The exact meaning of conjugate radical roots theorem 
Here, what are exactly $s$. $t$ and $u$? Are they rational numbers with $u$ being nonnegative? There is no exact explanation of the theorem. So I am confused..
 A: Let $u\in \Bbb Q \setminus \{q^2:q\in \Bbb Q\}.$ Let $\sqrt u$ denote some (fixed) $z\in \Bbb C$ such that $z^2=u.$
Let $\Bbb Q[z]=\{a=s+tz:s,t\in \Bbb Q\}.$ 
Exercise : $\Bbb Q \subset \Bbb Q[z],$ and  $\forall x,x'\in \Bbb Q[z]\;(x+x'\in \Bbb Q[z]\land xx'\in \Bbb Q[z]).$ 
Exercise: If $s,s',t,t'\in \Bbb Q$ then $s+tz=s'+t'z\iff (s=s'\land t=t').$ Corollary: For each $x\in \Bbb Q[z]$ there is a unique pair $(s,t)\in \Bbb Q^2$ such that $x=s+tz.$ 
For $s,t\in \Bbb Q$ and $x=s+tz$ let $f(x)=s-tz.$ Note that $f(x)$ is well-defined by the above corollary. 
Exercise: 
(i). $\forall q\in \Bbb Q\;(q=f(q)).$ 
(ii). $\forall x\in \Bbb Q[z]\;( f(x)=0\iff x=0)$.
(iii). $\forall x,x'\in \Bbb Q[z]\;(f(x+x')=f(x)+f(x')\land f(xx')=f(x)f(x')).$
(iv). By (iii) and by induction on $j\in \Bbb N$ we have $\forall j\in \Bbb N\;(f(z^j)=f(z)^j\;).$
From these exercises, if $x\in Q[z]$ then $p(x)\in \Bbb Q[z]$ and $$f(p(x))=f(\sum_{j=0}^na_jx^j) 
 =\sum_{j=0}^nf(a_jx^j) =$$ $$=\sum_{j=0}^nf(a_j)f(x^j)=\sum_{j=0}^n a_jf(x)^j=p(f(x)).$$ So for any $x \in \Bbb Q[z],$ we have $p(x)=0\iff p(f(x))=0.$
Some parts of the exercises are obvious but all of it is needed for the conclusion.
A: All three of $s,t,$ and $u$ are rational numbers. For a proof of the theorem, see here.
By this theorem (which is what comes to my mind first when I hear "the conjugate root theorem"), $u$ does not need to be non-negative.
A: In simple terms, the rational root theorem states:

For a function $f(x)$, if that function has a root in the form of $a+bi$, where $a$ and $b$ are coefficients, then that function also has a root at $a-bi$.
That is, if $f(a+bi)=0$, then $f(a-bi)$ must equal $0$.

