Factoring Problems: Solving for Inverse-Laplace Transformation

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The discussion centers on factoring the expression 0.25s/(s^2 + 0.25s + 0.25) for the purpose of performing an inverse Laplace transformation. Participants are trying to complete the square in the denominator, arriving at (s + 1/8)^2 + 15/64, but express confusion over the term 15/64 not being a perfect square. One contributor clarifies that 15/64 can be expressed as the square of √15/8, which resolves the issue. The overall goal is to simplify the expression into a form suitable for the inverse Laplace transform. The conversation highlights the importance of recognizing squares in algebraic manipulation.
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How can i factor this nicely, so that i can get a form to inverse-laplace transform it

\frac{0.25s}{s^2+0.25s+0.25}

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

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 (a+b)^2 = a^2+2ab+b^2

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

\frac{0.25s}{s^2+0.25s+0.25}

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

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

\frac{\sqrt {15}}{8}

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

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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