What Does the Solution Set for |x² - 4| < 1 Look Like?

[fishingspree2]

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]

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

 

[HallsofIvy]

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).