My mathematics have been rusty for a while, and I couldn’t find a way to solve this.
Given the formula X = N * (N-1) * t, express it in terms of X.
So far I have
N * (N-1) * t = X
N^2 - N * t = X
N^2 - N = X/t
And I am afraid I am stumped at this point. Is there something else I could do at this point?
That would be the quadratic formula. If you can’t see how to apply it and this isn’t some kind of homework problem, post again and I’m sure we can provide more details.
<hint> it’s a quadratic equation…
On edit, damn, beaten!
Multiply that out, and you get…
x = tN[sup]2[/sup] - tN (your second step is incorrect)
Apply the quadratic formula, and you get…
N = [t ± sqrt(t[sup]2[/sup] - 4t(0))] / 2t
N = [t ± t] / 2t
N = 2t / 2t , 0 / 2t
N = 1 , 0
Or, more easily done with factoring, since your “c” term = 0, you get
x = N (tN - t)
…N = 0
…tN - t = 0
…tN = t
…N = 1
N = 1 , 0
<hint> You’re doing it wrong… (“c” does not = 0).
tN[sup]2[/sup] - tN -X = 0
a = t
b = -t
c = -X
If the task is to describe N in terms of X, this is wrong.
oops. I am wrong. sorry.
Why is it called solving for N in terms of x when there are so many t’s involved? Seriously, the answer has four (either way you express it).
If you’re asking this out of sincere befuddlement, it’s because a single equation with terms involving three variables can’t be expressed without all three variables. So if you isolate one on one side of the equation, you must have the other two on the other side. (Okay, technically you can make a trivial example by having terms that cancel out, but that’s not the case here).
Since the title of the OP had N=, that would seem to be what’s desired. It’s possible that in their case, t is considered a known constant rather than a variable. Otherwise it is incorrect to say that this “in terms of X” instead of “in terms of X and t” (which, if you weren’t confused about the basic concept, was probably your point).
I think we are predisposed to think that t means time and is a variable, but it doesn’t have to be so. If the letter were a instead of t, it might look more like a constant and solving for X might seem more familiar.
Yeah, sorry for the confusion, t is a constant. I should have invert the casing of the alphabets (guess this shows long I have left school) as constants are typically capital letters.
I am still working out the formula. Trivial: The equation is how much experience points you need to reach the Nth level in Dungeons and Dragons.
Niiiiiiice. I was sure it was homework.
So, what, you want to be able to determine how many levels a certain amount of experience points will earn you?
I remember seeing solutions to this before, but can’t recall if this involve differential equations or Laplace Transforms or something.
The only method I can replicate is trial and error solution.
Take the case where you know
X = 3515
t = 45
(I have no idea if these values are anywhere near representative, but the method should work.)
Take
N^2 - N = X/t
Plug in X and t and find X/t = 78.111…
You want an N * (N-1) that equals 78.111…
Start with N^2 = 78.111…, then N = 8.838 Add that N to 78.111… = 86.949
Then square root that value = 9.32465
Now use 9.32465 as your “N”, add that to 78.111… = 87.93876
Square root = 9.3507
Repeat to desired level of accuracy, keeping desired level of sig digs at each calculation above.
So for a t of 45, 3515 experience points can earn you 9 levels (assuming partial levels don’t count, always round down).
Or you can solve the quadratic equation mozchron gave.
x = n*(n-1)*t
x/t = n^2 - n
n^2 - n - x/t = 0
n = ( 1 ± sqrt(1 + 4xt) ) / 2
n = (sqrt(1 + 4x/t) + 1) / 2
x = 3515
t = 45
|
v
n = 9.35
So 3515 points gets you to level 9.
Doing a check with the original formula we see that:
x = n*(n-1)*t
t = 45
n = 9 -> x = 3240
n = 10 -> x = 4050
Oops, you are correct, the quadratic formula can solve any individual case.
I was caught up in how to solve that kind of process. Not necessary for the simple formula provided.