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Math Help - Square root fractions.....

  1. #1
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    Square root fractions.....

    Hi guys, I've been stuck on this question for days!
    If x= √3/2
    Find the value of the expression
    1+x/1+√(1+x) + 1-x/1-√(1-x)
    Thanks!
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  2. #2
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    Re: Square root fractions.....

    Just to confirm, its x = \sqrt{\frac{3}{2}}

    And \displaystyle \frac{1+x}{1+\sqrt{1+x}}+\frac{1-x}{1-\sqrt{1-x}}

    ?
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  3. #3
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    Re: Square root fractions.....

    Yes
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  4. #4
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    Re: Square root fractions.....

    Actually no sorry x = √(3)/2
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  5. #5
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    Re: Square root fractions.....

    Maybe start by rationalising the denominator

    \displaystyle \frac{1+x}{1+\sqrt{1+x}}+\frac{1-x}{1-\sqrt{1-x}}

    \displaystyle \frac{1+x}{1+\sqrt{1+x}}\times \frac{1-\sqrt{1-x}}{1-\sqrt{1-x}}+\frac{1-x}{1-\sqrt{1-x}}\times \frac{1+\sqrt{1+x}}{1+\sqrt{1+x}}

    What do you get?
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  6. #6
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    Re: Square root fractions.....

    I've tried everything! The answer is ment to be 1, maybe ive gone wrong in my calculation somewhere
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  7. #7
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    Re: Square root fractions.....

    Quote Originally Posted by annam911 View Post
    Actually no sorry x = √(3)/2
    That's O.K, still follow the advice in post #2.
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  8. #8
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    Re: Square root fractions.....

    Quote Originally Posted by annam911
    The answer is ment to be 1, maybe ive gone wrong in my calculation somewhere
    Show me your attempts. Maybe I can pick up a mistake.
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  9. #9
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    Re: Square root fractions.....

    following on from what u had....
    (1+x)(1-√1-x)/)1+√1+x)(1-√1-x) + (1-x)(1+√1+x)/(1-√1-x)1+√1+x)

    = (1+x)(1-√1-x) + (1-x)(1+√1+x)/1+√1+x)(1-√1-x)

    =2-√1-x - x√1-x + √1+x -x1+x/(1+√1+x)(1-√1-x)

    =2-√1-x(1+x) + √1+x(1-x)/(1+√1+x)(1-√1-x)

    Thats as far as i could get :s
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  10. #10
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    Re: Square root fractions.....

    In the denominator apply the difference of two squares. (a+b)(a-b) = a^2-b^2
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  11. #11
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    Re: Square root fractions.....

    Quote Originally Posted by pickslides View Post
    Maybe start by rationalising the denominator

    \displaystyle \frac{1+x}{1+\sqrt{1+x}}+\frac{1-x}{1-\sqrt{1-x}}

    \displaystyle \frac{1+x}{1+\sqrt{1+x}}\times \frac{1-\sqrt{1-x}}{1-\sqrt{1-x}}+\frac{1-x}{1-\sqrt{1-x}}\times \frac{1+\sqrt{1+x}}{1+\sqrt{1+x}}

    What do you get?
    Opps, I meant to say

    \displaystyle \frac{1+x}{1+\sqrt{1+x}}\times \frac{1-\sqrt{1+x}}{1-\sqrt{1+x}}+\frac{1-x}{1-\sqrt{1-x}}\times \frac{1+\sqrt{1-x}}{1+\sqrt{1-x}}

    go from there,
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  12. #12
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    Re: Square root fractions.....

    would you be able to solve it and ill work from there haha im losing my patience with it
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  13. #13
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  14. #14
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    Re: Square root fractions.....

    hmmm i dont understand how that working out results in 1
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  15. #15
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    Re: Square root fractions.....

    You'll get the most simplified form of it by using the hint pickslides has given you:
    \frac{1+x}{1+\sqrt{1+x}}+\frac{1-x}{1-\sqrt{1-x}}=\frac{(1+x)(1-\sqrt{1-x})+(1-x)(1+\sqrt{1+x})}{(1+\sqrt{1+x})(1-\sqrt{1-x})}
    =\frac{1+x-\sqrt{1-x}-x\sqrt{1-x}+1-x+\sqrt{1+x}-x\sqrt{1+x}}{(1+\sqrt{1x})(1-\sqrt{1-x})}
    =\frac{2-\sqrt{1-x}-x\sqrt{1-x}+\sqrt{1+x}-x\sqrt{1+x}}{(1+\sqrt{1x})(1-\sqrt{1-x})}

    Now you just have to substitute the given x value in this expression which is not that easy.
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