We would like to represent the polynomial t^2+2*b*t-c as the characteristic polynomial of a 2x2 matrix. By translating the origin, we can get rid of b, which makes the following analysis more perspicuous. So we have p(t)=t^2-c as the polynomial we want to represent by the symmetric real matrix [x y] [y z] = M. This matrix M satisfies its own characteristic polynomial, so we have the two equations: y*(x+z)=0 x^2+y^2-c=0 The first equation gives z=-x, while the second gives x=+- sqrt(c-y^2). So, M is [-sqrt(c-y^2) y] [y sqrt(c-y^2)] Thus, so long as |y|<sqrt(c), M will represent p(t). In general, for 2x2 matrices, y^2 enters as an addition to the discriminant, so we have some freedom to choose y if the discriminant is positive. Unfortunately, we have effectively solved the polynomial p(t) already, prior to constructing M, so it would appear that constructing M is not easier than solving p(t).