Question

x²-lxl -2 > 0 then find value of x

Answer 1

case:1
when x≥0
  x^2 - x - 2 > 0
x^2 - 2x + x -2 >0
x(x -2) +1(x - 2)>0
(x - 2)(x + 1) >0
x >2 or x< -1 but x>0 so, x>2 is the solution in case:1

case:2
when x<0
x^2 +x -2 >0
x^2 +2x -x -2>0
x(x +2 ) -1(x +2 )>0
(x -1)(x + 2) >0
x>1 or x<-2 but x<0 so, x<-2 is the solution of case:2

hence final solution x∈(2,∞)∪(-∞,-2)