Question

若方程x平方+2(k+1)x+k平方-1=0有2个实根x1和x2,求x1平方+x2平方的最小值

Answer 1

△ = 4(k + 1)² - 4(k² - 1) ≥ 0

解得:k ≥ -1

根据韦达定理

x1 + x2 = -2(k + 1)

x1 * x2 = k² - 1

x1² + x2²

= (x1 + x2)² - 2x1 * x2

= 4(k + 1)² - 2(k² - 1)

= 4k² + 8k + 4 - 2k² + 2

= 2k² + 8k + 6

= 2(k + 2)² - 2

当 k = - 1 时,取最小值 0