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Math Help - surds iii

  1. #1
    Member GAdams's Avatar
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    surds iii

    I got 1/6 for (a) but (b) is beyond me. Thanks.
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    1/6 is right.

    For b, you just want to show that the left is equal to the right. We do this by substituting and simplifying as follows:

    (r-p)^2=9-6\sqrt{2}
    (\sqrt{6}-\sqrt{3})^2 \Rightarrow (\sqrt{6})^2-2\sqrt{6}\sqrt{3}+(\sqrt{3})^2 \Rightarrow 6-2\cdot 3\sqrt{2}+3 \Rightarrow 9-6\sqrt{2} which was to be shown.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by GAdams View Post
    I got 1/6 for (a) but (b) is beyond me. Thanks.
    (b) is actually not that hard, what did you try?

    begin with (r - p)^2

    (r - p)^2 = r^2 - 2rp + p^2

    = ( \sqrt {6} )^2 - 2 \sqrt {6} \sqrt {3} + ( \sqrt {3} )^2

    Can you take it from here?
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  4. #4
    Member GAdams's Avatar
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    This is what I did:

    (sq.rt6 - sq.rt3) (sq.rt6 - sq.rt3) = 9 - 6 sq.rt2

    (sq.rt6)^2 - sq.rt18 - sq.rt18 + sq.rt9

    ...I sorta died about there.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by GAdams View Post
    This is what I did:

    (sq.rt6 - sq.rt3) (sq.rt6 - sq.rt3) = 9 - 6 sq.rt2

    (sq.rt6)^2 - sq.rt18 - sq.rt18 + sq.rt9

    ...I sorta died about there.
    why?

    (sqrt(6))^2 = 6

    sqrt(9) = 3

    -sqrt(18) - sqrt(18) = -2sqrt(18) = -2sqrt(9*2) = -2sqrt(9)*sqrt(2) = -2(3)sqrt(2) = -6sqrt(2)

    put those all together and you have the answer
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by GAdams View Post
    This is what I did:

    (sq.rt6 - sq.rt3) (sq.rt6 - sq.rt3) = 9 - 6 sq.rt2

    (sq.rt6)^2 - sq.rt18 - sq.rt18 + sq.rt9

    ...I sorta died about there.
    (\sqrt{6} - \sqrt{3})(\sqrt{6} - \sqrt{3})

    = (\sqrt{6})^2 - \sqrt{18} - \sqrt{18} + \sqrt{9}

    = 6 - 2\sqrt{18} + 3

    = 9 - 2\sqrt{18}

    = 9 - 2\sqrt{9 \cdot 2}

    = 9 - 2\sqrt{3^2 \cdot 2}

    = 9 - 2 \cdot3 \sqrt{2}

    = 9 - 6\sqrt{2}

    -Dan
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  7. #7
    Member GAdams's Avatar
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    6 - 2 sq.rt18 + 3

    9 - 2 sq.rt 18

    sq.rt18 = sq.rt9 * sq.rt 2 = 3 sq.rt2

    but that gives 9 - 3 sq.rt2

    EDit: thanks rualin, 9 - sq.rt 2 * 3 sq.rt2

    9 - 6 sq.rt2
    Last edited by GAdams; July 7th 2007 at 05:36 AM.
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    Times the 2 that was already there = 6 sqrt(2)
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  9. #9
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    By the way, what's a surd? Seems like absurd...making no sense? Excuse my ignorance.
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  10. #10
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    Quote Originally Posted by rualin View Post
    By the way, what's a surd? Seems like absurd...making no sense? Excuse my ignorance.
    My first answer to a question.......

    A surd is, of course, a verb. It's the past tense of 'to sur' as in the make someone a knight, like what the queen does (here in the Uk anyway).
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by rualin View Post
    By the way, what's a surd? Seems like absurd...making no sense? Excuse my ignorance.
    a surd is an unresolved radical expression.

    example, \sqrt {9} is a surd, since we could simplify (or resolve) this to get 3, but we leave it in its radical form because it's convenient to do so in our problem. that's a surd
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  12. #12
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    Ahhh! That makes a lot more sense. Thanks
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