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Math Help - [SOLVED] question about writing roots in fractions?

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    [SOLVED] question about writing roots in fractions?

    I'm doing a calculus problem and i found that x =\frac{\sqrt3}{3}
    in the answer in the back, it says \frac{1}{\sqrt3}
    I put them both in the calculator and it comes out to be about 0.57....
    why did the book put the root in the denominator? i thought it was standard form to have it in the numerator?

    not sure where this topic belongs, sorry if in wrong category!
    Last edited by dorkymichelle; April 24th 2010 at 11:17 PM.
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    Quote Originally Posted by dorkymichelle View Post
    I'm doing a calculus problem and i found that x =\frac{\sqrt3}{3}
    in the answer in the back, it says \frac{1}{\sqrt3}
    I put them both in the calculator and it comes out to be about 0.57....
    why did the book put the root in the denominator? i thought it was standard form to have it in the numerator?

    not sure where this topic belongs, sorry if in wrong category!
    Note that

    \frac{1}{\sqrt{3}} = \frac{1}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{3}.

    It is considered to be more elegant to rationalise denominators, because it better represents what a fraction is - namely, dividing a length into a COUNTABLE number of pieces.

    However, rationalising the denominator is not usually done until the very end, as it could result in more work with whatever you happen to be doing. Perhaps your book intends you to use this answer for further calculations...
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    I found it odd because i didnt encounter 1/root 3 in my calculations at all.
    is there a way to solve for x in the equation -12x^2+4
    without using the quadratic formula?
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    Quote Originally Posted by dorkymichelle View Post
    I found it odd because i didnt encounter 1/root 3 in my calculations at all.
    is there a way to solve for x in the equation -12x^2+4
    without using the quadratic formula?
    is your equation -12x^2+4=0??
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    yes
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    Quote Originally Posted by dorkymichelle View Post
    yes
    if -12x^2+4=0, then

    4 = 12x^2

    so,

    x^2 = \frac{4}{12} = \frac{1}{3}

    \therefore x = \pm \frac{1}{\sqrt 3}
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    wow... i can't believe i didn't see that!!!
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