Can you solve a quadratic equation using matrices? I was wondering whether there are any alternatives or more efficient methods to finding a solution to a quadratic equation other than simply trial and error or by using the quadratic formula.
I was once told that it could be very easily done using matrices. How would this work?
Additionally, are there any other "better" alternatives?
I would really appreciate if you were to give me examples and explain how to solve them with the alternate method.
Thank you :)
 A: You can solve the quadratic equation $x^2 + ax+b=0$ by computing the eigenvalues of the companion matrix
$$
\pmatrix{ 0 & - b \\ 1  & -a}.
$$
Simple and efficient.

 Of course not! It requires to solve exact the same equation most of the time. The characteristic polynimial of the matrix is $x^2 + ax+b$.

A: You could use Newton's method.  If you draw a triangle, you see that for
$$\frac{f(x)}{x - x'} = f'(x)$$
that $f(x')$ is close to zero.  Since you want that $x'$, you can solve for $x'$:
$$x' = x - \frac{f(x)}{f'(x)}$$
Applying $f(x) = ax^2 + bx + c$, you get that 
$$x' = \frac{ax^2 - c}{2ax + b} \approx 0$$
So use any given starting $x$ to find a closer $x'$, and use that to find a closer $x''$ by applying the same formula again, etc.
If you want a rational approximation, you can use $x' = \frac{p'}{q'}$, and $x = \frac{p}{q}$:
$$\frac{p'}{q'} = \frac{a\left(\frac{p}{q}\right)^2 - c}{2a\left(\frac{p}{q}\right) + b} = \frac{ap^2 + cq^2}{2apq + bq^2}$$
So repeatedly applying:

$$\begin{cases} p' = ap^2 - cq^2 \\ q' = 2apq + bq^2 \end{cases}$$

to almost any starting value will let you quickly approximate a root of a quadratic, as long as $p/q$ never becomes exactly the min or max of the quadratic.  I guess you might use something like this in hardware if you have to quickly compute a root and don't care much about accuracy.  I'd still probably use the quadratic formula though.

If you encounter a quadratic formula of the form $x^2 + 2bx + c$, which is a bit of a standard form for some uses (like conics), then the quadratic formula actually becomes very simple:  $$\pm \sqrt{b^2 - c} - b$$
