Solving equation with two different variables

[gnome222]

(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?

 

[Mentallic]

(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?

What exactly are you trying to do?

If this is what you wrote:

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

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

(2-1)!=(x^2+x)\times1!\times0!
x^2+x=1

For n=4:

(4-1)!=(x^2+x)\times 2!\times1!

x^2+x=3

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

\frac{(n-1)!}{\left(\frac{n}{2}\right)!\left(\frac{n-2}{2}\right)!}

was a constant (which it isn't) then x would have an explicit solution (likely two solutions since it's a quadratic).

 

[statdad]

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

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) !}
or
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) !}

 

[mathman]

Since x is a solution to a quadratic, there are 2 solutions for each n. Take both square roots to get them.

 

[statdad]

Since x is a solution to a quadratic, there are 2 solutions for each n. Take both square roots to get them.

Both solutions are in my post.