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Math Help - Prove an Equation involving Square Roots

  1. #1
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    Prove an Equation involving Square Roots

    Proof  <br />
 \frac{a-1}{2 } \sqrt{\frac{a+1}{a-1 } } +\frac{a+1}{2 } \sqrt{\frac{a-1}{a+1 } } = \sqrt{a^2-1} <br />

    fankyou.
    Last edited by BabyMilo; June 10th 2011 at 11:06 AM.
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  2. #2
    Super Member Quacky's Avatar
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    Quote Originally Posted by BabyMilo View Post
    Proof  <br />
 \frac{a-1}{2 } \sqrt{\frac{a+1}{a-1 } } +\frac{a+1}{2 } \sqrt{\frac{a-1}{a+1 } } = \sqrt{a^-1} <br />

    fankyou.
    I believe that there is a horrible typo in the question: it should be  \frac{a-1}{2 } \sqrt{\frac{a+1}{a-1 } } +\frac{a+1}{2 } \sqrt{\frac{a-1}{a+1 } } =\sqrt{a^2-1}

    I don't know if there is a shorthand, but I would consider writing the LHS as:
     \frac{1}{2}\sqrt{\frac{(a+1)(a-1)^2}{a-1}} +\frac{1}{2}\sqrt{\frac{(a-1)(a+1)^2}{a+1}}

    Take it from here.

    Edit: I see you've corrected the typo in the OP
    Last edited by Quacky; June 10th 2011 at 11:11 AM. Reason: Clarification
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  3. #3
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    I was trying to correct it, but my internet is so slow.
    Thank you for your solution
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  4. #4
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    Hello, BabyMilo!

    Another approach . . .



    \text{Prove: }\;\frac{a-1}{2 } \sqrt{\frac{a+1}{a-1 } } +\frac{a+1}{2 } \sqrt{\frac{a-1}{a+1 } } \;=\; \sqrt{a^2-1}

    We have: . \frac{a-1}{2}\cdot\frac{(a+1)^{\frac{1}{2}}}{(a-1)^{\frac{1}{2}}} + \frac{a+1}{2}\cdot\frac{(a-1)^{\frac{1}{2}}}{(a+1)^{\frac{1}{2}}}

    . . . . . . . =\;\frac{(a-1)^{\frac{1}{2}}(a+1)^{\frac{1}{2}}}{2} + \frac{(a+1)^{\frac{1}{2}}(a-1)^{\frac{1}{2}}}{2}

    . . . . . . . =\;\frac{[(a-1)(a+1)]^{\frac{1}{2}}}{2} + \frac{[(a+1)(a-1)]^{\frac{1}{2}}}{2}

    . . . . . . . =\;\frac{(a^2-1)^{\frac{1}{2}}}{2} + \frac{(a^2-1)^{\frac{1}{2}}}{2}

    . . . . . . . =\;(a^2-1)^{\frac{1}{2}}

    . . . . . . . =\;\sqrt{a^2-1}

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  5. #5
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    Equation involving square roots.

    Quote Originally Posted by BabyMilo View Post
    Proof  <br />
 \frac{a-1}{2 } \sqrt{\frac{a+1}{a-1 } } +\frac{a+1}{2 } \sqrt{\frac{a-1}{a+1 } } = \sqrt{a^2-1} <br />

    fankyou.
    L.H.S:-
    \frac{a-1}{2} \sqrt{\frac{a+1}{a-1}} + \frac{a+1}{2}\sqrt{\frac{a-1}{a+1}}

    = \sqrt{\frac{(a+1)(a-1)(a-1)}{(a-1)4} } + \sqrt{\frac{(a-1)(a+1)(a+1)}{(a+1)4}}

    = \sqrt{\frac{a^2-1}{4}} + \sqrt{\frac{a^2-1}{4}} = 2 \times \frac{1}{ 2}\sqrt{a^2-1}

     = \sqrt{a^2 - 1}

    L.H.S. = R.H.S

    I hope you have understood. All the best.
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