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Math Help - Equivalent Forms

  1. #1
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    Equivalent Forms

    <br />
^3\sqrt{2}   (\sqrt{3)<br />
is equal to?


    an equivalent form of  (2\sqrt{2} - \sqrt{6})^2 is ?



    (This is also an excuse for me to get used to the php math code, which I never used before)
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by zoso
    <br />
^3\sqrt{2}   (\sqrt{3)<br />
is equal to?


    an equivalent form of  (2\sqrt{2} - \sqrt{6})^2 is ?



    (This is also an excuse for me to get used to the php math code, which I never used before)
    I see no way to simplify the first expression. Note: A slightly better way to code the cube root of 2 is: \sqrt[3]{2} (Click to see the code.)

    \left ( 2 \sqrt{2} - \sqrt{6} \right ) ^2 = \left ( 2 \sqrt{2} \right ) ^2 - 2 \cdot \left ( 2 \sqrt{2} \right ) \left ( \sqrt{6} \right ) + \left ( - \sqrt{6} \right ) ^2

     = 4 \cdot 2 - 2 \cdot 2 \sqrt{12} + 6 = 14 - 4 \sqrt{12}

     = 14 - 4 \cdot 2 \sqrt{3} = 14 - 8 \sqrt{3}

    -Dan
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  3. #3
    Super Member malaygoel's Avatar
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    Quote Originally Posted by zoso
    <br />
^3\sqrt{2}   (\sqrt{3)<br />
is equal to?


    an equivalent form of  (2\sqrt{2} - \sqrt{6})^2 is ?



    (This is also an excuse for me to get used to the php math code, which I never used before)
    \sqrt[3]{2}\sqrt{3}=\sqrt[6]{4}\sqrt[6]{27}=\sqrt[6]{108}

    Malay
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  4. #4
    MHF Contributor Quick's Avatar
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    Quote Originally Posted by zoso
    (This is also an excuse for me to get used to the php math code, which I never used before)
    Here's a tutorial if you haven't seen it yet.
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  5. #5
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    k, I'm going to go read the tutorial, then come back and try something else out.
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  6. #6
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    ok, lets try posting with fractions

    Simplfy \frac {1} {x + 1} - \frac {3} {x - 1} assuming x  \neq\pm 1


    \frac {1} {j + 20} - \frac {1} {j} = 400

    \frac {1} {j} - \frac {1} {j + 20} = 1

    \frac {400} {j + 20} - \frac {400} {j} = 1

    \frac {400} {j} - \frac {400} {j + 20} = 1

    M drives 400 in one hour less than J. If M's speed is 20 km/h faster than J's, what is the proper equation to determine the speed of J?
    Last edited by zoso; July 20th 2006 at 10:14 AM.
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  7. #7
    MHF Contributor Quick's Avatar
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    Quote Originally Posted by zoso
    ok, lets try posting with fractions

    Simplfy \frac {1} {x + 1} - \frac {3} {x - 1} assuming x = \pm 1 (What is the tag for non-equivalents? that should read "assuming, x is not equal to..")
    x \neq \pm 1
    or
    x \not =\pm 1

    something interesting:

    \frac{8}{4}\quad\Rightarrow\quad\frac{\not2\times \not2\times 2}{\not2\times \not2}\quad\Rightarrow\quad2
    Last edited by Quick; July 20th 2006 at 10:22 AM.
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  8. #8
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    Quote Originally Posted by zoso
    ok, lets try posting with fractions

    Simplfy \frac {1} {x + 1} - \frac {3} {x - 1} assuming x  \neq\pm 1


    \frac {1} {j + 20} - \frac {1} {j} = 400

    \frac {1} {j} - \frac {1} {j + 20} = 1

    \frac {400} {j + 20} - \frac {400} {j} = 1

    \frac {400} {j} - \frac {400} {j + 20} = 1

    M drives 400 in one hour less than J. If M's speed is 20 km/h faster than J's, what is the proper equation to determine the speed of J?
    Anyone going to work through these two, quickly?
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  9. #9
    MHF Contributor Quick's Avatar
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    Quote Originally Posted by zoso
    ok, lets try posting with fractions

    Simplfy \frac {1} {x + 1} - \frac {3} {x - 1} assuming x  \neq\pm 1
    find the common denominator...
    \frac {1(x-1)} {(x + 1)(x-1)} - \frac {3(x+1)} {(x - 1)(x+1)}

    multiply: \frac {x-1} {x^2 + 1} - \frac {3x+3} {x^2 + 1}

    subtract: \frac {x-1-3x-3} {x^2 + 1}

    group like terms: \frac {x-3x-1-3} {x^2 + 1}

    subtract: \frac {\neg 2x-4} {x^2 + 1}
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  10. #10
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    So,  \frac {-2(x + 2)} {x^2 - 1} ?
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  11. #11
    MHF Contributor Quick's Avatar
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    Quote Originally Posted by zoso
    So,  \frac {-2(x + 2)} {x^2 - 1} ?
    yes, I'm glad you realized that, it should be written as,  \neg2\frac {x + 2} {x^2 + 1} though
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  12. #12
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Quick
    yes, I'm glad you realized that, it should be written as,  \neg2\frac {x + 2} {x^2 + 1} though
     -2 \frac {x + 2} {x^2 + 1} =  - \frac {2(x + 2)} {x^2 + 1} =  \frac {-2(x + 2)} {x^2 + 1}

    I don't see why any one form should be more "correct" than any other. As far as I know there is no consensus on any particular form.

    Also, Quick: Why do you keep using " \neg" as opposed to simply "-"?

    -Dan
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  13. #13
    MHF Contributor Quick's Avatar
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    Quote Originally Posted by topsquark
     -2 \frac {x + 2} {x^2 + 1} =  - \frac {2(x + 2)} {x^2 + 1} =  \frac {-2(x + 2)} {x^2 + 1}

    I don't see why any one form should be more "correct" than any other. As far as I know there is no consensus on any particular form.

    Also, Quick: Why do you keep using " \neg" as opposed to simply "-"?

    -Dan
    I think they are looking for simplest form...

    BTW: I am starting to use the \neg for negative numbers, particularly fractions.
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  14. #14
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Quick
    I think they are looking for simplest form...

    BTW: I am starting to use the \neg for negative numbers, particularly fractions.
    But why use it? As far as I know, "-" is standard for at least complex numbers.

    -Dan
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