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Thread: 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 $\displaystyle AB = \sqrt{(\sqrt5+2)(\sqrt5-2)}=\sqrt{5-4}=1$

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

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

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

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

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

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

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

    This then simplifies to

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

    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; Jul 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 $\displaystyle x \:=\:A + B$...
    But we are asked to find $\displaystyle \sqrt{A} + \sqrt{B}$

    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 $\displaystyle AB = \sqrt{(\sqrt5+2)(\sqrt5-2)}=\sqrt{5-4}=1$

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

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

    Quote Originally Posted by Grandad View Post

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

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

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

    and the positive root of $\displaystyle X^2-4X^2-16$ is $\displaystyle \frac{4+\sqrt{16+4\times 16}}{2}=2+2\sqrt{5}$ so we get $\displaystyle x=+\sqrt{2+2\sqrt{5}}$ 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
    $\displaystyle x=+\sqrt{A+2+B}=\sqrt{\sqrt{5}+2+\sqrt{5}-2}=\sqrt{2+2\sqrt{5}},$
    there is no need for the quadratic formula.

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

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

    $\displaystyle \ne 2\sqrt5 +2$

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

    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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