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Math Help - Square root fractions.....

  1. #16
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    Re: Square root fractions.....

    Let a = 1+x = 1+sqrt(3)/2 and b = 1-x = 1-sqrt(3)/2 ; then expression becomes:

    [a(1 - sqrt(b)) + b(1 + sqrt(a))] / [(1 + sqrt(a))(1 - sqrt(b))]
    = [a(1 - sqrt(b)) + b(1 + sqrt(a))] / [1 + sqrt(a) - sqrt(b) - sqrt(ab)]

    Since sqrt(ab) = 1/2, above simplifies to:
    2[a(1 - sqrt(b)) + b(1 + sqrt(a))] / [1 + 2(sqrt(a) - sqrt(b))]

    Substitute a = 1+sqrt(3)/2 and b = 1-sqrt(3)/2 and you'll get a nice 1 as answer
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  2. #17
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    Re: Square root fractions.....

    Substituting (√3)/2 into √(1x), for x, may not look like it has much potential to be simplified, but it can be simplified quite nicely.

    \sqrt{1\pm\frac{\sqrt{3}\,}{2}\,}=\sqrt{\frac{4\pm  2\sqrt{3}\,}{4}\,}
    =\frac{\sqrt{4\pm2\sqrt{3}\,}}{2}}

    =\frac{\sqrt{3+1\pm2\sqrt{3}\,}}{2}}

    =\frac{\sqrt{(\sqrt{3})^2\pm2\sqrt{3}\,+1}}{2}}

    =\frac{\sqrt{\left(\sqrt{3}\pm1\right)^2}}{2}}

    =\frac{\sqrt{3}}{2}\pm1
    This should make plugging in \frac{\sqrt{3}}{2} for x much easier.
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  3. #18
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    Re: Square root fractions.....

    Quote Originally Posted by annam911 View Post
    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!
    Are you sure that the answer is 1??
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  4. #19
    MHF Contributor Siron's Avatar
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    Re: Square root fractions.....

    Yes, you can check it with wolphram alpha if you want.
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