# anyone know root 12?

• Jun 20th 2009, 05:33 AM
s_ingram
anyone know root 12?
Ok guys, here's another easy one that's got me foxed!

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

earlier I was asked to use the series to find $\displaystyle \sqrt 1.08$ 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
• Jun 20th 2009, 05:56 AM
yeongil
Quote:

Originally Posted by s_ingram
Ok guys, here's another easy one that's got me foxed!

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

earlier I was asked to use the series to find $\displaystyle \sqrt 1.08$ 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:
$\displaystyle (1 + x)^{1/2} = 1 + \frac{1}{2}x - \frac{1}{8}x^2$
or this:
$\displaystyle (1 + x)^{1/2} = 1 + \frac{1}{2x} - \frac{1}{8x^2}$?

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

01
• Jun 20th 2009, 06:07 AM
e^(i*pi)
Quote:

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

Regarding the left side, how about the following -- would this work?
\displaystyle \begin{aligned} \sqrt{12} &= (16 \cdot 0.75)^{1/2} \\ &= (16(1 - 0.25))^{1/2} \\ &= 4(1 + 1/2 x - 1/8 x^2)\;\;\text{(plug in -0.25 for x?)} \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: $\displaystyle \sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3} = 2\sqrt{1+2}$

$\displaystyle (1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3...$