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Math Help - anyone know root 12?

  1. #1
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    Wink anyone know root 12?

    Ok guys, here's another easy one that's got me foxed!

    given that  (1 + x)^{1/2} = 1 + 1/2 x - 1/8 x^2
    deduce \sqrt{12} \approx 3.464

    earlier I was asked to use the series to find  \sqrt 1.08 which is OK because it fits the (1 + x) pattern - but sqrt 12 ?

    I thought of sqrt 12 = sqrt (10 x 1.2) = sqrt 10. sqrt (1 + 0.2) but this means I have to know what sqrt 10 is and if I knew that, I wouldn't have a problem with sqrt 12!

    sqrt 12 = 2 sqrt 3 but this doesn't get me anywhere either. Can anyone see something I can't?

    regards
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  2. #2
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    Quote Originally Posted by s_ingram View Post
    Ok guys, here's another easy one that's got me foxed!

    given that  (1 + x)^{1/2} = 1 + 1/2 x - 1/8 x^2
    deduce \sqrt{12} \approx 3.464

    earlier I was asked to use the series to find  \sqrt 1.08 which is OK because it fits the (1 + x) pattern - but sqrt 12 ?

    I thought of sqrt 12 = sqrt (10 x 1.2) = sqrt 10. sqrt (1 + 0.2) but this means I have to know what sqrt 10 is and if I knew that, I wouldn't have a problem with sqrt 12!

    sqrt 12 = 2 sqrt 3 but this doesn't get me anywhere either. Can anyone see something I can't?

    regards
    On the right side, are the x's in the numerator or the denominator? In other words, is it this:
     (1 + x)^{1/2} = 1 + \frac{1}{2}x - \frac{1}{8}x^2
    or this:
     (1 + x)^{1/2} = 1 + \frac{1}{2x} - \frac{1}{8x^2} ?

    Regarding the left side, how about the following -- would this work?
    \begin{aligned}<br />
\sqrt{12} &= (16 \cdot 0.75)^{1/2} \\<br />
&= (16(1 - 0.25))^{1/2} \\<br />
&= 4(1 + 1/2 x - 1/8 x^2)^{1/2}\;\;\text{(plug in -0.25 for x?)}<br />
\end{aligned}
    so x = -0.25 (or -1/4)?


    01
    Last edited by yeongil; June 20th 2009 at 08:18 AM.
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  3. #3
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    Quote Originally Posted by yeongil View Post
    On the right side, are the x's in the numerator or the denominator? In other words, is it this:
     (1 + x)^{1/2} = 1 + \frac{1}{2}x - \frac{1}{8}x^2
    or this:
     (1 + x)^{1/2} = 1 + \frac{1}{2x} - \frac{1}{8x^2} ?

    Regarding the left side, how about the following -- would this work?
    \begin{aligned}<br />
\sqrt{12} &= (16 \cdot 0.75)^{1/2} \\<br />
&= (16(1 - 0.25))^{1/2} \\<br />
&= 4(1 + 1/2 x - 1/8 x^2)\;\;\text{(plug in -0.25 for x?)}<br />
\end{aligned}
    so x = -0.25 (or -1/4)?


    01
    It looks like the binomial expansion so it would be the top one.

    I would use the following method: \sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3} = 2\sqrt{1+2}

    Looks like I misread:

    The binomial expansion up to the third term is:

    (1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3...
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  4. #4
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    Yes, sorry about the formatting, I am still getting used to this LATEX app. It is a binomial expansion so the top version is correct!
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  5. #5
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    Thanks yeongil. 16 * 0.75 is the way to go. I knew 2 sqrt 3 wouldn't do it because x cannot be 2 for a binomial.
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