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Simultaneous equations

  1. Jan 4, 2008 #1
    Hi, I have the following two equations of the form:

    A^2 - B^2=hj(k^2) and 2AB=mjk

    and I need to get them into the form




    I've been trying in vain for a while now and its starting to annoy me coz I know it should be simple. Any ideas?
  2. jcsd
  3. Jan 4, 2008 #2


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    What about you just solve the second one for either A or B and plug it into the other one?
    Didn't check it, but if they are really equivalent it should work.
  4. Jan 4, 2008 #3
    Yeah I've done that, thats not the problem really,its just getting it into the form I need.
  5. Jan 4, 2008 #4


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    I'm not sure I'll get exactly what you got, but I'll give it a try.

    The second one gives B = mjk/2A. Plugging this into the first one gives
    [tex]A^2 - \frac{ (m j k / 2)^2 }{ A^2 } = h j k^2 [/tex]
    Calling [itex]\alpha = A^2[/itex] and multiplying through by [itex]\alpha[/itex], we get
    [tex]\alpha^2 - h j k^2 \alpha - (m j k / 2)^2 = 0[/tex]
    The solutions are
    [tex]\alpha = \frac12\left( h j k^2 \pm \sqrt{h^2 j^2 k^4 + j^2 k^2 m^2 \right)[/tex]
    so let's take the positive one. From now on I'll start taking things out of square roots by using [itex]\sqrt{X Y^2} = Y \sqrt{X}[/itex], so you need the assumption that they are positive. First rewrite alpha into
    [tex]\alpha = \cdot \frac{1}{2} \left(h j k^2 + h j k^2 \sqrt{1 + m^2 / (h^2 k^2)} \right)[/tex]
    and then take out h j k^2:
    [tex]\alpha = k^2 \cdot \frac{h j}{2} \left(1 + \sqrt{1 + m^2 / (h^2 k^2)} \right)[/tex]
    Solve [itex]\alpha = A^2[/itex] for A (pick the positive root again), take the k^2 outside the square root and use [itex]\sqrt{a b} = \sqrt{a} \cdot \sqrt{b}[/itex].
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