Inequalities and Absolute Values

The solution set for the inequality |x+2| < |4x+1|

(a) -3/5 < x < 1/3
(b) -1/3 < x < 3/5
(c) 0 < x < 1/3
(d) -1/3 < x < 1/3
(e) None of these

Correct Answer: (E)

Since the expressions on both sides of the inequality are under modulus, you can square on both sides without disturbing the inequality.

(x+2)² < (4x+1)²

x² + 4x + 4 < 16x² + 8x + 1

15x² + 4x – 3 > 0

(5x + 3)(3x – 1) > 0

Hence, x € (-∞, -3/5) U (1/3, +∞)