Parabola disjoint from a line as a parameter varies

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Find the values of the parameter $r$ for which the parabola $y=x^2+rx+r$ is external to the line $y=2x+1$ (i.e. does not meet it).

Solution

Intersect parabola and line: $x^2+rx+r=2x+1 \implies x^2+(r-2)x+(r-1)=0.$ “External” means no real intersection, so a negative discriminant: $\Delta=(r-2)^2-4(r-1)=r^2-8r+8<0.$ The zeros of $r^2-8r+8$ are $r=4\pm2\sqrt2$, and the parabola in $r$ is negative between them: $\boxed{4-2\sqrt2<r<4+2\sqrt2}$