设x1x2为方程x²-kx(x-2)+2-k=0的两个实数根.且x1平方+x1x2+x2平方=11/2求k

设x1x2为方程x²-kx(x-2)+2-k=0的两个实数根.且x1平方+x1x2+x2平方=11/2求k
设x1x2为方程x²-kx(x-2)+2-k=0的两个实数根.且x1平方+x1x2+x2平方=11/2求k

设x1x2为方程x²-kx(x-2)+2-k=0的两个实数根.且x1平方+x1x2+x2平方=11/2求k
x^2-kx(x-2)+2-k=0
(1-k)x^2+2kx+2-k=0
x1+x2=-2k/(1-k)=2k/(k-1)
x1x2=(2-k)/(1-k)=(k-2)/(k-1)
x1^2+x1x2+x2^2
=(x1+x2)^2-x1x2
=[2k/(k-1)]^2-(k-2)/(k-1)
=[4k^2-(k-1)(k-2)]/(k-1)^2
=(4k^2-k^2+3k-2)/(k-1)^2
=(3k^2+3k-2)/(k-1)^2=11/2
2(3k^2+3k-2)=11(k-1)^2
6k^2+6k-4=11k^2-22k+11
5k^2-28k+15=0
(5k-3)(k-5)=0
k=3/5 或k=5

用韦达定理K=5 δ》0 X=五分之3舍去？

方程x²-kx(x-2)+2-k=0可变形为（1-k）x²+2kx+（2-k)=0 (k≠1)
△=12k-8>0 .k>2/3
∴x1+x2=2k/(k-1) ,x1x2=(2-k)/(1-k)
∴(x1+x2)²=x1²+2x1x2+x2²=[2k/(k-1)]²
∵x1²+x1x2+x2&...

全部展开

方程x²-kx(x-2)+2-k=0可变形为（1-k）x²+2kx+（2-k)=0 (k≠1)
△=12k-8>0 .k>2/3
∴x1+x2=2k/(k-1) ,x1x2=(2-k)/(1-k)
∴(x1+x2)²=x1²+2x1x2+x2²=[2k/(k-1)]²
∵x1²+x1x2+x2²=11/2
∴[2k/(k-1)]²-x1x2=[2k/(k-1)]²-(2-k)/(1-k)=11/2
即：5k²-28k+15=0
解得：k=3/5(舍去)或k=5
∴k=5

收起