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Math Help - surds

  1. #1
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    surds

    I am having difficulty solving these surds problems

    (5+2 \sqrt{7})^2

    and

    \sqrt{8}+\sqrt{18}-\sqrt{50}
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  2. #2
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    (5+2\sqrt7)^2=25+28+20\sqrt{7}
    =53+20\sqrt{7}

    \sqrt{8}+\sqrt{18}-\sqrt{50}=2\sqrt{2}+3\sqrt{2}-5\sqrt{2}=0
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  3. #3
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    Quote Originally Posted by b3nj1d View Post
    I am having difficulty solving these surds problems

    (5+2 \sqrt{7})^2

    and

    \sqrt{8}+\sqrt{18}-\sqrt{50}
    Since we haven't learnt FOIL yet, I understand why you're having trouble with the first one.

    If you have two binomials like (a + b)(c + d), they are expanded by multiplying the Firsts, then the Outers, then the Inners, then the Lasts.

    So (a + b)(c + d) = ac + ad + bc + bd.


    So for your first one, you have

    (5 + 2\sqrt{7})^2 = (5 + 2\sqrt{7})(5 + 2\sqrt{7})

    Multiplying the firsts gives 5 \times 5 = 25.

    Multiplying the outers gives 5 \times 2\sqrt{7} = 10\sqrt{7}

    Multiplying the inners gives 5 \times 2\sqrt{7} = 10\sqrt{7}

    Multiplying the lasts gives 2\sqrt{7}\times 2\sqrt{7} = 4 \times 7 = 28.


    So (5 + 2\sqrt{7})^2 = 25 + 10\sqrt{7} + 10\sqrt{7} + 28

     = 53 + 20\sqrt{7}.



    For the second, you need to simplify all of the surds first. Then hopefully you'll have "like surds" that can be collected.

    \sqrt{8} = \sqrt{4 \times 2} = \sqrt{4} \times \sqrt{2} = 2\sqrt{2}.


    \sqrt{18} = \sqrt{9 \times 2} = \sqrt{9}\times\sqrt{2} = 3\sqrt{2}.


    \sqrt{50} = \sqrt{25 \times 2} = \sqrt{25}\times\sqrt{2} = 5\sqrt{2}.



    So \sqrt{8} + \sqrt{18} - \sqrt{50} = 2\sqrt{2} + 3\sqrt{2} - 5\sqrt{2}

     = 5\sqrt{2} - 5\sqrt{2}

     = 0\sqrt{2}

     = 0.



    Chookas for the exam tomorrow.
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