Question: > 
> Prove that there is no irreducible polynomial of degree 2 in $\mathbb C[x]$.
> 
> 
> 

I know this result is true from the Fundamental Theorem of Algebra, which states that any polynomial in $\mathbb C[x]$ with degree $n\ge 1$, has at least $1$ root in $\mathbb C$.

However, this question is asked in the textbook before the FTA is introduced. How can I go about proving it?

Answer: 
I'm going to use $t$ instead of $f$.

The points $A + tB$ lie on a line through the point $A$ in the direction of the vector $B$. You are trying to find the values of $t$ that give you the intersections of this line with a circle $C$ of radius $M$ centered at the origin. 

If the point $A$ lies inside the circle $C$, there will be two values of $t$, one negative and one positive. If the point $A+B$ lies outside the circle $C$, then the positive $t$ value will be less than $1$.

So how to find the desired value of $t$.

We have:

\begin{align}
&A + tB \;\text{ lies on the circle } C \\
\Leftrightarrow\;& \|A + tB\| = M \\
\Leftrightarrow\;& \|A + tB\|^2 = M^2 \\
\Leftrightarrow\;& (A + tB) \cdot (A + tB) = M^2 \\
\Leftrightarrow\;& A \cdot A + 2t(A \cdot B) + t^2(B.\cdot B) = M^2 
\end{align}
The desired solution of the quadratic is:
$$
t = \frac{-A \cdot B + \sqrt{ (A \cdot B)^2 - (B.B) (A \cdot A - M^2)} } { B \cdot B}
$$

The necessary code would be:

```
double AA = x*x + y*y;     // A dot A
double BB = u*u + v*v;     // B dot B
double AB = u*x + v*y;     // A dot B

double d = sqrt(AB*AB - BB*(AA - M*M));

double t = (d - AB)/BB;

```

This is roughly the same formula you derived, but in a rather tidier form (in my opinion), and with slightly less arithmetic. I don't think you can do any better than this. The only part that will take any time is the square root, and I think this is unavoidable. You should be able to do this calculation millions of times a second, even on a wristwatch. Iterative algorithms will be much slower, as you already discovered.
Iterative approximation to the required value of $f$ seems to fit to your needs, if I understood well.
You want$$
\left( {x + f\,u} \right)^{\,2} + \left( {y + f\,v} \right)^{\,2} - M^{\,2} \le 0
$$
i.e.
$$
\,f^{\,2} + 2\,{{\left( {x\,u\, + \,y\,v} \right)} \over {\left( {u^{\,2} + \,v^{\,2} } \right)}}\,f + {{x^{\,2} + y^{\,2} - M^{\,2} } \over {\left( {u^{\,2} + \,v^{\,2} } \right)}} \le 0
$$
where the left term is a vertical parabola, with axis at $
f = - \,{{\left( {x\,u\, + \,y\,v} \right)} \over {\left( {u^{\,2} + \,v^{\,2} } \right)}}
$ , etc.  

Now fix $f=1$ and calculate the ordinate: if negative that is ok, if positive apply the Newton-Raphson iteration and in a few steps you will get a good approximation of the required value for $f$.
Adjust the algorithm to fix for precision / max no. of steps, banal values, etc.