Solving inequalities

Jun 2015
34
0
ireland
Hi
i am trying to solve the following but i cant figure out the last line of the solution
\(\displaystyle 2x^2 \geq (x+1)^2\)


rearranging i get \(\displaystyle 2x^2 \geq x^2+2z+1\)

\(\displaystyle x^2-2x-1\geq 0\)

\(\displaystyle (x-1)^2 \geq 2\) This is the last step given in the solution i can understand how to get to this point but i dont know where\(\displaystyle \geq2\) comes from any help would be appreciated
 

skeeter

MHF Helper
Jun 2008
16,217
6,765
North Texas
$x^2-2x-1 \ge 0$

$x^2-2x-1+2 \ge 2$

$x^2-2x+1 \ge 2$

$(x-1)^2 \ge 2$
 
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Jun 2015
34
0
ireland
thanks for the response
im still unsure where the \(\displaystyle +2\geq2\) comes from sorry now its prob something simple that i cant see
 

romsek

MHF Helper
Nov 2013
6,836
3,079
California
thanks for the response
im still unsure where the \(\displaystyle +2\geq2\) comes from sorry now its prob something simple that i cant see
$(x-1)^2 = x^2 -2x +1$


$(x-1)^2 \neq x^2 - 2x -1$
 
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skeeter

MHF Helper
Jun 2008
16,217
6,765
North Texas
One more way to look at it ...

$x^2-2x-1 \ge 0$

$x^2-2x \ge 1$

Complete the square ...

$x^2-2x+1 \ge 1+1$

$(x-1)^2 \ge 2$
 
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Jun 2015
34
0
ireland
thanks for the response i couldnt see that completing the square was used in the question