已知关于x的方程x^2+2mx+2m+3=0的两个不等实根都在区间(0,2),则实数m的取值范围是_________.下面是我解的,为什么跟答案对不上?Δ=b²-4ac=4m²-8m-12>0f(0)=2m+3>0m>-3/2=-9/6f(2)>04+4m+2m+3>0m>
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![已知关于x的方程x^2+2mx+2m+3=0的两个不等实根都在区间(0,2),则实数m的取值范围是_________.下面是我解的,为什么跟答案对不上?Δ=b²-4ac=4m²-8m-12>0f(0)=2m+3>0m>-3/2=-9/6f(2)>04+4m+2m+3>0m>](/uploads/image/z/3713625-9-5.jpg?t=%E5%B7%B2%E7%9F%A5%E5%85%B3%E4%BA%8Ex%E7%9A%84%E6%96%B9%E7%A8%8Bx%5E2%2B2mx%2B2m%2B3%3D0%E7%9A%84%E4%B8%A4%E4%B8%AA%E4%B8%8D%E7%AD%89%E5%AE%9E%E6%A0%B9%E9%83%BD%E5%9C%A8%E5%8C%BA%E9%97%B4%280%2C2%29%2C%E5%88%99%E5%AE%9E%E6%95%B0m%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4%E6%98%AF_________.%E4%B8%8B%E9%9D%A2%E6%98%AF%E6%88%91%E8%A7%A3%E7%9A%84%2C%E4%B8%BA%E4%BB%80%E4%B9%88%E8%B7%9F%E7%AD%94%E6%A1%88%E5%AF%B9%E4%B8%8D%E4%B8%8A%3F%CE%94%3Db%26%23178%3B-4ac%3D4m%26%23178%3B-8m-12%26gt%3B0f%280%29%3D2m%2B3%26gt%3B0m%26gt%3B-3%2F2%3D-9%2F6f%282%29%26gt%3B04%2B4m%2B2m%2B3%26gt%3B0m%26gt%3B)
已知关于x的方程x^2+2mx+2m+3=0的两个不等实根都在区间(0,2),则实数m的取值范围是_________.下面是我解的,为什么跟答案对不上?Δ=b²-4ac=4m²-8m-12>0f(0)=2m+3>0m>-3/2=-9/6f(2)>04+4m+2m+3>0m>
已知关于x的方程x^2+2mx+2m+3=0的两个不等实根都在区间(0,2),则实数m的取值范围是_________.
下面是我解的,为什么跟答案对不上?
Δ=b²-4ac=4m²-8m-12>0
f(0)=2m+3>0
m>-3/2=-9/6
f(2)>0
4+4m+2m+3>0
m>-7/6
-m<2 => m>2
-m>0 => m<0
无解
f(-m)=m²-2m²+2m+3<0
=-m²+2m+3<0
b²-4ac=4-4*(-1)*3=16>0
(-2±4)/-2=1±2
-1<m<3
此时解集为m∈(-1,3)
哪里错了?
已知关于x的方程x^2+2mx+2m+3=0的两个不等实根都在区间(0,2),则实数m的取值范围是_________.下面是我解的,为什么跟答案对不上?Δ=b²-4ac=4m²-8m-12>0f(0)=2m+3>0m>-3/2=-9/6f(2)>04+4m+2m+3>0m>
-m-2,不是m>2.
另,印刷版的答案完美,f(-m)是不需要考虑的.
m自身就是小于零的,-m<2说明m大于-2
Δ=b²-4ac=4m²-8m-12>0 m<-1或m>3
f(0)=2m +3>0 m>-3/2
f(2)=6m+7>0 m>-7/6
f(-m)=-m.m+2m+3<0 m<-1或m>3
解得: -7/6