Help with basic calculus :) - Easy 10 Points.?

Question

Determine the possible values 'x' can take for;

|x^2 - 6x -1| < 6

the brackets are absolute signs. :)

Thanks in advance

Anonymous · Accepted Answer

-6 < x^2 - 6x - 1 < 6, so
(x-1)(x-5) >0 and (x-7)(x+1) <0 
(-1, 1) U (5,7) Answer

Redeemed · Answer

Mundilfo is correct...whenever u deal with absolute signs u must take the two conditions,
x^2-6x-1<6 and x^2-6x-1>-6
even if the question is |x^2 - 6x -1| = 6 u have to consider 
x^2 - 6x -1 = 6 and x^2 - 6x -1 = - 6 so for inequalities no different.

symaaa · Answer

-6<x^2-6x-1<6

x^2-6x-7<0 x=7, x=-1

x^2-6x+5>0 x=5, x=1

answer: (-1,7)

mick40010 · Answer

Ix^2 - 6x -1| < 6
or Ix^2 - 6x -7| < 0
or I (x-7)(x+1) I < 0
x = 7 or x = -1

Help with basic calculus :) - Easy 10 Points.?

4 Answers