anonymous
  • anonymous
need to find the solution to this equation.
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
\[y ^{-1}-22y -_{2}^{1}+12-=0\]
anonymous
  • anonymous
I couldn't get this to look right. it is supposed to be y with the -1/2 as the square root
anonymous
  • anonymous
Is it \[{1 \over y}-22y^{-1/2}+12=0?\]

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myininaya
  • myininaya
let u=y^(-1/2) so we have u^2=y^(-1)
myininaya
  • myininaya
u^2-22u+12=0
anonymous
  • anonymous
I got 12 and 10. is that correct?
anonymous
  • anonymous
or maybe they are negative?
myininaya
  • myininaya
one sec im scanning something
anonymous
  • anonymous
ok, thanks
myininaya
  • myininaya
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anonymous
  • anonymous
wow, thanks for doing that! I can see where I maybe slipped up. I am a bit confused as to what the answer is though. Is there still one more step in solving those 2 solutions?
myininaya
  • myininaya
i didn't simplify
anonymous
  • anonymous
for some reason i'm not getting a whole number
myininaya
  • myininaya
i'm not getting a whole number
anonymous
  • anonymous
i just realized that you had 12 in the problem when it's supposed to be 120. Sorry if I typed it wrong.
anonymous
  • anonymous
yep, I did... ahh. sorry.
myininaya
  • myininaya
lol. okay lets take another look at it
anonymous
  • anonymous
Thanks! :)
myininaya
  • myininaya
so we get u=10 and u=12 like you been talking about this whole time but u=y^(-1/2) y^(-1/2)=10 and y^(-1/2)=12 so y=10^(-2)=1/(10^2)=1/100 so y=12^(-2)=1/(12^2)=1/144 there you basically had it you just needed to solve for y
myininaya
  • myininaya
we got u=12 and 10 because u^2-22u+120=(u-12)*(u-10)=0 so very good
anonymous
  • anonymous
ok I just wasn't sure if they were supposed to stay positive.
anonymous
  • anonymous
Thanks for clarifying!
myininaya
  • myininaya
to get rid of that -1/2 exponent i just raised both sides to -2 because law of exponents tells me if we have (x^{-1/2)^{-2} to multiply the exponents but if i do this to one side i have to do it to the other
anonymous
  • anonymous
okk got it:)

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