27 Let $x_{1}, x_{2}, x_{3}, x_{4}$ denote the four roots of the equation
$$
x^{4}+k x^{2}+90 x-2009=0 \text {. }
$$

If $x_{1} x_{2}=49$, find the value of $k$.
27 Answer: (7)
From given equation, $x_{1}+x_{2}+x_{3}+x_{4}=0, x_{1} x_{2}\left(x_{3}+x_{4}\right)+x_{3} x_{4}\left(x_{1}+x_{2}\right)=-90$ and $x_{1} x_{2} x_{3} x_{4}=-2009$. Now $x_{1} x_{2}=49 \Rightarrow x_{3} x_{4}=-41$. So $x_{1}+x_{2}=1, x_{3}+x_{4}=-1$. Hence $k=x_{1} x_{2}+x_{1} x_{3}+x_{1} x_{4}+x_{2} x_{3}+x_{2} x_{4}+x_{3} x_{4}=49+(1) \times(-1)-41=7$.