need help finding the solution(s) to this equation..

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need help finding the solution(s) to this equation..

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\[y ^{-1}-22y ^{-1 \over 2}+120=0\]
let u=y^(-1/2)
so what does u^2=?

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x?
u^2=y^(-1)
ok, thanks. so where do I go from here?
do you know how to factor u^2-22u+120=0
i'm trying to figure out what works for 22 and 120..does anything besides 2 go into them both?
(u-10)(u-12)=0
so will the 2 solutions be positive 10 and 12 then?
y^(-1/2)=10 y^(-1/2)=12 since u=y^(-1/2)
solve for y from both of these equations and you are done
would it be 10 1/2 and 12 1/2 then? i didn't know the solutions could be anything besides a whole numbers
y=10^(-2) y=12^(-2) you can evaluate those
Sorry I have so many questions..I just want to make sure I understand completely. To evaluate these what exactly do i do?
do you know law of exponents?
possibly. I'm not sure exactly what that entails. I don't know the tnames for all these math terms
y=10^(-2)=1/(10^2) y=12^(-2)=1/(12^2) can you evaluate it now?
do they just end up being 10 and 12?
y=1/(10*10) y=1/(12*12) can you evaluate these now?
100 and 144?
I just didn't think it would be that large of number.
y=1/100 y=1/144
you can plug these in to check if you want
oh the 1 does stay on top? wasn't sure about that. Sorry i'm so incompetent at math.
Ok will do. thanks
np how about a medal? lol
i'd say you earned it!
yay! :)
thanks you i will give you one too
yay! Idk if i deserve one but thanks anyways! :)
lol ok deserve it here if we have the function f(n)=2 find f(3)=...
2..or n...er.. uhh
lol thats right
woohoo!
the 2 is right
its a horizontal line going through the y axis at 2 so no matter what you plug in fir x (or n) you will always get 2
have a goodnight

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