What is (|x-1|+|x-2|)(|1-x|-|2-x|)<= x^2-6 ?

The solution to (|x-1|+|x-2|)(|1-x|-|2-x|) = 3

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(|x-1|+|x-2|)(|1-x|-|2-x|)<= x^2-6

(∣ x - 1 ∣ + ∣ x - 2 ∣) (∣ 1 - x ∣ - ∣ 2 - x ∣) \leq x^{2} - 6

x \leq - 1 or x \geq 3

Interval Notation

(- \infty, - 1] \cup [3, \infty)

Solution steps

(∣ x - 1 ∣ + ∣ x - 2 ∣) (∣ 1 - x ∣ - ∣ 2 - x ∣) \leq x^{2} - 6

Find positive and negative intervals

x \leq 1, 1 < x \leq 2, x > 2

Solve the inequality for each interval

x \leq - 1 or No Solution or x \geq 3

Merge Overlapping Intervals

x \leq - 1 or x \geq 3

- 2.4 z - 6.6 \geq 5.4 (1 - z)

p^{3} r^{5} - p^{3} r

22 y 4 z + 3 x 4

4 n^{2} - 11 n + 6

(x + 2) (x - 3) > (x - 1) (x + 1)

What is (|x-1|+|x-2|)(|1-x|-|2-x|)<= x^2-6 ?
The solution to (|x-1|+|x-2|)(|1-x|-|2-x|)<= x^2-6 is x<=-1\lor x>= 3