已知实数a是x2﹣5x﹣14=0的根,不解方程,求(a﹣1)(2a﹣1)﹣(a+1)2+1的值.
解:∵实数a是x2﹣5x﹣14=0的根,
∴a2﹣5a﹣14=0,即a2﹣5a=14,
(a﹣1)(2a﹣1)﹣(a+1)2+1=2a2﹣a﹣2a+1﹣(a2+2a+1)+1
=2a2﹣3a+1﹣a2﹣2a﹣1+1
=a2﹣5a+1
=14+1<
若关于x的一元二次方程kx2+2x-1=0有实数根,则k的取值范围是( )
命题:①对顶角相等;②垂直于同一条直线的两直线平行;③相等的角是对顶角;④同位角相等.其中假命题有( )
-- Mum, can I go to the zoo with Jack?
-- When your homework , you can.