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Struggling to rearrange equation

  1. Nov 13, 2013 #1
    I'm trying to prove that

    $$\dfrac{l}{c+u}+\dfrac{l}{c-u}=\dfrac{2l}{\sqrt{c^2-u^2}}$$


    My workings so far:
    $$\dfrac{l}{c+u}+\dfrac{l}{c-u},$$
    put over common denominator,
    $$\dfrac{l(c-u)}{(c-u)(c+u)}+\dfrac{l(c+u)}{(c-u)(c+u)},$$
    $$\dfrac{l(c-u)+l(c+u)}{(c-u)(c+u)},$$
    expand out,
    $$\dfrac{cl-lu+cl+lu}{(c-u)(c+u)},$$
    group up and cancel out,
    $$\dfrac{2cl}{c^2-u^2},$$

    we need to make it equal to
    $$\dfrac{2l}{\sqrt{c^2-u^2}}$$
    HOW?!?
     
    Last edited: Nov 13, 2013
  2. jcsd
  3. Nov 13, 2013 #2

    vela

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    You can't because they're not equal. It's more apparent when you rewrite each slightly:
    $$\frac{2cl}{c^2-u^2} = \frac{2cl}{c^2[1-(u/c)^2]} = \frac{2l}{c}\frac{1}{1-(u/c)^2}$$ and
    $$\frac{2l}{\sqrt{c^2-u^2}} = \frac{2l}{\sqrt{c^2[1-(u/c)^2]}} = \frac{2l}{c}\frac{1}{\sqrt{1-(u/c)^2}}$$
     
  4. Nov 13, 2013 #3

    PeroK

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    There's no law against plugging some numbers in! If you're not sure whether two expressions are equal, try some numbers and see! l = 1, u = 1 and c = 2 proves a lack of equality here.
     
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