Solve for xxx: ∣x2−4∣+∣x2−1∣=3|x^2 - 4| + |x^2 - 1| = 3∣x2−4∣+∣x2−1∣=3.
x∈[−2,−1]∪[1,2]x \in [-2, -1] \cup [1, 2]x∈[−2,−1]∪[1,2]
x∈[−2,2]x \in [-2, 2]x∈[−2,2]
x∈[−1,1]x \in [-1, 1]x∈[−1,1]
x=±1,±2x = \pm 1, \pm 2x=±1,±2