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HARDCORE Math lol!

  1. Nov 30, 2004 #1
    solve (4x^2-9)^2-11(4x^2-9)+24=0
    I tried simplifying this and got 16x^4-44x^2+204=0 im not sure if this is right, and I dont know what to do to solve. Please help! :confused:
     
  2. jcsd
  3. Nov 30, 2004 #2

    Gokul43201

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    Is there not a simple substitution that will make things easier ?
     
  4. Nov 30, 2004 #3
    There is nothing but the question, what do I do?
     
  5. Nov 30, 2004 #4

    Gokul43201

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    Actually, I was making a suggestion there. Try a substitution...look for a repeating quantity and call it something.
     
  6. Nov 30, 2004 #5
    let W = (4x^2-9), solve for W using some method, and then replace W with (4x^2-9), and now solve for x

    :)
     
  7. Nov 30, 2004 #6
    Done needs to be checked plz!

    :smile: Thanks I get how to do it but want to make sure I did it right can you please check? Here is what I did:
    (4x^2-9)-11(4x^2-9)+24=0
    Let w=(4x^2-9)
    w^2-11w+24=0
    factored to become (w-8)(w-3)=0
    sub in w value (4x^2-9-8)(4x^2-9-3)=0
    (4x^2-17) (4x^2-12)=0

    so solving for x
    4x^2-17=0
    4x^2=17
    x=the square root of +or- 17/4

    or 4x^2-12=0
    4x^2=12
    x=the square root of +or-3
     
  8. Nov 30, 2004 #7
    off the subject aisha, did you get your name from the ebaumsworld video aisha, I am sure it is on otherwebsites as well, but it is the video fo the skinny dude singing about some girl in want to be backstreet boys way?
     
  9. Nov 30, 2004 #8

    Gokul43201

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    Correct...except that you mean x=+or- the square root of 17/4 NOT x=the square root of +or- 17/4

    And Tom : 'Aisha' is a reasonably common name; originally of Persian origin, it has spread into other cultures as well.
     
  10. Dec 1, 2004 #9
    Thanks EVERYONE!!! :smile:
     
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