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Math Help - Find simple form of algebraic expression

  1. #1
    Newbie Auger's Avatar
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    Find simple form of algebraic expression

    Question is in the link. Just zoom it in for a clearer view. I was trying to multiply the numerator and the denominator each separate term by the denominators of the other two terms, but it was becoming too complex.
    help is much appreciated!

    http://img19.imageshack.us/img19/5452/69306157.jpg

    Thanks,

    Auger
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  2. #2
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    \frac{ x - 1 }{\sqrt{x} - 1 } - \frac{ x^2 - 2x -3 }{ x + 1 } - \sqrt{x} =

    = \frac{(\sqrt{x} - 1)(\sqrt{x} + 1 )}{\sqrt{x} - 1} - \frac{(x -3)(x + 1 )}{ x + 1 } - \sqrt{x} =

    = \sqrt{x} + 1 - ( x - 3 ) - \sqrt{x}  =  4 - x
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  3. #3
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    Quote Originally Posted by Auger View Post
    Question is in the link. Just zoom it in for a clearer view. I was trying to multiply the numerator and the denominator each separate term by the denominators of the other two terms, but it was becoming too complex.
    help is much appreciated!

    http://img19.imageshack.us/img19/5452/69306157.jpg

    Thanks,

    Auger

    \frac{x-1}{\sqrt{x}-1}-\frac{x^2-2x-3}{x+1}-\sqrt{x}

    Multiply the numerator and denominator of first term by the conjugate of the denominator.
    \frac{(x-1)(\sqrt{x}+1)}{(\sqrt{x}-1)(\sqrt{x}+1)}=\frac{(x-1)(\sqrt{x}+1)}{x-1}=\sqrt{x}+1

    For the next term, we have \frac{x^2-2x-3}{x+1}=\frac{(x+1)(x-3)}{x+1}=x-3
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  4. #4
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    Quote Originally Posted by josipive View Post
    \frac{ x - 1 }{\sqrt{x} - 1 } - \frac{ x^2 - 2x -3 }{ x + 1 } - \sqrt{x} =

    = \frac{(\sqrt{x} - 1)(\sqrt{x} + 1 )}{\sqrt{x} - 1} - \frac{(x -3)(x + 1 )}{ x + 1 } - \sqrt{x} =

    = \sqrt{x} + 1 - ( x - 3 ) - \sqrt{x} = 4 - x
    Beat me by six minutes. lol
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