverse-laplace transform it

[tex]\frac{0.25s}{s^2+0.25s+0.25}[/tex]

so far i get this in denominator: [tex](s+\frac{1}{8})^2+\frac{15}{64}[/tex]

after completing the square, but then the last fraction is not a square of anything... and i need a square there, the way everything else looks...
 
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The denominator is already a square. I don't see how you got the other stuff. The answer should be a simple polynomial multiplied by an exponential.

Carl
 
I don't see a square there, if [tex](a+b)^2 = a^2+2ab+b^2[/tex]

this is a simple algebra thing... i guess I'm not seeing it :frown: could you show please?
 
EvLer said:
How can i factor this nicely, so that i can get a form to inverse-laplace transform it

[tex]\frac{0.25s}{s^2+0.25s+0.25}[/tex]

so far i get this in denominator: [tex](s+\frac{1}{8})^2+\frac{15}{64}[/tex]

after completing the square, but then the last fraction is not a square of anything... and i need a square there, the way everything else looks...
Do not panic. 15/64 is the square of

[tex]\frac{\sqrt {15}}{8}[/tex]

:smile:ehild
 
EvLer said:
I don't see a square there, if [tex](a+b)^2 = a^2+2ab+b^2[/tex]

this is a simple algebra thing... i guess I'm not seeing it :frown: could you show please?

My mistake. I don't see anything wrong with the way you're doing this problem, so far.

Carl
 

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