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Math Help - division of radicals.........

  1. #1
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    division of radicals.........

    Hi! I'm a canadian student in america.

    Please help me on division of radicals =)

    1.) 6√18 12√40

    2.) 8√19 4√38

    3.) 20√3 5√ 3

    4.) 42√ 6 3√6

    5.) -4√20 √2

    6.) 1 / 2+√5

    7.) 1 / 3-√11

    8.) 3 / √3-1

    9.) 5 / 2-√3

    10.) 2 / 4-√14

    thanks!
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  2. #2
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    Quote Originally Posted by quebec567 View Post
    Hi! I'm a canadian student in america.

    Please help me on division of radicals =)

    1.) 6√18 ÷ 12√40

    2.) 8√19 ÷ 4√38
    1.) 6*sqrt(9)*sqrt(2)/[2*sqrt(10)]

    18*sqrt(2)/[2*sqrt(10)]

    (9*sqrt(2))/(sqrt(10))

    (9*sqrt(5))/5

    2.) (8*sqrt(19))/((4*sqrt(38))

    (2*sqrt(2))/2

    sqrt(2)
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  3. #3
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    Quote Originally Posted by quebec567 View Post
    3.) 20√3 5√ 3

    4.) 42√ 6 3√6
    (20*sqrt(3))/(5*sqrt(3))

    (4*sqrt(3))/(sqrt(3))

    = 4

    4.) (42*sqrt(6))/(3*sqrt(6))

    (14*sqrt(6))/(sqrt(6))

    = 14

    And with that said, I have to get ready for work and such- I'll answer some of the other unanswered ones later.
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  4. #4
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    Quote Originally Posted by quebec567 View Post
    Hi! I'm a canadian student in america.

    Please help me on division of radicals =)

    1.) 6√18 12√40
    2.) 8√19 4√38
    3.) 20√3 5√ 3
    4.) 42√ 6 3√6
    5.) -4√20 √2
    6.) 1 / 2+√5
    7.) 1 / 3-√11
    8.) 3 / √3-1
    9.) 5 / 2-√3
    10.) 2 / 4-√14
    thanks!
    Hello,

    \frac{6\sqrt{18}}{12\sqrt{40}}=\frac{6 \cdot 3 \cdot \sqrt{2}}{12 \cdot 2 \cdot \sqrt{5}}=\frac{3\sqrt{2}}{4\sqrt{5}}=\frac{3\sqrt  {10}}{20}

    \frac{-4\sqrt{20}}{\sqrt{2}}=\frac{-4 \cdot 2 \sqrt{5}}{\sqrt{2}}=\frac{-8 \cdot \sqrt{10}}{2}=-4\sqrt{10}

    From #6 on I presume, that the denominators are sums(?):

    \frac{1}{2+\sqrt{5}} \cdot \frac{2-\sqrt{5}}{2-\sqrt{5}}=\frac{2-\sqrt{5}}{-1}=-2+\sqrt{5}

    \frac{1}{3-\sqrt{11}} \cdot \frac{3+\sqrt{11}}{3+\sqrt{11}}=\frac{3+\sqrt{11}}  {-2}

    \frac{3}{\sqrt{3} - 1} \cdot \frac{\sqrt{3} + 1}{\sqrt{3} + 1}=\frac{3 \sqrt{3} + 3}{2}

    \frac{5}{2-\sqrt{3}} \cdot \frac{2+\sqrt{3}}{2+\sqrt{3}}=\frac{10+5\sqrt{3}}{  1}=10+5\sqrt{3}

    \frac{2}{4-\sqrt{14}} \cdot \frac{4+\sqrt{14}}{4+\sqrt{14}}=\frac{8+2\sqrt{14}  }{2}=4+\sqrt{14}

    EB
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  5. #5
    Junior Member AlvinCY's Avatar
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    Quote Originally Posted by earboth View Post
    Hello,

    \frac{6\sqrt{18}}{12\sqrt{40}}=\frac{6 \cdot 3 \cdot \sqrt{2}}{12 \cdot 2 \cdot \sqrt{5}}=\frac{3\sqrt{2}}{4\sqrt{5}}=\frac{3\sqrt  {10}}{20}
    Don't you mean:

    \frac{6\sqrt{18}}{12\sqrt{40}} = \frac{6 \cdot 3 \cdot \sqrt{2}}{12 \cdot 2 \cdot \sqrt{10}} = \frac{3\sqrt{2}}{4\sqrt{10}} = \frac{3\sqrt{20}}{40} = \frac{3\sqrt{5}}{20}
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  6. #6
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    Quote Originally Posted by AlvinCY View Post
    Don't you mean:

    \frac{6\sqrt{18}}{12\sqrt{40}} = \frac{6 \cdot 3 \cdot \sqrt{2}}{12 \cdot 2 \cdot \sqrt{10}} = \frac{3\sqrt{2}}{4\sqrt{10}} = \frac{3\sqrt{20}}{40} = \frac{3\sqrt{5}}{20}
    Hi,

    of course I meant that - thanks for correcting.

    EB
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