Question

已知方程x²-2x-1=0,利用根与系数的关系求一个一元二次方程,使它的根事原方程各根的(1)平方；(2)相反

Answer 1

x1+x2=2 x1x2=-1

(1)平方:x1²+x2²=x1²+2x1x2+x2²-2x1x2=(x1+x2)²-2x1x2=2²-2(-1)=5+2=7

x1²x2²=(x1x2)²=(-1)²=1

方程为x²-7x+1=0

(2)相反数：-x1-x2=-(x1+x2)=-2

(-x1)(-x2)=x1x2=-1

方程为x²+2x-1=0