k取什麽值时,方程(k-2)x²-2kx+k-1=0有两个相等的实数根?并求出这时方程的根.
由△=4k²-4(k-2)(k-1)=4k²-4(k²-3k+2)=12k-8=0,得k=8/12=2/3.
x₁=x₂=2k/[2(k-2)]=(4/3)/[2(2/3-2)]=(4/3)/(-8/3)=-1/2.
k取什麽值时,方程(k-2)x²-2kx+k-1=0有两个相等的实数根?并求出这时方程的根.
由△=4k²-4(k-2)(k-1)=4k²-4(k²-3k+2)=12k-8=0,得k=8/12=2/3.
x₁=x₂=2k/[2(k-2)]=(4/3)/[2(2/3-2)]=(4/3)/(-8/3)=-1/2.