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Suppose I have the equation:

$\vec A x^2 + \vec Bx + \vec C = 0$

If $\vec A$, $\vec B$, and $\vec C$ where scalars I would use the quadratic formula:

$x = \frac{\vec B \pm \sqrt{\vec B^2 - 4 \vec A \vec C}}{2 \vec A}$

The problem is that this involves vector multiplication, and I am not sure what is the correct vector multiplication to use in this case if any......

So what is the general strategy to solve such equations?

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  • $\begingroup$ i think your problem is when you are dividing by a vector $\endgroup$ – Dr. Sonnhard Graubner Sep 16 '17 at 13:15
  • $\begingroup$ @Dr.SonnhardGraubner That is part of the thing that I cannot figure out....yes $\endgroup$ – DarthRubik Sep 16 '17 at 13:20
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$\newcommand{\vect}[1]{{\bf #1}}$

Assume you can represent your vectors in some basis: $\vect{A} = \sum_k A_k\hat{\vect{e}_k}$, with similar expression for $\vect{B}$ and $\vect{C}$, your equation then becomes

\begin{eqnarray} \left(\sum_k A_k \hat{\vect{e}_k} \right) x^2 + \left(\sum_k B_k \hat{\vect{e}_k} \right) x + \left(\sum_k C_k \hat{\vect{e}_k} \right) &=& 0 \\ \sum_k\left( A_k x^2 + B_k x + C_k \right) \hat{\vect{e}_k} &=&0 \\ A_k x^2 + B_k x + C_k &=& 0 ~~~\mbox{for}~~~ k = 1,2,\cdots \end{eqnarray}

Which can be solved for each component. A solution exists only if these solutions are consistent with each other

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