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Math Help - Radicals calcules

  1. #1
    Super Member dhiab's Avatar
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    Radicals calcules

    You have : ,
    Calculate :
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  2. #2
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    More information please

    Hello dhiab
    Quote Originally Posted by dhiab View Post
    You have : ,
    Calculate :
    B is not a real number. Are you sure you have the question correct?

    Grandad
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  3. #3
    Super Member dhiab's Avatar
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    Quote Originally Posted by Grandad View Post
    Hello dhiabB is not a real number. Are you sure you have the question correct?

    Grandad
    Hello : THANK YOU
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  4. #4
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    Hello dhiab

    I'm not sure what form you want the answer in - I assume you want it as a single radical. How does this look?

    Note first that AB = \sqrt{(\sqrt5+2)(\sqrt5-2)}=\sqrt{5-4}=1

    Let x = \sqrt{A} + \sqrt{B}

    Then x^2 = A + 2\sqrt{AB} + B  = A +2+B

    \Rightarrow (x^2)^2 =(A+B+2)^2= A^2 +2AB + B^2 + 4(A+B) + 4

    Noting that A+B = x^2 - 2, this simplifies to

    (x^2)^2 = 4x^2 +2\sqrt5 - 2

    \Rightarrow (x^2)^2 - 4x^2 -(2\sqrt5-2)=0

    \Rightarrow x^2 = \frac{4\pm\sqrt{16 + 4(2\sqrt5-2)}}{2}, using the quadratic formula

    This then simplifies to

    x=\sqrt{2+\sqrt{2+2\sqrt{5}}}

    Grandad
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  5. #5
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    Hello, everyone!

    Grandad is right . . . I misread the problem . . . *blush*

    As Emily Litella would say: "Never mind."

    Last edited by Soroban; July 10th 2009 at 05:51 AM.
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  6. #6
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    Hello Soroban
    Quote Originally Posted by Soroban View Post
    ...
    Let x \:=\:A + B...
    But we are asked to find \sqrt{A} + \sqrt{B}

    Grandad
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  7. #7
    Super Member flyingsquirrel's Avatar
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    Hello,
    Quote Originally Posted by Grandad View Post

    Note first that AB = \sqrt{(\sqrt5+2)(\sqrt5-2)}=\sqrt{5-4}=1

    Let x = \sqrt{A} + \sqrt{B}

    Then x^2 = A + 2\sqrt{AB} + B  = A +2+B
    ... hence
    x=+\sqrt{A+2+B}=\sqrt{\sqrt{5}+2+\sqrt{5}-2}=\sqrt{2+2\sqrt{5}},
    there is no need for the quadratic formula.

    Quote Originally Posted by Grandad View Post

    \Rightarrow (x^2)^2 =(A+B+2)^2= A^2 +2AB + B^2 + 4(A+B) + 4

    Noting that A+B = x^2 - 2, this simplifies to

    (x^2)^2 = 4x^2 +2\sqrt5 - 2
    I think the last equation is not correct. We have
    \begin{aligned}(x^2)^2&=(A+B+2)^2\\<br />
&=(A+B)^2 + 4(A+B) + 4\\<br />
&=(2\sqrt{5})^2+4(x^2-2)+4\\<br />
&=4x^2+20-8+4\\<br />
(x^2)^2&=4x^2+16\end{aligned}

    and the positive root of X^2-4X^2-16 is \frac{4+\sqrt{16+4\times 16}}{2}=2+2\sqrt{5} so we get  x=+\sqrt{2+2\sqrt{5}} which matches the solution given above.
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  8. #8
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    Not enough square root signs

    Hello everyone

    Sorry flyingsquirrel, but you haven't read the question correctly - you haven't got enough square root signs
    Quote Originally Posted by flyingsquirrel View Post
    Hello,

    ... hence
    x=+\sqrt{A+2+B}=\sqrt{\sqrt{5}+2+\sqrt{5}-2}=\sqrt{2+2\sqrt{5}},
    there is no need for the quadratic formula.

    ...
    A = \sqrt{\sqrt5+2}, B = \sqrt{\sqrt5-2}

     \Rightarrow A+ B+2 = \sqrt{\sqrt5+2}+ \sqrt{\sqrt5-2}+2

    \ne 2\sqrt5 +2

    It is A^2 + B^2 that equals 2\sqrt5, not A + B.

    So my original answer stands.

    Grandad
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  9. #9
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by Grandad View Post

    Sorry flyingsquirrel, but you haven't read the question correctly - you haven't got enough square root signs
    You're right Grandad, I'm sorry.
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