# Binomial expansion question

• Mar 7th 2011, 10:23 PM
MichaelLight
Binomial expansion question
Prove that , if x is so small that terms in x^3 and higher powers may be neglected, then Attachment 21081.

By substituting a suitable value of x in your result, show that (11)^1/2 is approximately equal to 663/200.

I can solve the first part, but i have difficulty in the second part (show that (11)^1/2 is approximately equal to 663/200). Can anyone help me? Thanks.
• Mar 7th 2011, 11:39 PM
Prove It
If $\displaystyle \sqrt{11} = \sqrt{\frac{1 + x}{1 - x}}$, then what is $\displaystyle x$?
• Mar 8th 2011, 02:07 AM
MichaelLight
Quote:

Originally Posted by Prove It
If $\displaystyle \sqrt{11} = \sqrt{\frac{1 + x}{1 - x}}$, then what is $\displaystyle x$?

By solving that, i get x= 5/6. Substituting x=5/6 into http://www.mathhelpforum.com/math-he...10787h09a9.gif, i get (11)^1/2=157/72 which is incorrect...
• Mar 8th 2011, 02:10 AM
Unknown008
Indeed, I was getting this too. I think there is something wrong with the question. The issue is surely due to the fact that 5/6 is not small enough for x^3 and higher powers to be neglected.

Either that, or I'm missing something. (Itwasntme)
• Mar 8th 2011, 03:39 AM
HallsofIvy
"Approximately equal" is a loose term. Yes, using x= 5/6 gives $1+ 5/6+ 25/72= 1.18$ while $\sqrt{11}= .32$ approximately. That's not very close but then, as Unknown008 said, 5/6 is not all that "small"! $\left(\frac{5}{6}\right)^3$ is larger than 1/2- hardly "negligible".
• Mar 8th 2011, 03:52 AM
Unknown008
$\sqrt{11} = 3.316...$

$1 + \dfrac56 + \dfrac12 \left(\dfrac{25}{36}\right) = 2.18...$

(Smile)

It must eirler be very late, or very early where you are HallsofIvy (Giggle)
• Mar 8th 2011, 03:54 PM
awkward
Hint: If you knew the square root of 11/9, you could easily compute the square root of 11.
• Mar 8th 2011, 08:37 PM
MichaelLight
Quote:

Originally Posted by awkward
Hint: If you knew the square root of 11/9, you could easily compute the square root of 11.

But how do i get 'square root of 11/9'? And how to compute 'square root of 11' after i get 'square root of 11/9'? Please enlighten me..
• Mar 9th 2011, 04:13 PM
awkward
Quote:

Originally Posted by MichaelLight
But how do i get 'square root of 11/9'?

The same way you found the square root of 11 previously.

Quote:

And how to compute 'square root of 11' after i get 'square root of 11/9'? Please enlighten me..
Simplify $\sqrt{\frac{11}{9}}$.