Question 860535
Analyze two separate inequalities.

{{{-x<x^2}}} AND {{{x^2<2x+1}}};  look for what is true for both at the same time.


{{{-x^2-x<0}}}
{{{x^2+x>0}}}
{{{x(x+1)>0}}}
Critical points 0 and -1.
Solutions:  {{{x<-1}}} OR {{{x>0}}}.


{{{x^2-2x-1<0}}}
discriminant, 4-4(-1)=8
roots or horiz intercepts will be the critical points:
{{{x=(2-2sqrt(2))/2=1-sqrt(2)}}}
OR
{{{x=1+sqrt(2)}}}.
The parabola opens upward, and the points below the level y=0 occur BETWEEN the horizontal intercepts.


The x values common for both inequalities are {{{highlight(0<x<1+sqrt(2))}}}, which is easier to see if you make a number line.