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Math Help - really hard fraction question HELP

  1. #1
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    really hard fraction question HELP

     \frac{8}{2 - \sqrt{2}}

    is this right,  \frac{8}{2 - \sqrt{2}} X \frac{2 + \sqrt{2}}{2 + \sqrt{2}} = 16+  8\sqrt{2} , as the numerator and 6 as the dinominator?
    Last edited by andyboy179; December 18th 2009 at 07:15 AM.
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  2. #2
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    Quote Originally Posted by andyboy179 View Post
     \frac{8}{2 - \sqrt{2}}

    is this right,  \frac{8}{2 - \sqrt{2}} \times \frac{2 + \sqrt{2}}{2 + \sqrt{2}} =  \frac{16}+ 3\sqrt{2}}
    = \frac {16 + 8\sqrt{2}}{2^2 - (\sqrt{2})^2} = . . .

    multiply it term by term and then simplify
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  3. #3
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    just did it but i can't work out how to write it so it comes up how i want it
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  4. #4
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    is what i put for the answer right? 16+, as the numerator and 6 as the denominator?
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  5. #5
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    Quote Originally Posted by andyboy179 View Post
    is what i put for the answer right? 16+, as the numerator and 6 as the denominator?
    the denominator is 2^2 - (\sqrt{2})^2 = . . .
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  6. #6
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    Quote Originally Posted by dedust View Post
    the denominator is 2^2 - (\sqrt{2})^2 = . . .

    how? could you please explain it?
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  7. #7
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    Quote Originally Posted by andyboy179 View Post
    how? could you please explain it?
    \frac{8}{2 - \sqrt{2}} \times \frac{2 + \sqrt{2}}{2 + \sqrt{2}} = \frac{8(2 + \sqrt{2})}{(2 - \sqrt{2})(2 + \sqrt{2})}
    = \frac{(16 + 8\sqrt{2})}{(2^2 + 2\sqrt{2} - 2\sqrt{2} + \sqrt{2}^2)} = \frac{(16 + 8\sqrt{2})}{(4-2)} = \frac{(16 + 8\sqrt{2})}{2} = 8 + 4\sqrt{2}
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  8. #8
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    i get told to do this at school for the denominator,

    really hard fraction question HELP-222221.jpg
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  9. #9
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    Quote Originally Posted by andyboy179 View Post
    i get told to do this at school for the denominator,

    Click image for larger version. 

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    the last entry in the 2nd column should be -2, because
    (-\sqrt{2}) \times (+\sqrt{2}) = -{2}
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  10. #10
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    thankyou so much
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