I'm self-teaching, so I'm going back and refreshing myself and I'm trying to figure something out.
I know that the answer is "no real solutions," but my question is why?
in general, with absolute value expressions, there are two cases:
1. the stuff inside the absolute value sign is negative, use the rule |x| = -x when x < 0
2. the stuff inside the absolute value sign is non-negative, use the rule |x| = x, when x ≥ 0.
let's look at your two problems in this light, to see if our method gives the same answer as before:
-3|x - 2| = 3.
case 1: x - 2 < 0. note this also means x < 2.
-3(2 - x) = 3 (replacing the stuff inside the absolute value sign by its negative)
-6 - (-3)x = 3
-6 + 3x = 3
3x = 9
x = 3. but 3 is not less than two, this is a contradiction.
case 2: |x - 2| ≥ 0, so x ≥ 2.
-3(x - 2) = 3 (removing the absolute value signs since we are assuming what's inside is non-negative)
-3x + 6 = 3
-3x = -3
x = 1. another contradiction, 1 is not greater than or equal to 2.
conclusion, no solution for x.
|x^2 + 2| = 4x + 2
case 1: x^2 + 2 < 0, so x^2 < -2, this is impossible, a square cannot be negative.
case 2: x^2 + 2 ≥ 0, so x^2 ≥ -2, true for any x.
x^2 + 2 = 4x + 2
x^2 = 4x
x = 0 will work. let's see if there are any non-zero solutions.
assuming x ≠ 0:
x = 4 (dividing by x).
so case 2 yields the solutions: x = 0,4
this method, although it can be long-winded (much like myself at times) is the easiest way to "keep the signs straight", and should always work.