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Math Help - Math help - square roots?

  1. #1
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    Math help - square roots?

    Hi!

    I'm having some trouble with square roots -
    I understand the square root of numbers like 49 (7), 9 (3) and all, though I'm very confused with finding the square root of numbers like 18, or 32.

    Here is the example the teacher had:



    http://i47.tinypic.com/23h0tbk.jpg

    Could anyone possibly explain this in some way to me? What I don't understand is how you would even start a problem - like what actual math steps do you take to get those results, and how do you know what numbers to choose, and/or start with?
    I'm very confused by this, if anyone could explain or show a better example - Like are there certain steps?
    Any help is appreciated, as Math is not my strong subject! : /
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  2. #2
    Senior Member Stroodle's Avatar
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    Maybe it's easier to see it as:

    \sqrt{18}=\sqrt{9\times2}=\sqrt{9}\times\sqrt{2}=3  \times\sqrt{2}=3\sqrt{2}

    and,

    \sqrt{72}=\sqrt{36\times2}=\sqrt{36}\times\sqrt{2}  =6\sqrt{2}

    With practice you'll become better at recognising square numbers (like 9, 25, 36, and so on) and the multiples of these.

    Larger square numbers can be harder to recognise than smaller ones, so you may sometimes find it easier to do it in steps:

    \sqrt{72}=\sqrt{4\times18}=2\sqrt{9\times2}=2\time  s3\times\sqrt{2}=6\sqrt{2}
    Last edited by Stroodle; December 10th 2009 at 06:05 AM. Reason: Added info
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  3. #3
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    Hi
    Remember !
    \sqrt{a^2}=\left | a \right |
    let's take \sqrt{18}
    decompose 18 :
    18=2\times3\times3.
    therefore,
    \sqrt{18}= \sqrt{2\times3\times3}=\sqrt{2}\times \sqrt{3^2}=3\sqrt{2}.
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  4. #4
    Super Member Bacterius's Avatar
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    There are some very important formulas you need to learn by heart in order to understand square roots. They are :

    \sqrt{a \times b} = \sqrt{a} \times \sqrt{b}

    \sqrt{a \times b^2} = b \times \sqrt{a}

    Note that the second one can be proved from the first one :

    \sqrt{a \times b^2} = \sqrt{a} \times \sqrt{b^2} = \sqrt{a} \times b = b \times \sqrt{a}

    This way, if the expression under the square root can be broken into factors, and that some factors can be written as squares, then they can be "extracted" from the square root.

    --------------------------------------

    Let's take, \sqrt{48} for the example. You want to simplify it.
    First, break 48 into its prime factors, to make it easier to spot squares that you can remove from the square root.

    Note that 48 = 24 \times 2 = 2 \times 12 \times 2 = 2 \times 2 \times 6 \times 2 = 2 \times 2 \times 2 \times 3 \times 2 = 2 \times 2 \times 2 \times 2 \times 3 = 2^4 \times 3

    So, \sqrt{48} = \sqrt{2^4 \times 3}.

    Now, remember 2^4 = (2^2)^2, so :

    \sqrt{48} = \sqrt{(2^2)^2 \times 3}

    And you see that there is a square. You can remove it :

    \sqrt{48} = \sqrt{(2^2)^2} \sqrt{3}

    And the (2^2)^2 square root is easily simplified :

    \sqrt{48} = \sqrt{16} \sqrt{3}

    \sqrt{48} = 4 \sqrt{3}

    Do you understand better now ?
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  5. #5
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    Quote Originally Posted by Bacterius View Post
    There are some very important formulas you need to learn by heart in order to understand square roots. They are :

    \sqrt{a \times b} = \sqrt{a} \times \sqrt{b}

    \sqrt{a \times b^2} = b \times \sqrt{a}

    Note that the second one can be proved from the first one :

    \sqrt{a \times b^2} = \sqrt{a} \times \sqrt{b^2} = \sqrt{a} \times b = b \times \sqrt{a}

    This way, if the expression under the square root can be broken into factors, and that some factors can be written as squares, then they can be "extracted" from the square root.

    --------------------------------------

    Let's take, \sqrt{48} for the example. You want to simplify it.
    First, break 48 into its prime factors, to make it easier to spot squares that you can remove from the square root.

    Note that 48 = 24 \times 2 = 2 \times 12 \times 2 = 2 \times 2 \times 6 \times 2 = 2 \times 2 \times 2 \times 3 \times 2 = 2 \times 2 \times 2 \times 2 \times 3 = 2^4 \times 3

    So, \sqrt{48} = \sqrt{2^4 \times 3}.

    Now, remember 2^4 = (2^2)^2, so :

    \sqrt{48} = \sqrt{(2^2)^2 \times 3}

    And you see that there is a square. You can remove it :

    \sqrt{48} = \sqrt{(2^2)^2} \sqrt{3}

    And the (2^2)^2 square root is easily simplified :

    \sqrt{48} = \sqrt{16} \sqrt{3}

    \sqrt{48} = 4 \sqrt{3}

    Do you understand better now ?
    Your answers are always well detailed,very well done indeed.
    Last edited by Raoh; December 10th 2009 at 09:59 AM. Reason: errr !
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  6. #6
    Super Member Bacterius's Avatar
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    Quote Originally Posted by Raoh View Post
    You're answers are always well detailed,very well done indeed.
    I try to make them as detailed and clear as possible because I don't like to repeat myself
    Thanks though !
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