l x^2 - 3x - 7 l < 3
Step 1: Remove the modulus sign by recognising
that the function could take on values from -3 to 3 such that the
modulus is less than 3.
-3 < x^2 - 3x - 7 < 3
Step 2: Solve the 2 sides - factorising,
etc.
0 < x^2 - 3x -
4
and x^2 - 3x
- 10 < 0
0 < (x - 4) (x +1)
and (x - 5)
(x + 2) < 0
Step 3: Determine the range of values of x
__\________/___
-1
\
/ 4
\
/
\
___/
For this inequality to be more than zero, x < -1 or x >
4
__\________/___
-2
\
/ 5
\
/
\
___/
For this inequality to be less than zero, -2 < x < 5
Combining these 2 inequalities, you get -2 < x < -1 or
4 < x <
5
I hope I'm correct.
