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Math Help - Grade 10 math, radicals (?)

  1. #1
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    Grade 10 math, radicals (?)

    Hello ive worked these out in my notebook but havbe no way to check if they are correct or not here they are

    a) ( √2 - √8) = 2
    b) ( -1/3√3 - 1 )
    c) ( √3 + √12 )

    I know I got to do the perfect square trinomial
    a+ 2ab + b I did that just would like the working out of it and answers in case I got them wrong

    ps , im not skipping on my homewrok just got a test on thurdays and want to see if im ok since i go t no more math classes till then thats why I ask for the work out of it

    THANKSS
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  2. #2
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    Tommy your answer for a) is correct.

    For the other questions

    b) \frac{4}{3} + \frac{2}{3} \sqrt{3}

    c) 27

    You are correct in saying you need to use a perfect square binomial expansion.
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  3. #3
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    can I see how a b and c are done? if its not too much trouble for anyone

    got a right but B and C wrong
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  4. #4
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    question c)

     (\sqrt{3} + \sqrt{12})^2

     = (\sqrt{3} + \sqrt{12}) (\sqrt{3} + \sqrt{12})

    now multiply the terms out using this general form.

     = (a + b) (c + d)

     = ac + ad + bc + bd

    this will given you

     = \sqrt{3} \sqrt{3}+ \sqrt{3} \sqrt{12} + \sqrt{12} \sqrt{3} + \sqrt{12}\sqrt{12}

    Then simplifying gives

     = 3+ 2 \sqrt{3} \sqrt{12} +12

     = 3+ 2 \sqrt{36} +12

     = 3+ 2 .6 +12

     = 3+ 12 +12

     = 27


    Follow similar steps for question b)
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  5. #5
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    Find value of X with radicals

    a) √2 (x - √3) + 5 √6 = - √6 / 2

    b) x+√5 (minus) X- √15 = 0 * both of the terms have a 2 as denominator
    ----- --------
    2 . 2

    c) (x + 2√10) (x - √90) = (cubed root)3√36

    ---

    This is a new unit lets say and i want to see if I can get ahead a little, having the thing worked out in full would be awsome so I have some background as to what our first class after the test will be taught
    Last edited by Tommy.; April 20th 2009 at 04:23 PM.
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  6. #6
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    Find value of X with radicals

    Double post, wasnt a bump im in argentina and these things get messy
    Last edited by Tommy.; April 20th 2009 at 04:17 PM. Reason: damn
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  7. #7
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    Quote Originally Posted by Tommy. View Post
    a) √2 (x - √3) + 5 √6 = - √6 / 2

    b) x+√5 (minus) X- √15 = 0 * both of the terms have a 2 as denominator
    ----- --------
    2 . 2

    c) (x + 2√10) (x - √90) = (cubed root)3√36

    ---

    This is a new unit lets say and i want to see if I can get ahead a little, having the thing worked out in full would be awsome so I have some background as to what our first class after the test will be taught

    so the new problems are:

    a)  \sqrt{2} (x-\sqrt{3}) + 5\sqrt{6} = \frac{-\sqrt{6}}{2}

    b)  \frac{x+\sqrt{5}}{2} - \frac{x-\sqrt{5}}{2} = 0

    c)  (x+2\sqrt{10})(x -\sqrt{90}) = \sqrt[3]{36}
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  8. #8
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    A and C are ok , this is how B should be

    b)  \frac{x+\sqrt{5}}{2} - \frac{x-\sqrt{15}}{2} = 0
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  9. #9
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    Quote Originally Posted by Tommy. View Post
    A and C are ok , this is how B should be

    b)  \frac{x+\sqrt{5}}{2} - \frac{x-\sqrt{15}}{2} = 0

    Are you sure this is correct?

     \frac{x+\sqrt{5}}{2} - \frac{x-\sqrt{15}}{2} = 0

    I would multiply both sides through by 2 giving

    (x+\sqrt{5})-(x-\sqrt{15}) = 0

    then,

    x+\sqrt{5}- x+\sqrt{15} = 0

    And now the xs cancel and gives the following non-sensical statement as

    \sqrt{5} +\sqrt{15} \neq 0
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