Question

已知集合A={x/x²-4x-2a=6=0}B={x/x<0},若A∩B≠空集,求实数a的取值范围

Answer 1

A={x|x²-4x-2a+6=0},
若A∩B≠空集,则 A≠ø 且方程只能有非负根
从而 △= 16+4(2a-6)≥ 0 即 8a ≥ 8 ，a ≥ 1
且两根之积 -2a+6≥ 0 即a≤3
所以 1≤a≤3

Answer 2

A={x|x²-4x-2a=6=0}, ???? =6=0
I suppose to be A={x|x²-4x-2a+6=0}
B={x|x<0},

if A∩B≠ø, then A≠ø

A={x|x²-4x-2a+6=0}, for real roots
△≥ 0
=> 16+4(2a-6)≥ 0
8a ≥ 8
a ≥ 1

AND
also root of equation x²-4x-2a+6=0
is less than 0

roots of equation
= 2+√(-2+2a) or 2+√(-2+2a)

ie
2+√(-2+2a) < 0 (rejected)
or
2-√(-2+2a) < 0
a > 3
ie a>3

combine (a > 3) and a ≥ 1

we get
a > 3 #