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Math Help - Recheck please

  1. #1
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    Recheck please

    I've done math problems on some paper, and there are no solutions given.
    1. Simplify \sqrt{3-\sqrt5}-\sqrt{3+\sqrt5}
    (My) solution: expression^2
    (3-\sqrt5)+(3+\sqrt5)-2\sqrt{(3-\sqrt5)(3+\sqrt5)}
    6-2\sqrt{4}
    6-4=2
    Two solutions?
    -\sqrt2
    \sqrt2
    Now, what if you've "gained" some (and which?) solution by squaring the original?

    2. Simplify
    a) \frac{1}{2}*\sqrt2*\sqrt[3]{4}:\sqrt[4]{8}\Rightarrow \frac{2^{\frac{1}{2}}*2^{\frac{2}{3}}}{2*2^{\frac{  3}{4}}}\Rightarrow 2^{-\frac{7}{12}}\Rightarrow {\frac{1}{\sqrt[12]{2^7}}}
    b) (For what a\in\mathbb{R} does the following expression exist?) a^{-1}*(1+a^{-2})^{-\frac{1}{2}}*(1+a^2)^{\frac{1}{2}} \Rightarrow \frac{(1+a^2)^{\frac{1}{2}}}{a*(1+a^{-2})^{\frac{1}{2}}} \Rightarrow \frac{(1+a^2)^{\frac{1}{2}}}{(a^2+1)^{\frac{1}{2}}  } = 1(If the solution is correct) what is now the answer (on "expression exists for what a?"), every a?

    3. Solve
    x+2\sqrt{1-x}=1 \Rightarrow expression^2 \Rightarrow 4*(1-x)=1+x^2-2x \Rightarrow x_1=-3, x_2=1
    Again, what if you've "gained" some solution by squaring the original? What are considered best practises in solving such?
    Last edited by courteous; September 18th 2008 at 05:07 AM. Reason: reformulating the 2. Q
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  2. #2
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    Quote Originally Posted by courteous View Post
    I've done math problems on some paper, and there are no solutions given.
    1. Simplify \sqrt{3-\sqrt5}-\sqrt{3+\sqrt5}
    (My) solution: expression^2
    (3-\sqrt5)+(3+\sqrt5)-2\sqrt{(3-\sqrt5)(3+\sqrt5)}
    6-2\sqrt{4}
    6-4=2
    Two solutions?
    -\sqrt2
    \sqrt2
    Now, what if you've "gained" some (and which?) solution by squaring the original?

    Mr F says: You have to go back to the original expression and test each potential solution. Clearly {\color{red}\sqrt{3-\sqrt5}-\sqrt{3+\sqrt5} < 0} therefore discard the positive solution.

    [snip]

    3. Solve
    x+2\sqrt{1-x}=1 \Rightarrow expression^2 \Rightarrow 4*(1-x)=1+x^2-2x \Rightarrow x_1=-3, x_2=1
    Again, what if you've "gained" some solution by squaring the original? What are considered best practises in solving such?

    Mr F says: You have to test each potential solution in the original equation. Both solutions satisfy the original equation so you keep both.
    ..
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