Sorry, but this question is not clear.  What is the original problem that you want to solve?  Do you perhaps wish to minimise or maximise the function x^TQx + c?  If so, are there any constraints that must be satisfied, such as non-negativity?

Thank you for your kind response. The original problem is Quadratic constrained Quadratic Programming. So, we can assume any quadratic constraint. But first my question is about matrix A. If det(A) is equal to the original quadratic equation, how come the three equations doesn't hold for some values of x1 and x2?

Quadratically constrained quadratic programming (QCQP) is NP-hard, so it can't be reduced to an LP.  If the objective and constraint functions are convex, then QCQP can be solved efficiently, e.g., by interior-point methods, but still cannot be reduced to an LP.

If you write out the first-order KKT conditions, then you get linear equations, but the complementarity condition is still quadratic (and non-convex).

Now, suppose you wish to minimise a quadratic term of the form x^TQx, possibly subject to linear and/or convex quadratic constraints.  If this term is convex, i.e., if Q is positive semidefinite, then Q can be factorised as A^TA. Thus, you can re-write the term as x^TA^TAx, which is equivalent to ||Ax||_2^2.  A popular transformation is then to add a new vector of variables, say y, and the equations y = Ax.  The quadratic term then simplifies to ||y||_2^2.  Although it is still non-linear, it can be handled efficiently using second-order cone (SOC) programming: minimise a continuous variable z, subject to the SOC constraint z >= ||y||_2 and the linear equations y = Ax (plus non-negativity if it is present).

If your objective function also contains a linear term, then there is a more complicated transformation to a SOC program.  Convex quadratic constraints can be handled in the same way.  See, e.g., this survey paper:

http://link.springer.com/article/10.1007/s10107-002-0339-5

Thank you for your thorough answer. From your answer I can deduce that, we cannot use the linear equations form from matrix A instead of the quadratic equation, right?

I have some more questions. Please answer it categorically.

If I have the Problem , assuming the Q matrix as positive semi definite.

minimize 2*x12+5*x22+3*x1*x2

subject to  x12+3*x22=1

1) Is it possible to write the quadratic equation of the constraint in linear terms? If yes, how? In other words, is it possible to transform this problem into Quadratic programming with linear constraints?

2)Is it possible to approximate the quadratic equation with linear terms? If yes, how?

3) Is there any relation of the eigenvalues of the constraint equation with the given function?

1. The equation x1^2 + 3 x2^2= 1 is non-convex, and therefore cannot be reformulated in terms of linear constraints.

2. The reformulation-linearization technique (RLT) of Sherali & Adams provides a way to approximate non-convex polynomial equations using linear constraints.  Alternatively, you could derive a semidefinite programming relaxation by defining the matrix variable X = xx^T and then relaxing it to the constraint that X - xx^T must be positive semidefinite (see papers by Shor, Lasserre, Parrilo etc.)

3. I'm not sure what you mean here.  Of course, a quadratic function is convex if and only if the matrix of quadratic terms is positive semidefinite, which in turn is true if and only if the matrix has non-negative Eigenvalues.   The same condition can be used to check if a quadratic *inequality* is convex.

So if the quadratic constraint is convex, then we can reformulate it as linear constraints?

If a quadratic inequality constraint is convex, it can be reformulated as a second-order cone constraint, not a linear constraint.  As for quadratic equality constraints, they are convex if and only if they are linear!

I recommend a good textbook on convex optimisation.  The one by Boyd & Vandenberghe is good, and it is available for free on the web.

 Thank you very much. Your feedback helped me a lot.