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Math Help - Radicals

  1. #1
    Senior Member vaironxxrd's Avatar
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    Radicals

    Hello forum, vaironxxrd here.
    I somehow con fussed myself trying to solve radicals I got many more questions but I will ask the simple ones first.


    Problem: \sqrt12
    What I believe is the solution is 3\sqrt2

    Procedure: 12/3=4 , 4/2= 2
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  2. #2
    Super Member TheChaz's Avatar
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    Re: Radicals

    Use the property that, for a > 0 and b > 0,
    √(ab) = √(a) * √(b)

    So √(12), which is √(4*3) = √(4) * √(3) = 2√3
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    Re: Radicals

    Quote Originally Posted by vaironxxrd View Post
    Hello forum, vaironxxrd here.
    I somehow con fussed myself trying to solve radicals I got many more questions but I will ask the simple ones first.


    Problem: \sqrt12
    What I believe is the solution is 3\sqrt2

    Procedure: 12/3=4 , 4/2= 2
    Your method is definitely wrong. (Square the original number an your result and you'll get 12 = 18 which is very seldom).

    To simplify a radical you should factor the radicand:

    \sqrt{12} = \sqrt{2 \cdot 2 \cdot 3} = \sqrt{2^2} \cdot \sqrt{3} = 2\sqrt{3}
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  4. #4
    Senior Member vaironxxrd's Avatar
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    Re: Radicals

    Quote Originally Posted by TheChaz View Post
    Use the property that, for a > 0 and b > 0,
    √(ab) = √(a) * √(b)

    So √(12), which is √(4*3) = √(4) * √(3) = 2√3
    Quote Originally Posted by earboth View Post
    Your method is definitely wrong. (Square the original number an your result and you'll get 12 = 18 which is very seldom).

    To simplify a radical you should factor the radicand:

    \sqrt{12} = \sqrt{2 \cdot * 2 \cdot 3} = \sqrt{2^2} \cdot \sqrt{3} = 2\sqrt{3}
    Thanks for clarifying that.
    To make sure I'm understand this here are two problems .

    Problem: \sqrt50


    Solve: \sqrt{25 \cdot 2 }= 5\sqrt2

    A more complicated problem for me.

    Problem: \sqr48=
    I was testing the results and this works, but can you guys explain why should I place the 4 out and aware the 9.

    Solution: \sqrt{3 \cdot 3 \cdot 4[
    4\sqrt3
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    Re: Radicals

    Quote Originally Posted by vaironxxrd View Post
    Thanks for clarifying that.
    To make sure I'm understand this here are two problems .

    Problem: \sqrt50


    Solve: \sqrt{25 \cdot 2 }= 5\sqrt2

    A more complicated problem for me.

    Problem: \sqr48=
    I was testing the results and this works, but can you guys explain why should I place the 4 out and aware the 9.

    Solution: \sqrt{3 \cdot 3 \cdot 4[
    4\sqrt3
    Your problem here is that 3(3)(4)= 4(9)= 36, not 48! 48= 3(16) so \sqrt{48}= \sqrt{3(16)}= \sqrt{3}\sqrt{16}= 4\sqrt{3}
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  6. #6
    Senior Member vaironxxrd's Avatar
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    Re: Radicals

    Quote Originally Posted by HallsofIvy View Post
    Your problem here is that 3(3)(4)= 4(9)= 36, not 48! 48= 3(16) so \sqrt{48}= \sqrt{3(16)}= \sqrt{3}\sqrt{16}= 4\sqrt{3}
    That is clear, I figure it out after watching a couple of videos but thanks this explanation is even better.
    What when working with variables ? Like...

    \sqrt{16n^2}

    and
    \sqrt{81x^3y^4}
    Last edited by vaironxxrd; October 13th 2011 at 01:07 PM.
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  7. #7
    Super Member Quacky's Avatar
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    Re: Radicals

    Quote Originally Posted by vaironxxrd View Post
    That is clear, I figure it out after watching a couple of videos but thanks this explanation is even better.
    What when working with variables ? Like...

    \sqrt16n^2

    and
    \sqrt81x^3y^4
    Why would it be any different?

    \sqrt{16n^2}
    =\sqrt{4\times 4\times n\times n}
    =\sqrt{(4n)^2}
    =4n
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  8. #8
    Super Member TheChaz's Avatar
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    Re: Radicals

    Quote Originally Posted by Quacky View Post
    Why would it be any different?

    \sqrt{16n^2}
    =\sqrt{4\times 4\times n\times n}
    =\sqrt{(4n)^2}
    =4n
    .... or 4|n|
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  9. #9
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    Re: Radicals

    Quote Originally Posted by vaironxxrd View Post
    That is clear, I figure it out after watching a couple of videos but thanks this explanation is even better.
    What when working with variables ? Like...

    \sqrt16n^2
    What you wrote is actually \sqrt{1}(6n^2)= 6n^2!!! Put everything you want in the square root in { }: \sqrt{16n^2} gives \sqrt{16n^2}= 4n if n is 0 or positive and -4n if n is negative. More generally \sqrt{16n^2}= 4|n| where |n| is the "absolute value".




    and
    \sqrt81x^3y^4
    \sqrt{81x^3y^4}= \sqrt{9^2x^2(y^2)^2x}= 9|x|y^2\sqrt{x}
    I don't need the absolute value for y because |y^2| is never negative.
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  10. #10
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    Re: Radicals

    \sqrt{81x^3y^4}= \sqrt{9^2x^2(y^2)^2x}= 9|x|y^2\sqrt{x}
    I don't need the absolute value for y because |y^2| is never negative.
    \sqrt{81x^3y^4} = 9xy^2\sqrt{x}

    I was told by someone much more knowledgeable than myself that you don't need the absolute value for x either, because x must be greater than or equal to zero in the original expression.
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