Problem source

Prove that both $x^{4}-2x^{3}+x^{2}$ and $x^{2}-8x+17$ are non-negative for all real $x$. By considering the intervals $x\leqslant0$, $0 < x\leqslant2$ and $x > 2$ separately, or otherwise, prove that the equation 
\[
x^{4}-2x^{3}+x^{2}-8x+17=0
\]
has no real roots. 
Prove that the equation $x^{4}-x^{3}+x^{2}-4x+4=0$ has no real roots.

Solution source

\begin{align*}
x^4 - 2x^3+x^2 &= x^2(x^2-2x+1) \\
&= x^2(x-1)^2 > 0
\end{align*}

Since $x$and $x-1$ can't both be zero, and square cannot be negative.

\begin{align*}
x^2 - 8x+17 &= (x-4)^2 +1 \geq 1 > 0
\end{align*}

If $x \leq 2$ then $x^4 - 2x^3+x^2 > 0$ and $17-8x \geq 1$ so $x^4-2x^3+x^2-8x+17 > 0$

If $x > 2$ then $x^4-2x^3 = x^3(x-2) \geq 0$ and $x^2-8x+17 > 0$ so $x^4-2x^3+x^2-8x+17 > 0$, so for all real numbers our polynomial is positive and therefore cannot have any roots.

Note that: $x^4-x^3+x^2 = x^2(x^2-x+1) > 0$ and $x^2-4x+4 =(x-2)^2 \geq 0$

If $x < 1$ then $x^4-x^3+x^2 > 0$ and $4(1-x) > 0$ so $x^4-x^3+x^2-4x+4 > 0$.

If $x > 1$ then $x^4-x^3 = x^3(x-1) > 0$ and $x^2-4x+4 \geq 0$ therefore  $x^4-x^3+x^2-4x+4 > 0$.

Therefore  $x^4-x^3+x^2-4x+4 > 0$ for all real $x$ and hence there are no real roots.