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Math Help - square root problem

  1. #1
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    square root problem

    hello guys, i got a problem in solving this

    if
    x=\tfrac{2ab}{{b_{}}^{2}+1}  <br />

    calculate :
    \tfrac{\sqrt{a+x} - \sqrt{a-x}}{\sqrt{a+x} + \sqrt{a-x}}

    thanks
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  2. #2
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    Awetuouncsygg
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    Uhm, just multiply by \sqrt{a+x}-\sqrt{a-x} and then calculate. Where is the problem? o.o
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  3. #3
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    i know that, i got stucked on this :

    \frac{2a-2\sqrt{a^{2}-x^{2}}}{2x}
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  4. #4
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    And then calculate, remember? Replace x by 2ab/(b^2+1).
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  5. #5
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    Quote Originally Posted by disharmonie View Post
    hello guys, i got a problem in solving this

    if
    x=\tfrac{2ab}{{b_{}}^{2}+1}  <br />

    calculate :
    \tfrac{\sqrt{a+x} - \sqrt{a-x}}{\sqrt{a+x} + \sqrt{a-x}}

    thanks
    Why not just calculate the 2 relevant terms first to see what you are dealing with ?

    a+x=a+\frac{2ab}{b^2+1}=\frac{a\left(b^2+1\right)}  {b^2+1}+\frac{2ab}{b^2+1}

    =\frac{ab^2+a+2ab}{b^2+1}=\frac{a\left(b^2+2b+1\ri  ght)}{b^2+1}

    =\frac{a(b+1)^2}{b^2+1}

    Then

    \sqrt{a+x}=(b+1)\sqrt{\frac{a}{b^2+1}}

    Similarly

    \sqrt{a-x}=(b-1)\sqrt{\frac{a}{b^2+1}

    which gives

    \sqrt{a+x}-\sqrt{a-x}=2\sqrt{\frac{a}{b^2+1}}

    \sqrt{a+x}+\sqrt{a-x}=2b\sqrt{\frac{a}{b^2+1}}

    Then it's quite straightforward to perform the division.
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  6. #6
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    just replacing x with will lead into a complex calculation. thanks to Archie, that's a better method to solve this
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