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Math Help - Fractional Square Roots w/ unperfect squares...

  1. #1
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    Fractional Square Roots w/ unperfect squares...

    I'm having some difficulty in understanding the following square root "procedure".

    When finding the square root of a fraction, it's pretty easy when both the numerator and denominator are perfect squares: \sqrt\frac{m^4n^8}{81p^2} which I beleive = \frac{m^2n^4}{9p}

    I'm having some trouble in understanding the procedure when the numerator and or denominator are not perfect squares...such as \sqrt\frac{3}{8}. The way I understand it is, I need to multiply the numerator by the number that would make the number in the denominator a perfect square. Since 8 is the square root of 64, I need to multiply the numerator, 3, by 8, which gives 24. So I then have the following \sqrt\frac{24}{8}. This I understand, what I don't understand is the following (I don't know if it would be called simplification or not?) \sqrt\frac{24}{8} = \frac{2\sqrt6}{8} I do not understand 2\sqrt6 in the numerator. How do they come to this conclusion and why can't you leave it as \sqrt\frac{24}{8}?

    Any help would be greatly appreciated as always!
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    It is generally accepted that "simplest form" is the goal.

    \sqrt{\frac{24}{8}} is almost universally accepted as simplest in the form \sqrt{3}

    \sqrt{\frac{49}{8}} is a bit trickier, but still requires extraction of all perfect squares before anythone thins it is "simplest".

    \sqrt{\frac{49}{8}}\;=\;\sqrt{\frac{7^{2}}{2^{2}\c  dot 2}}\;=\;\frac{7}{2}\sqrt{\frac{1}{2}}\;=\;\frac{7}  {2}\frac{1}{\sqrt{2}}

    For me, this is done, but there remains another convention that says redicals in the denominator are evil. I don't tend to agree with this convention, but if it is in your curriculum as "simplification", your teacher will be justified in marking it wrong if you don't do it. It takes the idea, a very useful idea, of rationalizing the denominator.

    \frac{1}{\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}}=\  frac{\sqrt{2}}{2}

    It is a purely cosmetic effort. Check carefully to see if you are required to do it. It is also a carry-over from the days before calculators. The last expression may nto be simpler than others at all from the point of view of your calculator. It also takes longer to do this. If you have panic exam time pressure, just don't do it.
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  3. #3
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by ejanderson View Post
    I'm having some difficulty in understanding the following square root "procedure".

    When finding the square root of a fraction, it's pretty easy when both the numerator and denominator are perfect squares: \sqrt\frac{m^4n^8}{81p^2} which I beleive = \frac{m^2n^4}{9p}

    I'm having some trouble in understanding the procedure when the numerator and or denominator are not perfect squares...such as \sqrt\frac{3}{8}. The way I understand it is, I need to multiply the numerator by the number that would make the number in the denominator a perfect square. Since 8 is the square root of 64, I need to multiply the numerator, 3, by 8, which gives 24. So I then have the following \sqrt\frac{24}{8}. This I understand, what I don't understand is the following (I don't know if it would be called simplification or not?) \sqrt\frac{24}{8} = \frac{2\sqrt6}{8} I do not understand 2\sqrt6 in the numerator. How do they come to this conclusion and why can't you leave it as \sqrt\frac{24}{8}?

    Any help would be greatly appreciated as always!
    While getting \sqrt\frac{24}{8}, you have only multiplied the numerator by 8, you also need to multiply the denominator. so, according to what you said, to make the denominator a perfect square, you multiple the numerator as well as the denominator. that is,

    \sqrt\frac{3}{8} * \sqrt\frac{8}{8}

    = \sqrt\frac{24}{64}

    = \frac{\sqrt24}{8}

    further,

    \sqrt{24} = \sqrt{2*2*6}
    = \sqrt{{2^2}*6}
    = 2 \sqrt{6}
    Last edited by harish21; March 15th 2010 at 10:44 PM.
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