Solution for Problem 1.8.1

Solve for x:  | x 2 | < 3   

Solve

Step

| x 2 | < 3  

-3 < x 2 < 3

-3 + 2 < x 2 + 2 < 3 + 2

-1 < x < 5

[-1, 5]

Use this technique when 
   |x| < a   
-a < x < a 
0r
|x| < a
-a < x <

Inequality Notation

Interval Notation

Graph

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