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Math Help - Difficult simplification problem

  1. #1
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    Difficult simplification problem

    I am trying to simplify this to express a as a function of t,r,p. It is killing me. Any help would be most appreciated.

    a^2=[(at)^2 + r^2]/p

    Edit: This problem is trivial, but I am also looking for a solution given the additional constraint,

    t=ab,
    Last edited by frustrated; September 26th 2012 at 02:42 AM.
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  2. #2
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    Re: Difficult simplification problem

    We are given to solve for a:

    a^2=\frac{(at)^2+r^2}{p}

    The first step is to multiply through by p:

    a^2p=(at)^2+r^2

    Now, subtract through by (at)^2:

    a^2p-(at)^2=r^2

    a^2p-a^2t^2=r^2

    Factor the left side:

    a^2(p-t^2)=r^2

    Can you proceed from here?
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  3. #3
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    Re: Difficult simplification problem

    Okay, but (at)^2 does not equal a^2t^2, it equals a^2 +2at + t^2

    I am stuck trying to get rid of the 2at term.
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  4. #4
    MHF Contributor MarkFL's Avatar
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    Re: Difficult simplification problem

    (at)^2=a^2t^2 however (a+t)^2=a^2+2at+t^2, but this is not what you posted.

    Which do you mean?
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  5. #5
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    Re: Difficult simplification problem

    Sorry, I was writing and reading (at)^2 and thinking (a+t)^2. The former is the correct expression. Thank you for our help.
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  6. #6
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    Re: Difficult simplification problem

    Okay, what if we introduce a further identity,

    t=ba

    giving

    A^2p - B^2A^4 = r^2

    Is this soluble ?
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  7. #7
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    Re: Difficult simplification problem

    Quote Originally Posted by frustrated View Post
    Okay, what if we introduce a further identity,

    t=ba

    giving

    A^2p - B^2A^4 = r^2

    Is this soluble ?
    1. I assume

    a) that you mean: a^2 \cdot p - b^2 \cdot a^4 = r^2

    b) that you want to solve for a.

    If so:

    2. Re-arrange the given equation to

    b^2 \cdot a^4 - p \cdot a^2 + r^2 = 0

    Replace a^2 = y ~ \implies ~ |a| = \sqrt{y}

    b^2 \cdot y^2 - p \cdot y + r^2 = 0

    is a quadratic in y which will give the solutions:

    y= \frac{p \pm \sqrt{p^2-4b^2 r^2}}{2b^2}

    3. That means |a| = \sqrt{\frac{p \pm \sqrt{p^2-4b^2 r^2}}{2b^2}}


    Btw: If you have a new question start a new thread!
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  8. #8
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    Re: Difficult simplification problem

    I am confused, because your solution gives a=zero when b=0, but if we simply let b=0 and solve from the initial equation, we get a simple solution that is only zero when r=0, i.e.

    a=SQRT(r^2/p)

    Edit: I see that the quadratic formula does not apply when b=0.
    Last edited by frustrated; September 26th 2012 at 03:05 AM.
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