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Math Help - Important question plzzzzzzz help

  1. #1
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    Important question plzzzzzzz help

    If a = (3^1/2 + 2^1/2)^-3 and b = ((3^1/2 - 2^1/2)^-3 then the value of (a+1)^-1 + ( b+1)^-1 is

    a)48 root 2

    b)50 root 3

    c) 1

    d) 5



    PLZZZZZZZZZZZZZZZZZZ EXPLAIN
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  2. #2
    MHF Contributor ebaines's Avatar
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    Re: Important question plzzzzzzz help

    Not sure if there's an easier way to do this, but if  a = \frac 1 {(\sqrt 3 + \sqrt 2)^3} then multiplying it out yields

     a = \frac 1 {9 \sqrt 3 + 11 \sqrt 2}

    Similarly

     b = \frac 1 {9 \sqrt 3 - 11 \sqrt 2}

    For ease let's use notation  x = 9 \sqrt 3 and  y = 11 \sqrt 3, so that:

     a = \frac 1 {x + y}; \ b = \frac 1 {x-y}

    Now:

    \frac 1 {a+ 1} + \frac 1 {b+1} = \frac 1 {\frac 1 {x+y} + 1} + \frac 1 {\frac 1 {x-y} + 1} = \frac {x+y}{1 + x + y} + \frac {x-y}{ 1 + x -y}

    This gets a little messy, but it boils down to:

     \frac {2x + 2x^2 -2y^2}{1 + 2x +x^2 - y^2}

    To finish you just need to sub in the values for x and y and see what you get.
    Thanks from topsquark
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  3. #3
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    Re: Important question plzzzzzzz help

    Hello, sscgeek!

    a \:=\: (\sqrt{3}+\sqrt{2})^{-3},\;\;b \:=\:(\sqrt{3}- \sqrt{2})^{-3}

    \text{Find: }\:\frac{1}{a+1} + \frac{1}{b+1}

    . . . (a) \;48\sqrt{2} \qquad (b)\;50\sqrt{3} \qquad (c)\;1 \qquad (d)\;5

    \frac{1}{a+1} \;=\;\frac{1}{(\sqrt{3}+\sqrt{2})^{-3} + 1} \;=\;\frac{(\sqrt{3}+\sqrt{2})^3}{1 + (\sqrt{3}+\sqrt{2})^3}

    \frac{1}{b+1} \;=\;\frac{1}{(\sqrt{3}-\sqrt{2})^{-3} + 1} \;=\;\frac{(\sqrt{3}-\sqrt{2})^3}{1 + (\sqrt{3}-\sqrt{2})^3}


    \frac{1}{a+1} + \frac{1}{b+1} \;=\; \frac{(\sqrt{3}+\sqrt{2})^3}{1 + (\sqrt{3}+\sqrt{2})^3} + \frac{(\sqrt{3}-\sqrt{2})^3}{1 + (\sqrt{3}-\sqrt{2})^3}

    . . . . . . . . . . =\;\frac{(\sqrt{3}+\sqrt{2})^3\left[1 + (\sqrt{3}-\sqrt{2})^3\right] + (\sqrt{3}-\sqrt{2})^3\left[1 (\sqrt{3}+\sqrt{2})^3\right]} {\left[1+(\sqrt{3}+\sqrt{2})^3\right]\left[1+(\sqrt{3}-\sqrt{2})^3\right]}

    . . . . . . . . . . =\;\frac{(\sqrt{3}+\sqrt{2})^3 + (\sqrt{3}+\sqrt{2})^3(\sqrt{3}-\sqrt{2})^3 + (\sqrt{3}-\sqrt{2})^3 + (\sqrt{3}-\sqrt{2})^3(\sqrt{3}+\sqrt{2})^3}{1 + (\sqrt{3}-\sqrt{2})^3 + (\sqrt{3}+\sqrt{2})^3 + (\sqrt{3}+\sqrt{2})^3(\sqrt{3}-\sqrt{2})^3}

    . . . . . . . . . . =\;\frac{(\sqrt{3}+\sqrt{2})^3 + (3-2)^3 + (\sqrt{3}-\sqrt{2})^3 + (3-2)^3}{1 + (\sqrt{3}-\sqrt{2})^3 + (\sqrt{3}+\sqrt{2})^3 + (3-2)^3}

    . . . . . . . . . . =\;\frac{(\sqrt{3}+\sqrt{2})^3 + 1 + (\sqrt{3}-\sqrt{2})^3 + 1}{1 + (\sqrt{3}-\sqrt{2})^3 + (\sqrt{3}+\sqrt{2})^3 + 1}

    . . . . . . . . . . =\;\frac{(\sqrt{3}+\sqrt{2})^3 + (\sqrt{3}-\sqrt{2})^3 + 2}{(\sqrt{3}+\sqrt{2})^3 + (\sqrt{3}-\sqrt{2})^3 + 2}

    . . . . . . . . . . =\; 1\;\text{ . . . answer (c)}
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  4. #4
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    Re: Important question plzzzzzzz help

    (a+1)^-1 + ( b+1)^-1 = (a + b + 2) / (a + b + ab +1)
    and ab=1
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  5. #5
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    Re: Important question plzzzzzzz help

    Quote Originally Posted by ebaines View Post
    Not sure if there's an easier way to do this, but if  a = \frac 1 {(\sqrt 3 + \sqrt 2)^3} then multiplying it out yields

     a = \frac 1 {9 \sqrt 3 + 11 \sqrt 2}

    Similarly

     b = \frac 1 {9 \sqrt 3 - 11 \sqrt 2}

    For ease let's use notation  x = 9 \sqrt 3 and  y = 11 \sqrt 3, so that:

     a = \frac 1 {x + y}; \ b = \frac 1 {x-y}

    Now:

    \frac 1 {a+ 1} + \frac 1 {b+1} = \frac 1 {\frac 1 {x+y} + 1} + \frac 1 {\frac 1 {x-y} + 1} = \frac {x+y}{1 + x + y} + \frac {x-y}{ 1 + x -y}

    This gets a little messy, but it boils down to:

     \frac {2x + 2x^2 -2y^2}{1 + 2x +x^2 - y^2}

    To finish you just need to sub in the values for x and y and see what you get.
    Important question plzzzzzzz help-14-jun-13.png
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  6. #6
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    Re: Important question plzzzzzzz help

    HI BUDDY .CAN U PLZ EXPLAIN HW ab becomes 1
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  7. #7
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    Re: Important question plzzzzzzz help

    \left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)=1
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  8. #8
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    Re: Important question plzzzzzzz help

    thanks buddy
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  9. #9
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    Re: Important question plzzzzzzz help

    Remember (a+b)(a-b)=a^2-b^2
    Just apply this formula and you will get the result. for we have ab = (9sqrt3)^2-(11sqrt2)^2= 81*3 -121*2=243-242=1
    Ok
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