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anonymous

  • 5 years ago

I lied, I have one more :)

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  1. anonymous
    • 5 years ago
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    \[y ^{4}-3y ^{2}-88=0\]

  2. anonymous
    • 5 years ago
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    Again, substitute something for y^2 to get a normal quadratic you can solve

  3. anonymous
    • 5 years ago
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    It's just like the others you were doing

  4. anonymous
    • 5 years ago
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    hmm..

  5. anonymous
    • 5 years ago
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    Give it a try

  6. anonymous
    • 5 years ago
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    i substituted y^4 for y^2 and the y^2 turned into just a y?.. ahh. Idk why this is so difficult for me. I'm sure it's simple.

  7. anonymous
    • 5 years ago
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    i meant it the other way around y 2 for the y 4

  8. anonymous
    • 5 years ago
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    Use u instead of y for the second variable

  9. anonymous
    • 5 years ago
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    it'll be less confusing. Let \(u = y^2\) So \(?\ =y^4\)

  10. anonymous
    • 5 years ago
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    u^2?

  11. anonymous
    • 5 years ago
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    Yep. So now in your original equation replace all the \(y^2\) with u and all the \(y^4\) with \(u^2\)

  12. anonymous
    • 5 years ago
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    the do i want to move everything else to the other side, where it's 0?

  13. anonymous
    • 5 years ago
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    No, you need the 0 there in order to use the quadratic equation.

  14. anonymous
    • 5 years ago
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    \[au^2 + bu + c = 0\] Means: \[u = \frac{-(b) \pm \sqrt{b^2-4(a)(c)}}{2(a)}\]

  15. anonymous
    • 5 years ago
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    Look familiar?

  16. anonymous
    • 5 years ago
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    yes it does!

  17. anonymous
    • 5 years ago
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    Ok, so plug in your a, b, and c and solve for u, what do you get?

  18. anonymous
    • 5 years ago
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    3+/- \[\sqrt{-167}\]

  19. anonymous
    • 5 years ago
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    so if it's in the root it would actually be positive. i'm trying to remember.. do I incorporate the 3 with the -167 at all? or did I accurately solve this even?

  20. anonymous
    • 5 years ago
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    That's not the right square root

  21. anonymous
    • 5 years ago
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    \[\sqrt{(-3)^2 - 4(1)(-88)} = \sqrt{9 +352} = \sqrt{361} \]

  22. anonymous
    • 5 years ago
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    So what do you get with that correction?

  23. anonymous
    • 5 years ago
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    361

  24. anonymous
    • 5 years ago
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    No, I mean for the whole equation.

  25. anonymous
    • 5 years ago
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    It came down to 361=19 and those divided =19.. so would it just come to 19 or is there more numbers to that? Everything seemed to cancel out

  26. anonymous
    • 5 years ago
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    2a is not 19, it's 2. Look, a= 1, b= -3, c= -88 \[u = \frac{-(-3) \pm \sqrt{(-3)^2-4(1)(-88)}}{2(1)}\] \[ = \frac{3 \pm \sqrt{361}}{2} = \frac{3 \pm 19}{2} \]

  27. anonymous
    • 5 years ago
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    So what are the two values we get for u?

  28. anonymous
    • 5 years ago
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    11 and -11. 361 squared is 19.. so how are there 4 answers?

  29. anonymous
    • 5 years ago
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    Because those are the answers for u. 11, and -8. But \(u = y^2\) So that means y can be?

  30. anonymous
    • 5 years ago
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    I already typed the other one in and got it wrong. ha. oops. but thanks

  31. anonymous
    • 5 years ago
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    Actually, since u = y^2 u cannot be -8

  32. anonymous
    • 5 years ago
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    oh. limited tries?

  33. anonymous
    • 5 years ago
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    i have one more problem I have to get right. it's a different problem if you want me to type it on here. otherwise I'll post it to the left

  34. anonymous
    • 5 years ago
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    So there's only 2 solutions. u = \(\pm \sqrt{11}\)

  35. anonymous
    • 5 years ago
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    yep, thats what it said. thanks!

  36. anonymous
    • 5 years ago
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    I have to go to bed I'm afraid. Good luck

  37. anonymous
    • 5 years ago
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    THanks for all your help! night

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