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#1
Jun 4 2014 07:06pm


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#2
Jun 4 2014 07:39pm
use the fact that a^2-b^2 = (a-b)(a+b)

delta = 1/10
if 0<|x-1| < 1/10 then we can write, x being positive (otherwise |x-1| > 1 > 1/10:
|x-1| = |sqrt(x)-1|*|sqrt(x)+1| < 1/10
i.e: |sqrt(x)-1|<1/[10*(sqrt(x)+1)] but sqrt(x)+1 > 1 therefore 1/[10*(sqrt(x)+1)] < 1/10
that is : |sqrt(x)-1| < 1/10
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#3
Jun 4 2014 07:56pm
Quote (HbSoe @ Jun 4 2014 09:39pm)
use the fact that a^2-b^2 = (a-b)(a+b)

delta = 1/10
if 0<|x-1| < 1/10 then we can write, x being positive (otherwise |x-1| > 1 > 1/10:
|x-1| = |sqrt(x)-1|*|sqrt(x)+1| < 1/10
i.e: |sqrt(x)-1|<1/[10*(sqrt(x)+1)] but sqrt(x)+1 > 1 therefore 1/[10*(sqrt(x)+1)] < 1/10
that is : |sqrt(x)-1| < 1/10


So the final answer is 1/10? I just figured it out. And got 1/10.
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#4
Jun 4 2014 08:28pm
Quote (desertwolf @ Jun 4 2014 09:56pm)
So the final answer is 1/10? I just figured it out. And got 1/10.


Forgot to square root...got my final answer as 2/11 now.

dgl;dj;gjsd, nvm. Now I have 19/100 as my answer now. Ugh, I'm getting sleep deprived.


EDIT: NVM!


I finally figured it out. Wow, it really is 1/10!

This post was edited by desertwolf on Jun 4 2014 08:44pm
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