Find minimum value of $a^2+b^2$ Given that $a$ and $b$ are real constants and that the equation $x^4+ax^3+2x^2+bx+1=0$ has at least one real root, find the minimum possible value of $a^2+b^2$.
I began this way: Let the polynomial be factorized as $(x^2+\alpha x + 1)(x^2+\beta x +1)$. Then expanding and comparing coefficients we get $\alpha\beta=0$, meaning either $\alpha=0$ or $\beta=0$. Suppose $\alpha=0$. Then we see that $(x^2+\beta x+1)$ should have real roots, from which we get $\beta^2 \geq 4$. But we get $a=b=\beta$ from the comparison above. So $a^2+b^2 = 2\beta^2 \geq 8$.
Is it correct? Or is there any mistake? Any other solution is also welcome.

$$x^4+ax^3+2x^2+bx+1=$$
$$=(x^2+\frac{a}{2}x)^2-\frac{a^2}{4}x^2+2x^2+bx+1=$$
$$=(x^2+\frac{a}{2}x)^2-\frac{a^2}{4}x^2+2x^2-\frac{b^2}{4}x^2+(\frac{b}{2}x+1)^2=$$
$$=(x^2+\frac{a}{2}x)^2+x^2(2-\frac{a^2+b^2}{4})+(\frac{b}{2}x+1)^2$$
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 $8$.
