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Multi-variable quadratic question

  1. Dec 12, 2008 #1
    1. The problem statement, all variables and given/known data
    At the end of a much longer problem, I'm asked to find a second height that will satisfy a formula found for a height in the first part of the problem where:

    2. Relevant equations

    3. The attempt at a solution
    I know the answer I should get: h3=h2-h1

    But I cannot figure how to manipulate the variables to get that. I've tried using the quadratic equation but get lost under the square root sign. This has got to be easier than I'm making it!
  2. jcsd
  3. Dec 12, 2008 #2


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    Homework Helper

    Welcome to PF.

    Multiply it out both sides and rearrange:

    h2h1 -h1² = h2h3 - h3²

    h2h1 - h2h3 = h2(h1 - h3) = h1² - h3² = (h1 + h3)(h1 - h3)

    divide by (h1-h3)
  4. Dec 12, 2008 #3
    Wow, thank you. It's been awhile since I've done this stuff!
  5. Dec 12, 2008 #4


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    Staff Emeritus
    Science Advisor
    Gold Member

    You want to get a quadratic with h3 as the variable and the other heights as constants (because you want h3 *in terms of* those other two heights). So expand the right hand side (but not the left) and rearrange:

    [tex] h_3^2 - h_2 h_3 + h_1 (h_2 -h_1) = 0 [/tex]

    There's your quadratic. a = 1, b = h2, c = left hand side of the original equation.

    Now, it's messy, but the two solutions you'll get are the one you're expecting, and another one, namely h3 = h1 (which is obviously a solution, by inspection).

    Hint: Your discriminant is:

    [tex] h_2^2 - 4h_1(h_2 - h_1) [/tex]

    [tex] = h_2^2 - 4h_1h_2 + 4h_1^2 [/tex]

    This is a *perfect square*, making things really easy.
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