Absolute value equation

We solve the absolute-value equation x2=3|x - 2| = 3 by reading it as the intersection of y=x2y = |x - 2| and y=3y = 3.

Reading it as a distance

Read the absolute value as a distance and the answer appears without any computation. The expression x2|x - 2| is the gap between xx and 22 on the number line, so x2=3|x - 2| = 3 simply asks which numbers lie 33 away from 22. Three to the right gives 55; three to the left gives 1-1. The solutions are therefore x=5x = 5 and x=1x = -1, and the two intersection points are (5,3)(5, 3) and (1,3)(-1, 3). An absolute-value equation is a ruler laid against the number line.

Solving by cases

RangeEquation without the absolute valueSolutionFits its range
x2x \geq 2x2=3x - 2 = 3x=5x = 5yes
x<2x < 2(x2)=3-(x - 2) = 3x=1x = -1yes

Each candidate fits the range it came from, so both are kept.

Reading the graph

y=x2y = |x - 2| is the V of y=xy = |x| shifted 22 to the right, with its corner at (2,0)(2, 0). The height 00 of that corner is the boundary that decides how often a horizontal line meets the graph.

LineNumber of intersections
y=ky = k with k>0k > 022
y=0y = 011, the corner itself
y=ky = k with k<0k < 000

An equation such as x2=1|x - 2| = -1 has no solution precisely because the V never dips below the xx-axis.

Moving on to inequalities

The same reading handles inequalities.

InequalityMeaning as a distanceSolution
x2<3|x - 2| < 3the gap from 22 is less than 331<x<5-1 < x < 5
x2=3|x - 2| = 3the gap is exactly 33x=1,5x = -1, 5
x2>3|x - 2| > 3the gap is more than 33x<1x < -1 or x>5x > 5

On the graph these are the stretches where the V lies below the line y=3y = 3 and where it lies above. Writing a measurement tolerance as xad|x - a| \leq d rests on the same idea of a gap from a center aa. The large dots on the graph are the intersections (5,3)(5, 3) and (1,3)(-1, 3).