Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Staff Emeritus
    Science Advisor

    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


    User Avatar
    Staff Emeritus
    Science Advisor

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook