|x-6|>|x2-5x+9|

Question

how to solve.. its very urgent

Answer 1

|x-6|>|x²-5x+9|

x-6>0 when x>6.

x²-5x+9 has only complex zeroes and is always greater than zero. So the abs can be removed.

|x-6|>x²-5x+9.

When x>6: x-6>x²-5x+9 has to be solved. x²-6x+15<0 is false because this quadratic has only complex zeroes and is always greater than zero. Therefore we need to consider x-6<0.

6-x>x²-5x+9, x²-4x+3<0, (x-1)(x-3)<0 only when 1<x<3, which is the solution.