若方程2x²-3x-4=0的两根是x1,x2求(x1+1)(x2+1)的值 2.求x1²+x2²

问题描述:

若方程2x²-3x-4=0的两根是x1,x2求(x1+1)(x2+1)的值 2.求x1²+x2²

由韦达定理
x₁+x₂ = 3/2
x₁x₂ = - 2
1、(x₁+1)(x₂+1)
= x₁x₂ + (x₁+x₂) +1
= - 2 + 3/2 + 1
= 1/2
2、x₁² + x₂²
= (x₁+x₂)² - 2x₁x₂
= 9/4 + 4
= 25/4