Find minimum value of $a^2+b^2$

$$x^4+ax^3+2x^2+bx+1=$$ $$=(x^2+\frac{a}{2}x)^2-\frac{a^2}{4}x^2+2x^2+bx+1=$$ $$=(x^2+\frac{a}{2}x)^2-\frac{a^2}{4}x^2+2x^2-\frac{b^2}{4}x^2+(\frac{b}{2}x+1)^2=$$ $$=(x^2+\frac{a}{2}x)^2+x^2(2-\frac{a^2+b^2}{4})+(\frac{b}{2}x+1)^2$$

If $2-\frac{a^2+b^2}{4}>0$, then we have a problem; the sum of three nonnegative terms equals zero (at our real root), which can only happen if $x=0$ to make the second term zero. But then $(\frac{b}{2}x+1)^2>0$. So if $a^2+b^2<8$ there is no real root. Hence $a^2+b^2\ge 8$.

Now, if $a^2+b^2=8$, then the middle term is zero. Can we make the other two terms zero as well? We need $x=-\frac{a}{2}$ and $x=-\frac{2}{b}$, to make the first and last terms zero. This can be achieved, with $a=b=2$ and $x=-1$.

Hence, the minimum possible value of $a^2+b^2$ is indeed $8$.