jmc

Taking a step back, we notice that the given equation is of the form $$\sqrt{a}+\sqrt{b}=\sqrt{c},$$where $a=x^2-2x+2,$ $b=-x^2+6x-2,$ and $c=4x.$ Furthermore, we have $$a+b=(x^2-2x+2)+(-x^2+6x-2)=4x=c.$$So we square the equation $\sqrt{a}+\sqrt{b}=\sqrt{c}$ to get $$a+b+2\sqrt{ab}=c.$$Since $a+b=c,$ we have $2\sqrt{ab}=0,$ so either $a=0$ or $b=0.$ That is, either $x^2-2x+2=0$ or $-x^2+6x-2=0.$ The first equation has no real solutions because it is equivalent to $(x-1{)}^2+1=0.$ The second equation has two real roots $$x=\frac{6\pm \sqrt{6^2-4\cdot 1\cdot 2}}{2}=3\pm \sqrt{7}.$$Because both of these roots are positive, they both satisfy the original equation. The smaller root is $x=\boxed{3-\sqrt{7}}.$