Question

已知x1,x2是方程x²-2px-(6p+5)=0的两个实数根,问实数p为何值时,表达式x1²+x2

Answer 1

∵方程x²-2px-(6p+5)=0有两个实数根,

∴(-2p)²-4[-(6p+5)]≥0,即p²+6p+5≥0,解得p≤-5或p≥-1,

∵方程的两根x1、x2满足x1+x2=2p,x1x2=-(6p+5),

∴x1²+x2²=(x1+x2)²-2x1x2=4p²+12p+10=(2p+3)²+1,

设f(p)=(2p+3)²+1,则该函数在区间(-∞,-5]上单减,在区间[-1,+∞)上单增,

且f(-5)=50,f(-1)=2,

∴f(p)的最小值是2,即x1²+x2²的最小值是2.