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Math Help - Surds

  1. #1
    Member GAdams's Avatar
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    Surds

    I understand the basic idea of getting the sq.rt off the bottom by multiplying top and bottom by the sq.rt but it doesn't always seemto work, is there more to it?

    2/sq.rt8 -I get sq.rt8/4 , but answer says sq.rt2/2

    a/ sq.rt40/2 -I get sq.rt4/8 -answer says sq.rt10a/10

    And I am not sure how to tackle this one:

    2/ 1+sq.rt6

    Thanks.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by GAdams View Post
    I understand the basic idea of getting the sq.rt off the bottom by multiplying top and bottom by the sq.rt but it doesn't always seemto work, is there more to it?

    2/sq.rt8 -I get sq.rt8/4 , but answer says sq.rt2/2

    a/ sq.rt40/2 -I get sq.rt4/8 -answer says sq.rt10a/10

    And I am not sure how to tackle this one:

    2/ 1+sq.rt6

    Thanks.
    For the first one:
    \frac{2}{\sqrt{8}} \cdot \frac{\sqrt{8}}{\sqrt{8}} = \frac{2 \sqrt{8}}{8}

    = \frac{\sqrt{8}}{4}

    So far, so good. But:
    \frac{ \sqrt{8}}{4} = \frac{ \sqrt{4 \cdot 2}}{4} = \frac{2 \sqrt{2}}{4} = \frac{ \sqrt{2}}{2}

    For the second one: " a/ sq.rt40/2"
    \frac{a}{\frac{\sqrt{40}}{2}} = \frac{2a}{\sqrt{40}} = \frac{2a}{\sqrt{40}} \cdot \frac{\sqrt{40}}{\sqrt{40}}

    = \frac{2a\sqrt{40}}{40} = \frac{a\sqrt{40}}{20}

    = \frac{a \sqrt{4 \cdot 10}}{20} = \frac{2a\sqrt{10}}{20} = \frac{a\sqrt{10}}{10}

    For the last one: "2/ 1+sq.rt6"
    \frac{2}{1 + \sqrt{6}}

    We need to multiply the top and bottom by the "conjugate." (The works on the ones above too.) It works like this:
    (a + b)(a - b) = a^2 - b^2

    So
    (a + \sqrt{c})(a - \sqrt{c}) = a^2 - c
    which no has no square root in the final expression.

    Thus if we have a denominator with a a \pm \sqrt{c} in it we multiply the top and bottom by a \mp \sqrt{c}, which is called the conjugate of a \pm \sqrt{c}.

    So.
    \frac{2}{1 + \sqrt{6}} \cdot \frac{1 - \sqrt{6}}{1 - \sqrt{6}} = \frac{2 - 2 \sqrt{6}}{1^2 - 6} = \frac{2 - 2\sqrt{6}}{-5}

    = - \frac{2 - 2\sqrt{6}}{5} = \frac{-2 + 2\sqrt{6}}{5}

    -Dan
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  3. #3
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    Hello, GAdams!

    \frac{2}{\sqrt{8}} . . . I get \frac{\sqrt{8}}{4}, but answer says \frac{\sqrt{2}}{2} . . . . Reduce your answer

    Here's something I figured out when I was a student
    . . and I've taught it to all my students ever since.
    Yet I have never seen it mentioned in any textbook
    . . nor have I ever met a teacher who is familiar with it.

    The denominator has a square root so we're expected to "rationalize" it.
    . . Very well, how do we do this?

    We modify the fraction (legally) so that the square root contains a perfect square.


    The standard method is to multiply top and bottom by the square root:

    . . \frac{2}{\sqrt{8}}\cdot\frac{\sqrt{8}}{\sqrt{8}} \;=\;\frac{2\sqrt{8}}{\underbrace{\sqrt{64}}_{\tex  t{square}}} \;=\;\frac{2\sqrt{8}}{8}

    It is rationalized, but we are (of course) expected to reduce the fraction.
    . . Obviously, we can "cancel 2's" and get \frac{\sqrt{8}}{4}

    But \sqrt{8} can be reduced. .(We're supposed to be aware of this.)

    Since \sqrt{8} \:=\:\sqrt{4\cdot2} \:=\:\sqrt{4}\!\cdot\!\sqrt{2}\:=\:2\sqrt{2}, we have: . \frac{\sqrt{8}}{4}\:=\:\frac{2\sqrt{2}}{4} \:=\:\frac{\sqrt{2}}{2}


    But wouldn't be simplier to multiply top and bottom by \sqrt{2} ?

    . . \frac{2}{\sqrt{8}}\cdot\frac{\sqrt{2}}{\sqrt{2}} \;=\;\frac{2\sqrt{2}}{\underbrace{\sqrt{16}}_{\tex  t{square}}} \;=\;\frac{2\sqrt{2}}{4} \:=\:\frac{\sqrt{2}}{2}


    My point: Multiply top and bottom by the smallest amount
    . . . . . . . to make the denominator a square.


    Example: . \frac{1}{\sqrt{12}}

    Instead of automatically multiplying top and bottom by \sqrt{12},
    . . pause and give it some thought.

    If we introduce a "3", we will have a square: . \frac{1}{\sqrt{12}}\cdot\frac{\sqrt{3}}{\sqrt{3}} \:=\:\frac{\sqrt{3}}{\sqrt{36}} \:=\:\frac{\sqrt{3}}{6}

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    This problem arises because the square in the denominator can be simplified.

    An alternate approach is to simplify the denominator first:

    . . \frac{1}{\sqrt{12}} \;=\;\frac{1}{2\sqrt{3}} . . . then rationalize: . \frac{1}{2\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}} \;=\;\frac{\sqrt{3}}{2\sqrt{9}}\;=\;\frac{\sqrt{3}  }{6}

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  4. #4
    Member GAdams's Avatar
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    Thanks for all the previous replies, very very useful.

    I had a go at: 6/ 3 + sq.rt3

    = 6.(3-sq.rt3 / (3+sq.rt3)(3-sq.rt3)

    = 18 - 6sq.rt3 / 9 - 3

    = 3-6sq.rt3

    However, the answer says 3 -sq.rt3 ??

    Another one:

    sq.rt18 - sq.rt9

    = (sq.rt9 * sq.rt2) -3

    = 3sq.rt2 - 3 -is this right?
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  5. #5
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    Hello, GAdams!

    I had a go at: . \frac{6}{3 + \sqrt{3}}

    = \;\frac{6(3 -\sqrt{3})}{(3 + \sqrt{3})(3 - \sqrt{3})}

    = \;\frac{18 - 6\sqrt{3}}{9 - 3} . . . . Why distribute the 6?

    = \;3 - 6\sqrt{3} . . . . How?

    Your third line would have been: . \frac{6(3-\sqrt{3})}{9-3} \;=\;\frac{\not{6}(3-\sqrt{3})}{\not{6}} \;=\;3-\sqrt{3}



    Another one: . \sqrt{18} - \sqrt{9} \;=\;\sqrt{9}\cdot\sqrt{2} - 3 \;=\;3\sqrt{2} - 3

    Is this right?
    . . . Yes!

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  6. #6
    Forum Admin topsquark's Avatar
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    [quote=GAdams;586200

    I had a go at: 6/ 3 + sq.rt3
    [/quote]
    Just a quick note. Please try to use parenthesis. What you intended was
    \frac{6}{3 + \sqrt{3}}

    but according to order of operations, division comes before addition, so what you wrote was
    \frac{6}{3} + \sqrt{3}

    This could be the source of some confusion later on. If you don't learn the LaTeX then you should write it like this: 6/(3 + sq.rt3)

    -Dan
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