
设x1、x2是一元二次方程2x²-3x-5=0的两根,不解方程求x1²+3 x2²-3 x2的值
展开全部
2x²-3x-5=0
(2x-5)(x+1)=0
故
x1+x2=1.5
x1x2=-2.5
x1²+3 x2²-3 x2
=x1²+ x2²+(2x²-3x-5)+5
=x1²+ x2² +5
=(x1+x2)²-2x1x2+5
=2.25+5+5
=12.25
(2x-5)(x+1)=0
故
x1+x2=1.5
x1x2=-2.5
x1²+3 x2²-3 x2
=x1²+ x2²+(2x²-3x-5)+5
=x1²+ x2² +5
=(x1+x2)²-2x1x2+5
=2.25+5+5
=12.25
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询