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anonymous

  • 5 years ago

need to find the solution to this equation.

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  1. anonymous
    • 5 years ago
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    \[y ^{-1}-22y -_{2}^{1}+12-=0\]

  2. anonymous
    • 5 years ago
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    I couldn't get this to look right. it is supposed to be y with the -1/2 as the square root

  3. anonymous
    • 5 years ago
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    Is it \[{1 \over y}-22y^{-1/2}+12=0?\]

  4. myininaya
    • 5 years ago
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    let u=y^(-1/2) so we have u^2=y^(-1)

  5. myininaya
    • 5 years ago
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    u^2-22u+12=0

  6. anonymous
    • 5 years ago
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    I got 12 and 10. is that correct?

  7. anonymous
    • 5 years ago
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    or maybe they are negative?

  8. myininaya
    • 5 years ago
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    one sec im scanning something

  9. anonymous
    • 5 years ago
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    ok, thanks

  10. myininaya
    • 5 years ago
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  11. anonymous
    • 5 years ago
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    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?

  12. myininaya
    • 5 years ago
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    i didn't simplify

  13. anonymous
    • 5 years ago
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    for some reason i'm not getting a whole number

  14. myininaya
    • 5 years ago
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    i'm not getting a whole number

  15. anonymous
    • 5 years ago
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    i just realized that you had 12 in the problem when it's supposed to be 120. Sorry if I typed it wrong.

  16. anonymous
    • 5 years ago
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    yep, I did... ahh. sorry.

  17. myininaya
    • 5 years ago
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    lol. okay lets take another look at it

  18. anonymous
    • 5 years ago
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    Thanks! :)

  19. myininaya
    • 5 years ago
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    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

  20. myininaya
    • 5 years ago
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    we got u=12 and 10 because u^2-22u+120=(u-12)*(u-10)=0 so very good

  21. anonymous
    • 5 years ago
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    ok I just wasn't sure if they were supposed to stay positive.

  22. anonymous
    • 5 years ago
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    Thanks for clarifying!

  23. myininaya
    • 5 years ago
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    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

  24. anonymous
    • 5 years ago
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    okk got it:)

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