已知方程x²-2mx+m+6=0的两实根是x1,x2,求f(m)=x1²+x2²的最小值.
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/28 12:41:49
![已知方程x²-2mx+m+6=0的两实根是x1,x2,求f(m)=x1²+x2²的最小值.](/uploads/image/z/14606191-55-1.jpg?t=%E5%B7%B2%E7%9F%A5%E6%96%B9%E7%A8%8Bx%26%23178%3B-2mx%2Bm%2B6%3D0%E7%9A%84%E4%B8%A4%E5%AE%9E%E6%A0%B9%E6%98%AFx1%2Cx2%2C%E6%B1%82f%28m%29%3Dx1%26%23178%3B%2Bx2%26%23178%3B%E7%9A%84%E6%9C%80%E5%B0%8F%E5%80%BC.)
已知方程x²-2mx+m+6=0的两实根是x1,x2,求f(m)=x1²+x2²的最小值.
已知方程x²-2mx+m+6=0的两实根是x1,x2,求f(m)=x1²+x2²的最小值.
已知方程x²-2mx+m+6=0的两实根是x1,x2,求f(m)=x1²+x2²的最小值.
x²-2mx+m+6=0的两实根是x1,x2,
x1x2=m+6
x1+x2=2m
f(m)=x1²+x2²
=(x1+x2)²-2x1x2
=4m²-2m-12
=4(m-1/4)²-49/4
又Δ=4m²-4(m+6)≥0
m²-m-6≥0
(m+2)(m-3)≥0
m≥3或m≤-2
3-1/4=11/4
|-2-1/4\=9/4
所以
m=-2时取最小值=4×4+2×2-12=8
判别式≥0,则m的取值范围为 -2≤m≤3
x1²+x2²
=(x1+x2)²- 2x1x2
=(2m)² - 2(m+6)
= 4m² - 2m - 12
= 4(m - 0.25)² - 12.25
当m=0.25的时候取最小值为 -12.25