Answer:
x ∈ {2, 3, 4, 5, 6}
Explanation:
You want the positive integer solutions of ||x²-2x|-|3x-20|| = |x²+x-20|.
Domains
Each of the absolute value functions has turning points where the argument is zero. Those turning points are ...
x²-2x ⇒ x(x-2) = 0 ⇒ turning points at x=0 and x=2
3x-20 ⇒ x -20/3 = 0 ⇒ turning point at x=20/3
x²+x-20 ⇒ (x+5)(x-4) = 0 ⇒ turning points at x=-5 and x=4
In numerical order, the turning points are ...
x ∈ {-5, 0, 2, 4, 20/3}
These divide the domain of the equation into 6 intervals. Since we are concerned only with positive integer solutions, we are only interested in the intervals ...
[0, 2), [2, 4), [4, 20/3), [20/3, ∞)
Domain [0, 2)
In this domain, x²-2x < 0, 3x-20 < 0, and x²+x-20 < 0. This means the equation becomes ...
|-(x² -2x) +(3x -20)| = -(x² +x -20)
The difference in the absolute value bars is negative, so we get the equation ...
-(-x² +2x +3x -20) = -x² -x +20
2x² -4x = 0 . . . . . . . add x²+x+20
x = 0 . . . . . x = 2 is not in the domain
There are no positive integer solutions in this domain.
Domain [2, 4)
In this domain, x²-2x > 0, 3x-20 < 0, and x²+x-20 < 0. This means the equation becomes ...
|(x² -2x) +(3x -20)| = -(x² +x -20)
The difference in the absolute value bars is negative, so we get the equation ...
-(x² -2x +3x -20) = -x² -x +20
0 = 0 . . . . . . . . . . . . add x² +x -20
2 ≤ x < 4 . . . . . . . . all x-values in the domain are solutions
The positive integer solutions are x = 2, x = 3.
Domain [4, 6 2/3)
In this domain, x²-2x > 0, 3x-20 < 0, and x²+x-20 > 0. This means the equation becomes ...
|(x² -2x) +(3x -20)| = x² +x -20
The difference in the absolute value bars is positive, so we get the equation ...
(x² -2x +3x -20) = x² +x -20
0 = 0 . . . . . . . subtract x² +x -20
4 ≤ x < 6 2/3 . . . . . . . . . all values of x in the domain are solutions
The positive integer solutions are x = 4, x = 5, x = 6.
Domain [6 2/3, ∞)
In this domain, x²-2x > 0, 3x-20 > 0, and x²+x-20 > 0. This means the equation becomes ...
|(x² -2x) -(3x -20)| = x² +x -20
The difference in the absolute value bars is positive, so we get the equation ...
x² -2x -3x +20 = x² +x -20
-6x +40 = 0 . . . . . . . subtract x² +x -20
x = 6 2/3 . . . . . . . add 6x, divide by 6
There are no positive integer solutions in this domain.
The positive integer values of x are {2, 3, 4, 5, 6}.