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Math Help - A couple of problems, fract's, squares & potencies

  1. #1
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    A couple of problems, fract's, squares & potencies

    Hello everybody

    I am rehearsing my skills with fractions, square roots and potencies. I have got a couple of problems and simplifications I'd like you to check and see if I am correct.

    (first try with Latex...)

    \frac{2(frac{4}{3}

    Ok another go...

    Problem 1
    \frac{2(-(4\div3)\times5\div2)}{-(8\div3)}

    I hope you can make out what I try to put in Latex language:
    http://www.imageupload.se/files/070906/8778problem%201.JPG

    Note: Space in url, copy and paste recommended.

    Problem 2

    \frac{5\div6}{-(1\div6)} (I can't solve this one)

    Problem 3

    \frac{2x+3}{4x^2-9} (ditto)

    Problem 4 (another simplification)

     4\sqrt12 + 3\sqrt 18 - \sqrt 50 - 2\sqrt 48

    Problem 5

    \sqrt\frac{2}{3}(\sqrt6-\sqrt3)

    All replies are welcome

    EDIT: NOTE. Problem 4 is modified, I wrote it incorrectly

    P.S.
    I'll add my solutions later, writing these problems took enough time.
    Last edited by λιεҗąиđ€ŗ; September 7th 2007 at 12:58 AM. Reason: Correction
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by λιεҗąиđ€ŗ View Post
    Problem 1
    \frac{2(-(4\div3)\times5\div2)}{-(8\div3)}

    Problem 2

    \frac{5\div6}{-(1\div6)} (I can't solve this one)
    1)
    \frac{2(-(4\div3)\times5\div2)}{-(8\div3)} = 2 \left ( \frac{ -\frac{4}{3}\cdot \frac{5}{2}}{ -\frac{8}{3} } \right )

    = 2 \left ( \frac{ -\frac{20}{6} }{ -\frac{8}{3} } \right )

    This is called either a complex or compound fraction, I forget which. The key is that we want to multiply the numerator and denominator of the "larger" fraction by something that will clear out the "lesser" fractions. In this case we want the LCM of 6 and 3, which is 6. So:
    = 2 \left ( \frac{ -\frac{20}{6} }{ -\frac{8}{3} } \cdot \frac{6}{6} \right )

    = 2 \left ( \frac{ -\frac{20}{6} \cdot 6 }{ -\frac{8}{3} \cdot 6} \right )

    = 2 \left ( \frac{-20}{-16} \right )

    = 2 \left ( \frac{5}{4} \right )

     = \frac{10}{4} = \frac{5}{2}

    2)
    \frac{5\div6}{-(1\div6)} = \frac{ \frac{5}{6} }{ -\frac{1}{6} }

    Again we want to multiply by the number that removes the fractions from the numerator and denominator. Again, in this case, we multiply by 6:
    = \frac{ \frac{5}{6} }{ -\frac{1}{6} } \cdot \frac{6}{6}

    = \frac{5}{-1}

    = -5

    -Dan
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by λιεҗąиđ€ŗ View Post
    Problem 3

    \frac{2x+3}{4x^2-9} (ditto)
    Note that a^2 - b^2 = (a + b)(a - b)

    So the denominator is
    4x^2 - 9 = (2x)^2 - 3^2 = (2x + 3)(2x - 3)

    Thus
    \frac{2x+3}{4x^2-9} = \frac{2x + 3}{(2x + 3)(2x - 3)}

    Now, we can cancel the common factor of 2x + 3 in the numerator and denominator, but only if 2x + 3 is not 0: we must have x \neq -\frac{3}{2}.

    So:
    = \frac{1}{2x - 3};~x \neq -\frac{3}{2}

    -Dan
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by λιεҗąиđ€ŗ View Post
    Problem 4

    \sqrt [4]12 + \sqrt [3]18 - \sqrt 50 - \sqrt [2]48
    The trick here is to write everything under the radical in terms of the exponent given, as much as possible. For example:
    \sqrt[2]{48} = \sqrt[2]{3 \cdot 16} = \sqrt[2]{3 \cdot 4^2} = \sqrt[2]{3} \cdot \sqrt[2]{4^2} = \sqrt[2]{3} \cdot 4 = 4\sqrt[2]{3}

    By the way \sqrt[2]{x} = \sqrt{x}.

    So by the same process we can get \sqrt{50} = 5\sqrt{2}.

    What I see as being slightly confusing is that you can do nothing to either \sqrt[4]{12} \text{ or }\sqrt[3]{18} to simplify them.

    -Dan
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by λιεҗąиđ€ŗ View Post
    Problem 5

    \sqrt\frac{2}{3}(\sqrt6-\sqrt3)
    \sqrt\frac{2}{3}(\sqrt6-\sqrt3) = \sqrt{\frac{2}{3}} \cdot \sqrt{6} - \sqrt{\frac{2}{3}} \cdot \sqrt{3}

    = \sqrt{\frac{2}{3} \cdot 6} - \sqrt{\frac{2}{3} \cdot 3}

    = \sqrt{\frac{12}{3}} - \sqrt{\frac{6}{3}}

    = \sqrt{4} - \sqrt{2}

    = 2 - \sqrt{2}

    -Dan
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  6. #6
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    Feedback, thanks topsquark

    Thank you for the help, I appreciate it very much.

