Hope it is a right place to ask how to solve the equation on $\mathbf x$: $$\mathbf x^T \mathbf A\mathbf x + \mathbf x^T \mathbf b + c = 0.$$ where:
$\mathbf x$ is an $n\times 1$ column vector
$\mathbf A$ is an $n\times n$ matrix
$\mathbf b$ is an $n\times 1$ vector
$c$ is a scalar
Thanks

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Welcome to MSE! The set of solutions of your equation forms a quadratic hypersurface. The procedure of finding it is not too difficult, but it takes a little bit of space, so I'm not going to try to describe it to you. May be somebody else will? The principal axis theorem plays a key role (it has an $n$-dimensional analogue). Hopefully you have seen images of spheres, ellipsoids, parabaloids, cones, hyperboloids, parabolic hyperboloids (=saddle surface) et cetera in the case $n=3$. Something very similar will happen here. –  Jyrki Lahtonen Jul 20 '11 at 19:55
If $A$ happens to be symmetric, then the solution of your equation is equivalent to finding all points on a quadric in $\mathbb{R}^n$. –  Shaun Ault Jul 20 '11 at 20:00
@Shaun: Does the equation not stay the same, if we replace $A$ with $(A+A^T)/2$? So we can always assume that $A$ is symmetric? –  Jyrki Lahtonen Jul 20 '11 at 20:19
I cleaned up the math notation, and a notice says that my edits await "peer review". –  Michael Hardy Jul 20 '11 at 20:40
@Michael: They've been reviewed in the meantime. You can edit with reputation $\ge2000$. BTW, are you the Michael Hardy from Wikipedia? If so, I'm glad you found your way here :-) (I'm also joriki from Wikipedia.) –  joriki Jul 20 '11 at 21:02

Hint 1: We can assume ${\bf A}$ is symmetric. If it's not, we can replace it by ${\bf B}=({\bf A}+{\bf A}^T)/2$, because ${\bf x}^T {\bf A}{\bf x} = {\bf x}^T {\bf B}{\bf x}$ (check it). (That's why quadratic equations are usually expressed using symmetric matrices; we don't lose generality).
Hint 2: This is the generalization of the (scalar) quadratic $$a^2x + b x + c = 0$$ Do you know how to solve it (completing the square)? If so, try to generalize the procedure. If not, learn it.
Hint 3: Consider the special case ${\bf x}^T {\bf x} = v$. If $v$ is positive, the solutions lie on a sphere. Now, if ${\bf x^T A x} = v$ , if we can write ${\bf A} = {\bf P \Lambda P^T }$ (we can if A is symmetric), we make a change of variable ${\bf z} = {\bf P^T x}$ (a rotation of axis) and we get the equation of a (hyper)ellipse, if $v$ is positive and ${\bf \Lambda }$ is diagonal with positive entries.
Not sure why you're making $v$ bold when it's a scalar... –  anon Jul 22 '11 at 13:07