jmc

Let $\alpha$ and $\beta$ be the roots of $x^2+ux+49,$ which is a factor of $x^4+k{x}^2+90x-2009.$ Then the other factor must be of the form $x^2+vx-41.$ Thus, $$(x^2+ux+49)(x^2+vx-41)=x^4+k{x}^2+90x-2009.$$Expanding, we get $$x^4+(u+v)x^3+(uv+8)x^2+(-41u+49v)-2009=x^4+k{x}^2+90x-2009.$$Matching coefficients, we get $$\begin{array}{rl}u+v& =0,\\ uv+8& =k,\\ -41u+49v& =90.\end{array}$$Solving the system $u+v=0$ and $-41u+49v=90,$ we find $u=-1$ and $v=1.$ Therefore, $k=uv+8=\boxed{7}.$