Step 1: Use the Disjunction Property of Absolute Values
For any positive constant c>0, the inequality ∣A∣>c splits into two disjoint possibilities:
A>corA<−c
Substituting A=x2−3x+2 and c=4:
x2−3x+2>4orx2−3x+2<−4
Step 2: Solve Branch 1 (>4)
x2−3x+2x2−3x−2>4>0
Find the roots of the corresponding equation x2−3x−2=0:
x=2(1)−(−3)±(−3)2−4(1)(−2)=23±9+8=23±17
Numerically: 23−4.123≈−0.562 and 23+4.123≈3.562.
Because the leading coefficient is positive (a=1>0), the upward parabola is strictly positive outside the roots:
x∈(−∞,23−17)∪(23+17,∞)
Step 3: Solve Branch 2 (<−4)
x2−3x+2x2−3x+6<−4<0
Compute the discriminant of x2−3x+6:
Δ=(−3)2−4(1)(6)=9−24=−15<0
Since Δ<0 and a=1>0, the parabola y=x2−3x+6 lies entirely above the x-axis and is strictly positive for all x∈R. It can never be negative.
Thus, Branch 2 produces no solutions (∅).
Step 4: Union of Branches
The complete solution set is given entirely by Branch 1:
x∈(−∞,23−17)∪(23+17,∞)
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