As usual with absolute values, split into cases:
Case 1: x^2 - 3 < 0x2−3<0
If x^2 - 3 < 0x2−3<0 then abs(x^2-3) = -(x^2-3) = -x^2+3∣∣x2−3∣∣=−(x2−3)=−x2+3
and our (corrected) inequality becomes:
-x^2+3 < 2−x2+3<2
Add x^2-2x2−2 to both sides to get 1 < x^21<x2
So x in (-oo,-1) uu (1, oo)x∈(−∞,−1)∪(1,∞)
From the condition of the case we have
x^2 < 3x2<3, so x in (-sqrt(3), sqrt(3))x∈(−√3,√3)
Hence:
x in (-sqrt(3), sqrt(3)) nn ((-oo,-1) uu (1, oo))x∈(−√3,√3)∩((−∞,−1)∪(1,∞))
= (-sqrt(3), -1) uu (1, sqrt(3))=(−√3,−1)∪(1,√3)
Case 2: x^2 - 3 >= 0x2−3≥0
If x^2 - 3 >= 0x2−3≥0 then abs(x^2-3) = x^2+3∣∣x2−3∣∣=x2+3 and our (corrected) inequality becomes:
x^2-3 < 2x2−3<2
Add 33 to both sides to get:
x^2 < 5x2<5, so x in (-sqrt(5), sqrt(5))x∈(−√5,√5)
From the condition of the case we have
x^2 >= 3x2≥3, so x in (-oo, -sqrt(3)] uu [sqrt(3), oo)x∈(−∞,−√3]∪[√3,∞)
Hence:
x in ((-oo, -sqrt(3)] uu [sqrt(3), oo)) nn (-sqrt(5), sqrt(5))x∈((−∞,−√3]∪[√3,∞))∩(−√5,√5)
= (-sqrt(5), -sqrt(3)] uu [sqrt(3), sqrt(5))=(−√5,−√3]∪[√3,√5)
Combined:
Putting case 1 and case 2 together we get:
x in (-sqrt(5), -sqrt(3)] uu (-sqrt(3), -1) uu (1, sqrt(3)) uu [sqrt(3), sqrt(5))x∈(−√5,−√3]∪(−√3,−1)∪(1,√3)∪[√3,√5)
=(-sqrt(5), -1) uu (1, sqrt(5))=(−√5,−1)∪(1,√5)
