Graphical Solution of Absolute Value Inequality |x-2| + |2x+1| ≥ 3

[reubenmanzo]

Personally, I find it easier to solve these sorts of problems algebraically:
|x-2|+|2x+1|>=3
(squaring both sides)
5x^2+5+2|2x^2-3x-2|>=9
|2x^2-3x-2|>=(4-5x^2)/2
(2x^2-3x-2)^2>=(4-5x^2)^2/4 (squaring both sides again, to fully remove the absolute values)


9x^4+48x^3-44x^2-48x<=0
x(9x^3+48x^2-44x-48)<=0
x(x+6)(9x^2-6x-8)<=0 (polynomial division)

9x^2-6x-8=0
x=[6 +/- 18)]/18
x=4/3 or -2/3


Therefore, the roots of this polynomial are -6, -2/3, 0, 4/3. However, given the nature of the original function, we're only interested in the middle roots.
Therefore x<=-2/3, x>=0.

[fasd]

Awesome! Thanks a bunch. This helps me help my high school son.

[karabomashabane]

Good Day
what would happen if the one absolute value was subtracting the other?

Your response would be highly appreciated.

[kshitizz]

I once dreamt to be taught by such teacher.
My school techers are just Bookish fellow, ruining life.🤮🤮

[dhritimondal]

This graph is a bit wrong. The lines y=2-x and y=-1-2x won't intersect because they are parallel to each other.