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Math Help - Rearrange Formula Past Exam Question

  1. #1
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    Rearrange Formula Past Exam Question

    I'm really sorry for this because I know it's not the hardest of formulas, but I have a complete blank on how to do it.

    I need to make t the subject of the formula.

    F= 8 / root(t^2-6)

    ^2 being squared.

    Don't worry, it's not for any homework or assignment, I'm revising for my exam in 3 weeks time. I know it's silly that I can't do this at such a late stage, but can anyone help?

    I thought I had to times both sides by root(t^2-6) and divide both sides by F first, but I could easily be wrong.
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  2. #2
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    If this is your formula:

    F=\frac{8}{\sqrt{t^2-6}}

    then multiplying both sides by \sqrt{t^2-6} and deviding both sides by F gives you:

    \sqrt{t^2-6}=\frac{8}{F}

    continueing:

    t^2=(\frac{8}{F})^2 + 6

    t=(\sqrt((\frac{8}{F})^2 + 6))
    Last edited by technopersia; April 13th 2011 at 03:28 AM.
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  3. #3
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    Yeah I thought I should do that, but I didn't know what to do after to get t on it's own. But thanks

    EDIT.
    Sorry, the bottom bits didn't show up when I first read your post. Ignore this one.
    Last edited by dementedsquirrel; April 13th 2011 at 02:57 AM.
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  4. #4
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    This has been superceded
    Last edited by Carlow52; April 13th 2011 at 04:25 AM. Reason: technopersia corrected his first post
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  5. #5
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    Quote Originally Posted by Carlow52 View Post
    Is the move from first line to 2nd line correct here?

    Just wondering as it looks like the LHS has been squared and the RHS 'rooted'
    That is a point. Are you not supposed to square it? Or am I just talking baloney?
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  6. #6
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    You are right, my bad!

    If this is your formula:

    F=\frac{8}{\sqrt{t^2-6}}

    then multiplying both sides by \sqrt{t^2-6} and deviding both sides by F gives you:

    \sqrt{t^2-6}=\frac{8}{F}

    continueing:

    t^2=(\frac{8}{F})^2 + 6

    t=(\sqrt((\frac{8}{F})^2 + 6))
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  7. #7
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    Again superceded
    Last edited by Carlow52; April 13th 2011 at 04:21 AM. Reason: technopersia corrected his first post
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  8. #8
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    Perfect!
    Thank you very much.
    I can't believe I couldn't get it in the first place
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  9. #9
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    sorry about the first mistake.
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  10. #10
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    No problem.
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  11. #11
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    Don't forget your negative root!

    t= \pm \sqrt\left({\dfrac{8}{F}\right)^2 + 6}. Alternatively |t|= \sqrt\left({\dfrac{8}{F}\right)^2 + 6}


    To fit the original domain then t - 6 > 0 and from your rearranging F \neq 0
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  12. #12
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    Thank you.
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