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AMC 10 & 12

Absolute Value

Piecewise reasoning and distance interpretations turn absolute value from a scary symbol into a straightforward case split.

x|x| means “the distance from xx to 0 on the number line,” which is always nonnegative. That single idea — distance — resolves almost every absolute value problem you'll see, if you translate the symbols into a picture before you start solving.

The piecewise definition

x={xx0xx<0|x| = \begin{cases} x & x \geq 0 \\ -x & x < 0 \end{cases}

More generally, xa|x - a| is the distance between xx and aa. That reading makes equations and inequalities involving absolute value much easier to picture:

  • xa=r|x - a| = r means xx is exactly rr away from aa: so x=arx = a - r or x=a+rx = a + r.
  • xa<r|x - a| < r means xx is within rr of aa: the interval (ar, a+r)(a-r,\ a+r).
  • xa>r|x - a| > r means xx is farther than rr from aa: two rays, x<arx < a-r or x>a+rx > a+r.

When the inside isn't simple

Once the expression inside the bars is more complicated than a single variable, the distance picture stops being immediate, and you fall back on the case split: consider the case where the inside is0\geq 0 (bars come off unchanged), then the case where it's negative (bars come off with a sign flip). Solve each case as its own equation, restricted to the region where that case is actually valid — and always plug candidate solutions back into the original equation, since squaring or case-splitting can introduce extraneous roots.

Worked example

Problem: Solve 2x3=x+3|2x - 3| = x + 3.

Case 1 (2x302x-3 \geq 0): the equation becomes 2x3=x+32x - 3 = x + 3, so x=6x = 6. Check: 2(6)3=902(6)-3=9\geq0 ✓, and this is valid in this case.

Case 2 (2x3<02x - 3 < 0): the equation becomes (2x3)=x+3-(2x-3) = x+3, i.e.32x=x+33 - 2x = x + 3, so x=0x = 0. Check: 2(0)3=3<02(0) - 3 = -3 < 0 ✓.

Both candidates check out in the original equation (the right side x+3x+3 must also be nonnegative, which it is in both cases). So the solutions are x=0x = 0 and x=6x = 6.

Practice problems

1.Solve x+4=7|x + 4| = 7.

2.Solve and write in interval notation: 3x6<12|3x - 6| < 12.

3.Solve x1=2x+4|x - 1| = |2x + 4|.

4.Solve x5=2x|x - 5| = 2x.