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- #1

- Feb 14, 2012

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- Thread starter anemone
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- #1

- Feb 14, 2012

- 3,909

- Mar 31, 2013

- 1,346

so $x^2 + x + 1= (x + 1/2)^2 + 3/4$

= $(x+1/2)^2( 1+ 3/(4(x + 1/2)^2)$

so square root = $(x+1/2) ( 1 + 3/(8(x+1/2)^2) + ....)$

the term $3/(8(x+1/2)^2)$ is extremley small so << .1

so square root is x + 1/2 or 5 is the 1st digit after decimal

Last edited:

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- #3

- Feb 14, 2012

- 3,909

Hey

so $x^2 + x + 1= (x + 1/2)^2 + 3/4$

= $(x+1/2)^2( 1+ 3/(4(x + 1/2)^2)$

so square root = $(x+1/2) ( 1 + 3/(8(x+1/2)^2) + ....)$

the term $3/(8(x+1/2)^2)$ is extremley small so << .1

so square root is x + 1/2 or 5 is the 1st digit after decimal

- Mar 31, 2013

- 1,346

this edited post is more accurate as it defines the reason. As you have made a reference to un edited post I mention the unedited post( exact words I do not remeber so in lines as below) which is highly informalHeykaliprasad, thanks for participating! Well done! Your answer is correct... but I think this edited version of the solution isn't quite straightforward than the before edited post.

so $x^2+x+1=(x+1/2)^2+3/4$ and as 3/4 is too small we can igmore so

square root =x + 1/2 so 1st digit after decimal = 5

Last edited: