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Thread: math problem  for me!

20080128, 17:44 #1
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math problem  for me!
trying to solve for n on an equation with 2 vars (n and x). It's been a while since my algebra and I'm stuck. if you have ((x+2)/(n+4))*(1((x+2)/(n+4))) the first term is obviously ((x+2)/(n+4)) but how do you square (x+2)/(n+4)?

20080128, 20:28 #2
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Re: math problem  for me!
ok. I think the result should be x^2+4x+4/n^2+8n+16. But I did manage to dredge up computing a quadratic from gradeschool algebra at least I got that right. the equation itself is more complicated and in the original form a var was substituted for (x+2)/(n+4); I plugged that in as I want to solve for n.
Plugging in the expanded x and n into the equation after squaring gets (x+2/n+4)  ((x^2+4x+2)/(n^2+8n+16)); with the subtraction the result is  since the minus carries thru, right? (x^23x2)/(n^27n12). Kinda ugly. All that has to be divided by n+4 and, besides that, the thing is actually a square root.
If you're curious, the actual problem is attached. it's been a while since I've done this.

20080128, 20:33 #3
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Re: math problem  for me!
Edited by HansV to correct typo
The square of a/b is a^2/b^2, so the square of (x+2)/(n+4) is (x+2)^2 / (n+4)^2 = (x^2+4*x+4)/(n^2+8*n+16)
Depending on what your equation looks like, it might be better to indicate (x+2)/(n+4) with a letter, say y, so the expression becomes y*(1y). If you can solve for y, you can then go on to find x.

20080128, 20:34 #4
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Re: math problem  for me!
Yep, the 8*x was obviously a typo.

20080128, 20:36 #5
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Re: math problem  for me!
You can't just subtract the numerators and subtract the denominators. You have to make the denominators equal, then you can subtract the numerators.

20080128, 20:49 #6
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Re: math problem  for me!
oh  that rings faint bells. will work on it. It's easy if you just keep "p" as a var for (x+2)/(n+4) bit.

20080128, 22:11 #7
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Re: math problem  for me!
here's another shot at it  the problem I am running into now is, when subtracting fractions, I need to create the same denominator on both sides, which means multiply one side with n+4/n+4 and that gets me a twovariable result since (x + 2)*(n+4) = xn + 4x + 2n + 8... (see equation 6)  not sure how to get 'n' out at this point...

20080128, 22:30 #8
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Re: math problem  for me!
You can't do it that way (you can't simply cancel the (n+4)'s against each other), but there is no equation to solve  unless the sigma is allowed to remain in the solution. In that case, you end up with a third degree equation for n. Although thirddegree equations can be solved explicitly, the formulas needed are *very* complicated. Somehow I doubt that this is the intention.

20080128, 22:54 #9
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Re: math problem  for me!
Well, I thought I might have been a little too clever with cancelling out the n+4's. I wanted to simplify it!
If I treat the sigma as a variable in a solve for n it's a simple manipulation: n = (p(1p)/sigma^2)4.
But I was asked to "solve for n" but I am going to get x and n's somewhere along the line, from what I understand of it. I don't understand what you mean by "there is no equation to solve" but the major issue I have with this is there are two variables to handle and I can't see how to solve for n once I look at things like xn + 4x + 2n + 8, from (x+2)*(n+4), which I run into when trying to get common denominators on trying to subtract two fractions....
Well, thanks anyway!

20080128, 23:18 #10
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Re: math problem  for me!
Where does the request come from? Is it a course assignment, or ...?

20080128, 23:27 #11
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Re: math problem  for me!
Our statistician said he's hiring another statistician and asked me to solve this problem for n  I'm not sure why he asked. If he's trying to see what my math limits are he certainly has done so! If you don't mind, is this latest attempt on the right track? especially the arrival of a twovariable "quadratic" that I don't know what else to do with.
since the equation has to do with probabilities, I wonder if certain values of n or x need to be specified.

20080128, 23:46 #12
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Re: math problem  for me!
See attached version. I have "reduced" the equation to a cubic equation for (n+4).
See Cubic function on Wikipedia for Cardano's solution of cubic equations. <img src=/S/evilgrin.gif border=0 alt=evilgrin width=15 height=15>

20080128, 23:57 #13
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Re: math problem  for me!
ok  thanks  don't understand quite what happens when you cube n+4 to simplify it. perhaps I am not fated to, but hey not everyone has perfect pitch <img src=/S/bagged.gif border=0 alt=bagged width=22 height=22>

20080129, 00:00 #14
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Re: math problem  for me!
Darn it Hans, I had just settled into this and had brought in the cube of (n+4), look around and you post the solution.... <img src=/S/cheers.gif border=0 alt=cheers width=30 height=16>
PS I was just seeing if I could go further than S^2(n+4)^3=(x+2)(n+4)(x+2)^2.......but alas no <img src=/S/sad.gif border=0 alt=sad width=15 height=15>Jerry

20080129, 00:10 #15
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Re: math problem  for me!
Ok, I think I understand this: by adding the cube of (n+4) on both sides, you eliminate the three divisors on the right side (x+2/n+4)((x+2)(x+2)/(n+4)(n+4))  but where does the middle term (x+2)(n+4) come from?