    @ topsquark: I misinterpreted problem 4. The square roots are all simple, with various factors.

    Concerning problem 1, is 2 \left ( \frac{ -\frac{4}{3}\cdot \frac{5}{2}}{ -\frac{8}{3} } \right ) correct? The denominator shouldn't be multiplied by 2.
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by λιεҗąиđ€ŗ View Post
    Concerning problem 1, is 2 \left ( \frac{ -\frac{4}{3}\cdot \frac{5}{2}}{ -\frac{8}{3} } \right ) correct? The denominator shouldn't be multiplied by 2.
    Yes, it is correct. (Assuming there were no typos in the original.) What do you mean by "The denominator shouldn't be multiplied by 2."? Which 2 are you referring to? The one on the outside of the expression or the one under the 5?

    -Dan
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  8. #8
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by λιεҗąиđ€ŗ View Post
    Problem 4 (another simplification)

     4\sqrt12 + 3\sqrt 18 - \sqrt 50 - 2\sqrt 48
    Okay, another go at it!

     4\sqrt{12} + 3\sqrt{18} - \sqrt{50} - 2\sqrt{48}

    = 4 \sqrt{3 \cdot 4} + 3 \sqrt{2 \cdot 9} - \sqrt{2 \cdot 25} - 2 \sqrt{3 \cdot 16}

    = 4 \sqrt{3 \cdot 2^2} + 3 \sqrt{2 \cdot 3^2} - \sqrt{2 \cdot 5^2} - 2 \sqrt{3 \cdot 4^2}

    = 4 \sqrt{3} \cdot 2 + 3 \sqrt{2} \cdot 3 - \sqrt{2} \cdot 5 - 2 \sqrt{3} \cdot 4

    = 8 \sqrt{3} + 9 \sqrt{2} - 5 \sqrt{2} - 8 \sqrt{3}

    = (8 - 8) \sqrt{3} + (9 - 5) \sqrt{2}

    = 4 \sqrt{2}

    -Dan
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  9. #9
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    On Problem 1

    I mean the big denominator, -\frac {8}{3}, shouldn't be multiplied by two.

    Anyway you have been a great help topsquark. *Thumbs up*

    P.S. Have a look at the problem IRL, copy and paste this link.
    http://www.imageupload.se/files/070906/8778problem%201.JPG
    Last edited by λιεҗąиđ€ŗ; September 7th 2007 at 10:33 AM. Reason: P.S.
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  10. #10
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by λιεҗąиđ€ŗ View Post
    I mean the big denominator, -\frac {8}{3}, shouldn't be multiplied by two.

    Anyway you have been a great help topsquark. *Thumbs up*

    P.S. Have a look at the problem IRL, copy and paste this link.
    http://www.imageupload.se/files/070906/8778problem%201.JPG
    You are welcome.

    Um, I can't get the link to work...

    Anyway, I'm wondering if you might be misunderstanding something that's going on. Let me break this down a bit:
    2 \left ( \frac{ -\frac{4}{3}\cdot \frac{5}{2}}{ -\frac{8}{3} } \right ) = \frac{2}{1} \cdot \left ( \frac{ -\frac{4}{3}\cdot \frac{5}{2}}{ -\frac{8}{3} } \right )
    so only the numerator is being multiplied by the 2.

    However there is another 2 that shows up later on when I'm clearing the fractions of the fractions and maybe this is where you are seeing that extra 2:

    2 \left ( \frac{ -\frac{4}{3}\cdot \frac{5}{2}}{ -\frac{8}{3} } \cdot \frac{6}{6} \right )

    = 2 \left ( \frac{ -\frac{4}{3}\cdot \frac{5}{2} \cdot 6}{ -\frac{8}{3} \cdot 6 }\right )

    Now the denominator -\frac{8}{3} \cdot 6 = -8 \cdot 2
    is this perhaps why you thought I was multiplying the denominator by 2 at some point?

    -Dan
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  11. #11
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    Quote Originally Posted by topsquark View Post
    \sqrt\frac{2}{3}(\sqrt6-\sqrt3) = \sqrt{\frac{2}{3}} \cdot \sqrt{6} - \sqrt{\frac{2}{3}} \cdot \sqrt{3}

    = \sqrt{\frac{2}{3} \cdot 6} - \sqrt{\frac{2}{3} \cdot 3}

    = \sqrt{\frac{12}{3}} - \sqrt{\frac{6}{3}}

    = \sqrt{4} - \sqrt{2}

    = 2 - \sqrt{2}
    Another attempt

    \sqrt {\frac{2}<br />
{3}} \left( {\sqrt 6 - \sqrt 3 } \right) = \frac{{\sqrt 2 }}<br />
{{\sqrt 3 }}\left( {\sqrt 2 \cdot \sqrt 3 - \sqrt 3 } \right) = 2 - \sqrt 2

    Cheers,
    K.
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