# Absolute value inequality

## Main Question or Discussion Point

Hello,

abs(x^2 - 4) < 1

implies that:
x^2 - 4 < 1
and
4 - x^2 < 1

solving first equation for x gives:
-sqrt(5) < x < sqrt(5)

solving second equation for x gives:
-sqrt(3) < x < sqrt(3)

Now, my question is, what does that mean??
How do I give the solution set, without a graphing utility and, if possible, without trial and error?

Thank you

rock.freak667
Homework Helper
Draw the inequalities on two separate number lines. Then just find the intersection of the two(the region which is common to both).

HallsofIvy
Homework Helper
Hello,

abs(x^2 - 4) < 1

implies that:
x^2 - 4 < 1
and
4 - x^2 < 1

solving first equation for x gives:
-sqrt(5) < x < sqrt(5)

solving second equation for x gives:
-sqrt(3) < x < sqrt(3)
NO! if 4- x^2< 1, then 3< x^2 so either x> sqrt{3} or x< -sqrt{3}.

Now, my question is, what does that mean??
How do I give the solution set, without a graphing utility and, if possible, without trial and error?

Thank you
In order that |x^2- 4|< 1, you must have both x< -sqrt(3) or x> sqrt(3) and -sqrt(5)< x< sqrt(5). As rockfreak667 said, that is the intersection of the two sets:
-sqrt(5)< x< -sqrt(3) or sqrt(3)< x< sqrt(5).

I would prefer to do this problem in quite a different way: look at the corresponding equation |x^2- 4|= 1. Then either x^2- 4= 1 which leads to x^2= 5 and so x= +/- sqrt(5) or x^2- 4= -1 which leads to x^2= 3 and so x= +/- sqrt(3). But since the function is continuous, the only places it can change from "less than 1" to "greater than 1" is at one of those 4 points where it is "equal to 1".

Those four points divide the line into 5 parts. Check one value of x< -sqrt(5), say 3: If x= 3, |3^2-4|= 5> 1. Check one value of x between -sqrt(5) and -sqrt(3): x= 2. If x= 2 |4- 4|= 0< 1. Check one point between -sqrt(3) and sqrt(3): x= 0. If x= 0 |0- 4|= 4> 1. Check one point between sqrt(3) and sqrt(5): 2. if x= 2, |4- 4|= 0< 1. Finally, check one point larger than sqrt(5): 3. If x= 3, |9- 4|= 5> 1. |x^2- 4|< 1 is satisfied for -sqrt(5)< x< -sqrt(3) and for sqr(3)< x< sqrt(5).