x的一元二次方程x方-mx+2m-1=0的两个实数根分别是x1,x2,且x1+x2=7,则(x1-x2)的括号平方等于多少

问题描述:

x的一元二次方程x方-mx+2m-1=0的两个实数根分别是x1,x2,且x1+x2=7,则(x1-x2)的括号平方等于多少

在 x ² - m x + 2 m - 1 = 0 中,
a = 1,b = - m ,c = 2 m - 1
x1 + x2 = - b / a = - (- m)/ 1 = m = 7
x1x2 = c / a = (2 m - 1)/ 1 = 2 m - 1
∵ (x1 - x2)²
= x1² - 2 x1x2 + x2²
= x1² + 2 x1x2 + x2² - 2 x1x2 - 2 x1x2
= (x1 + x2)² - 4 x1x2
又∵ x1 + x2 = m = 7
∴ (x1 + x2)² - 4 x1x2
= 7 ² - 4 (2 m - 1)
= 49 - 4 (2 × 7 - 1)
= 49 - 4 (14 - 1)
= 49 - 56 + 4
= - 7 + 4
= - 3