∵方程2x2+kx-2k+1=0有两个实数根，∴△=k2-4×2（-2k+1）≥0，解得，k≥62-8或k＜-62-8．设方程2x2+kx-2k+1=0两个实数根为x1、x2．则x1+x2=-k2，x1•x2=-k+12．∴x12+x22=（x1+x2）2-2x1x2=k24+2k-1=294，即k2+8k-3...