For which values of $m$ we get: $x_1>4$ and $x_2<1$?

Question

given this equation:
$$x-\sqrt{x}(3+m)-2(1-m^2)=0$$
with roots $x_1$,$x_2$.

For which values of $m$ we get: $x_1>4$ and $x_2<1$?

Answer 1

We need to solve the following system: $$4-2(3+m)-2(1-m^2)<0$$ and $$1-1(3+m)-2(1-m^2)<0.$$

Indeed, let $\sqrt{x}=t$ and $f(t)=t^2-(3+m)t-2(1-m^2).$

Thus, the graph of $f$ it's parabola and we need $f(2)<0$ and $f(1)<0$.

Draw it!

I got $$-1<m<\frac{1+\sqrt{33}}{4}.$$