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Homework Help: Setting up Quadratic Equation

  1. Feb 24, 2009 #1
    1. The problem statement, all variables and given/known data
    A light bulb is wired in series with a 144Ohm resistor, and they are connected across a 120.0-V source. The power delivered to the light bulb is 23.4W. What is ther resistance of the light bulb? Note that there are two possible answers.


    2. Relevant equations
    P = I^2R and I = V/R


    3. The attempt at a solution
    Solve for I to find Rbulb. I = (V / Rb + Rr)^2 x Rb.

    Rearranged 23.4(Rb^2 + 288Rb + 144^2) = 120Rb

    This is where I'm stuck, I need to rearrange that so I can get ax^2 + bx + c = 0

    My math skills are a little weak since I have been out of school for a few years.

    Thanks a lot!:smile:
     
  2. jcsd
  3. Feb 24, 2009 #2

    cristo

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    Expand the brackets and subtract 120Rb from each side...
     
  4. Feb 24, 2009 #3
    So, pardon me, but I'd get the following?

    23.4Rb^2 + (23.4)288Rb + (23.4)144^2 - 120Rb = 0
     
  5. Feb 25, 2009 #4

    cristo

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    Yes. Now, since you are dealing with numbers, you can simplify 23.4*288 and subtract 120 to obtain an equation in your required form.
     
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