I'm self-teaching, so I'm going back and refreshing myself and I'm trying to figure something out.
-3|x-2|=3
I know that the answer is "no real solutions," but my question is why?
Thanks!
When you have to solve equations with absolut value signes then you have to work with the definition mostly (do you know the definition?)
In this case, what can you say about $\displaystyle |x^2+2|$? (are the absolut value signs necessary? Why? Why not?)
Yes, but for which conditions for $\displaystyle x$? We can say that $\displaystyle |x|=x$ if $\displaystyle x\geq0$ and $\displaystyle |x|=-x$ if $\displaystyle x<0$.
That means for example $\displaystyle |x-3|=x-3$ if $\displaystyle x\geq3$ and $\displaystyle |x-3|=-(x-3)$ if $\displaystyle x<3$
But in this case, what can you say about $\displaystyle |x^2+2|$?
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.