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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 :)

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    $\begingroup$ what's inefficient about using the formula? $\endgroup$ – danimal May 12 '15 at 11:44
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    $\begingroup$ I would complete the square rather than use the formula. Just as quick but so much more informative. $\endgroup$ – Paul May 12 '15 at 11:50
  • $\begingroup$ @danimal I am not saying that it is inefficient I was just wondering whether there was anything more efficient. $\endgroup$ – anonymous May 12 '15 at 11:50
  • $\begingroup$ @Paul Yes, I guess that would work but it still isn't very different form what usually do. I was hoping that there were a complete alternative. For example, we can solve simultaneous equations using elimination, substitution or even by using matrices. $\endgroup$ – anonymous May 12 '15 at 11:52
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    $\begingroup$ elimination and matrix methods are simply flavours of the same thing really, as are using the equation or completing the square for a quadratic. $\endgroup$ – Paul May 12 '15 at 12:46
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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 $x^2 + ax+b$.

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  • $\begingroup$ Generates the same quadratic?? $\endgroup$ – Paul May 12 '15 at 12:39
  • $\begingroup$ .........................lol $\endgroup$ – Paul May 12 '15 at 12:47
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    $\begingroup$ I love the spoiler!!! I am waiting for the next episode :)! $\endgroup$ – Chinny84 May 12 '15 at 12:48
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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: $$\pm \sqrt{b^2 - c} - b$$

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