Problem source

Let  $\f(x)=x^2+px+q$ and $\g(x)=x^2+rx+s\,$.  
Find an expression for $\f ( \g (x))$  and hence find a necessary and sufficient condition on $a$, $b$ and $c$  for it to be possible to write the quartic expression $x^4+ax^3+bx^2+cx+d$ in the form $\f ( \g (x))$, for some choice of values of $p$, $q$, $r$ and $s$.  
 
Show further that this condition holds  if and only if it is possible to write  the quartic expression  $x^4+ax^3+bx^2+cx+d$ in the form $(x^2+vx+w)^2-k$,  for some choice of values of $v$, $w$ and $k$. 
 
Find the roots of the quartic equation $x^4-4x^3+10x^2-12x+4=0\,$.

Solution source

\begin{align*}
&& f(g(x)) &= (g(x))^2 + p(g(x)) + q \\
&&&= (x^2+rx+s)^2 + p(x^2+rx+s) + q \\
&&&= x^4 + 2rx^3 + (2s+r^2+p)x^2 +(2rs+pr)x + (s^2+ps+q) 
\end{align*}

So we need $2r=a ,2s+r^2+p = b, r(2s+p) = c$. (We have full control over $d$ since we can always chance $q$ only affecting $d$.

\begin{align*}
&& r &= \frac{a}{2} \\
&& b-r^2 & =rc \\
&& b - \frac{a^2}{4} & =\frac{ac}{2} \\
\Rightarrow && 4b-a^2&= 2ac
\end{align*}
Clearly this condition is necessary. It is sufficient since if it is true the equations are solveable.

$(x^2+vx+w)^2 = x^4 + 2vx^3 + (2vw+v^2)x^2+2vw x + w^2$. We don't care about the constant term since we can control this with $k$, so we just need to check $4(2vw+v^2) - (2v)^2 = 8wv$ so this does satisfy the condition. The reverse is also clear.

\begin{align*}
&& 0 &= x^4-4x^3+10x^2-12x+4 \\
&&&= (x^2-2x+3)^2-5 \\
\Rightarrow && 0 &= x^2 - 2x+3 \pm \sqrt{5} \\
&& x &= \frac{2 \pm \sqrt{4 - 4(3 \pm \sqrt{5})}}{2} \\
&&&= 1 \pm \sqrt{\mp \sqrt{5} -2} \\
&&& = 1 \pm \sqrt{\sqrt{5}-2}, 1 \pm i\sqrt{\sqrt{5}+2}
\end{align*}