# Solving inequalities

#### stemmaths

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

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

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

$(x-1)^2 \ge 2$

1 person

#### stemmaths

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

$(x-1)^2 \neq x^2 - 2x -1$

2 people

#### skeeter

MHF Helper
One more way to look at it ...

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

$x^2-2x \ge 1$

Complete the square ...

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

$(x-1)^2 \ge 2$

1 person

#### stemmaths

thanks for the response i couldnt see that completing the square was used in the question