若方程2x²-3x-4=0的两根是x1,x2求(x1+1)(x2+1)的值 2.求x1²+x2²
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![若方程2x²-3x-4=0的两根是x1,x2求(x1+1)(x2+1)的值 2.求x1²+x2²](/uploads/image/z/3135610-10-0.jpg?t=%E8%8B%A5%E6%96%B9%E7%A8%8B2x%26%23178%3B%EF%BC%8D3x%EF%BC%8D4%EF%BC%9D0%E7%9A%84%E4%B8%A4%E6%A0%B9%E6%98%AFx1%2Cx2%E6%B1%82%EF%BC%88x1%EF%BC%8B1%EF%BC%89%EF%BC%88x2%2B1%EF%BC%89%E7%9A%84%E5%80%BC+2.%E6%B1%82x1%26%23178%3B%EF%BC%8Bx2%26%23178%3B)
若方程2x²-3x-4=0的两根是x1,x2求(x1+1)(x2+1)的值 2.求x1²+x2²
若方程2x²-3x-4=0的两根是x1,x2求(x1+1)(x2+1)的值 2.求x1²+x2²
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
韦达定理