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Solving equation with two different variables

  1. Apr 19, 2015 #1
    (n-1)! = (x2 + x)( (n/2)! (n-2/2)!
    Any idea how to solve this? I tried to solve for x but got stuck. Could you multiply out the factorials?
     
  2. jcsd
  3. Apr 19, 2015 #2

    Mentallic

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    What exactly are you trying to do?

    If this is what you wrote:

    [tex](n-1)!=\left(x^2+x\right)\left(\frac{n}{2}\right)!\left(\frac{n-2}{2}\right)![/tex]

    Then x can be many values, for example, if we take n=2 then

    [tex](2-1)!=(x^2+x)\times1!\times0![/tex]
    [tex]x^2+x=1[/tex]

    For n=4:

    [tex](4-1)!=(x^2+x)\times 2!\times1![/tex]

    [tex]x^2+x=3[/tex]

    So as you can see, x can take on various values depending on n, and vice versa. If you could show that

    [tex]\frac{(n-1)!}{\left(\frac{n}{2}\right)!\left(\frac{n-2}{2}\right)!}[/tex]

    was a constant (which it isn't) then x would have an explicit solution (likely two solutions since it's a quadratic).
     
  4. Apr 20, 2015 #3

    statdad

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    If I have not screwed something up (and since I am recovering from knee surgery and have some pain meds in me, I might have) you can write

    [tex]x=-\frac{\sqrt{{\left( \frac{n-2}{2}\right) !}^{2}\,{\left( \frac{n}{2}\right) !}^{2}+4\,\left( \frac{n-2}{2}\right) !\,\left( n-1\right) !\,\left( \frac{n}{2}\right) !}+\left( \frac{n-2}{2}\right) !\,\left( \frac{n}{2}\right) !}{2\,\left( \frac{n-2}{2}\right) !\,\left( \frac{n}{2}\right) !}[/tex]
    or
    [tex]x=\frac{\sqrt{{\left( \frac{n-2}{2}\right) !}^{2}\,{\left( \frac{n}{2}\right) !}^{2}+4\,\left( \frac{n-2}{2}\right) !\,\left( n-1\right) !\,\left( \frac{n}{2}\right) !}-\left( \frac{n-2}{2}\right) !\,\left( \frac{n}{2}\right) !}{2\,\left( \frac{n-2}{2}\right) !\,\left( \frac{n}{2}\right) !}[/tex]
     
  5. Apr 20, 2015 #4

    mathman

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    Since x is a solution to a quadratic, there are 2 solutions for each n. Take both square roots to get them.
     
  6. Apr 20, 2015 #5

    statdad

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    Both solutions are in my post.
     
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