Solve the strict absolute-value inequality.

01 / Standard#Your turn

|x(x-5)|<6

Official paper · jm01-2025 · I.6 · PDF 2

Skills and prerequisite lessons

Option A $(-1,6)$
Option B $(-2,1)$
Option C $(-1,1]\cup[4,5]$
Option D $(-\infty,-1]\cup[6,\infty)$
Option E $(-1,2)\cup(3,6)$

Working and explanation

BUILD THE REASONING

Hint 1

Replace |u|<6 by −6<u<6.

Hint 2

Intersect the solutions of two quadratic inequalities.

Worked solution

The upper bound gives an interior interval.
$x^2-5x-6<0\iff-1<x<6$
The lower bound excludes the closed interval [2,3].
$x^2-5x+6>0\iff x<2\ \text{or}\ x>3\implies x\in(-1,2)\cup(3,6)$

E: (−1,2)∪(3,6).

Checks and common pitfalls: All four boundary values are excluded.

Reasoning checklist · self / teacher assessment

Teaching assessment checklist, independently authored. Use the original paper for official marks.
The upper bound gives an interior interval.
$x^2-5x-6<0\iff-1<x<6$
The lower bound excludes the closed interval [2,3].
$x^2-5x+6>0\iff x<2\ \text{or}\ x>3\implies x\in(-1,2)\cup(3,6)$

Think first. Reveal a hint when the class is ready.

2025 JM01