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Suppose $x=[x_1, x_2, \cdots, x_n]^t$, $b=[b_1,b_2,\cdots,b_n]^t$ with $b_i\in K$ and $A\in M_n(K)$, where $K$ is a field. There are well known criteria for the system of equations $Ax=b$, by considering rank of $A$ and $[A|b]$.

If we consider the equation $x^tAx=\lambda$, for $\lambda \in K$, what are the criteria for the existance of solution and simple methods to solve it?

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I would expect nothing nice in general, we can run into number-theoretic issues. For fields closed under square root, there is almost nothing to do. For ordered fields closed under square root of positive elements, calculation should be easy. – André Nicolas Aug 22 '11 at 6:38
@user10889: added tags to reflect the actual complexity of the question. There are also algebraic geometry difficulties when $K$ is of positive transcendence degree, and undecidability (similar to Hilbert's 10th problem). – zyx Aug 22 '11 at 6:58
@zyx Why have you added algebraic-k-theory, actually? – Grigory M Aug 22 '11 at 7:45
@Grigory: to raise the probability an expert will see the question. Quadratic symbols are related to K_2, local-global principles to localization sequences in K-theory, and questions about vector bundles appear when the field (also called K in the question) is the function field of a positive-dimensional variety. Milnor K-theory of fields is certainly relevant and possibly Quillen's higher algebraic K-theory as well. – zyx Aug 22 '11 at 7:54
From the perspective of linear algebra, $\mathrm{vec}(x^TAx)=\mathrm{vec}(A)\cdot x\otimes x=\lambda$. Write $\mathrm{vec}A=a$. Then $x\otimes x=a(a^Ta)^{-1}\lambda+x_0$ with $a^T x_0=0$. If you get $x\otimes x$, $x$ can be determined. – Shiyu Aug 22 '11 at 9:02

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