1. Absolute value as distance
For any real number \(u\), the absolute value \(|u|\) measures its distance from \(0\)
on the number line:
\[
|u| =
\begin{cases}
\;\;\;u & \text{if } u \ge 0,\\[4pt]
-u & \text{if } u < 0.
\end{cases}
\]
More generally, an expression like \(|x - h|\) represents the distance between the
point \(x\) and the point \(h\) on the real line.
\[
|x - h| = \text{distance between } x \text{ and } h.
\]
2. Inequalities of the form \(\lvert a x + b\rvert \,\square\, c\)
2.1. Rewriting as a distance
Suppose \(a \ne 0\). We can factor the linear expression \(a x + b\) as
\[
a x + b = a\!\left(x + \frac{b}{a}\right)
= a\!\left(x - \left(-\frac{b}{a}\right)\right).
\]
Taking absolute values and using \(|a x + b| = |a|\,\bigl|x - h\bigr|\) with
\(h = -\dfrac{b}{a}\), we obtain
\[
|a x + b|
= |a|\,\bigl|x - h\bigr|,
\qquad
h = -\frac{b}{a}.
\]
Thus \(|a x + b|\) is a scaled distance from the point \(h\) on the real line.
2.2. Non-negative right-hand side
Because absolute values are always non-negative, the constant on the right should
usually satisfy \(c \ge 0\).
-
If \(c < 0\), then \(|a x + b| \le c\) or \(|a x + b| < c\) has
no solution.
-
If \(c < 0\), then \(|a x + b| \ge c\) or \(|a x + b| > c\) is
true for all real \(x\).
2.3. Two-sided inequalities (\(<\) or \(\le\))
Consider
\[
|a x + b| \le c
\quad\text{or}\quad
|a x + b| < c,
\qquad c \ge 0.
\]
Using \(|a x + b| = |a|\,|x - h|\) with \(h = -\dfrac{b}{a}\), we can divide both sides
by \(|a|\) (which is positive) and obtain an inequality involving a pure distance:
\[
|x - h| \,\square\, r,
\qquad
r = \frac{c}{|a|}.
\]
When \(\square\) is \(<\) or \(\le\), this means that \(x\) lies
within a closed or open interval around \(h\):
\[
|x - h| \le r
\;\Longleftrightarrow\;
h - r \le x \le h + r,
\]
\[
|x - h| < r
\;\Longleftrightarrow\;
h - r < x < h + r.
\]
In interval notation this becomes
\[
x \in [h - r,\; h + r] \quad\text{or}\quad x \in (h - r,\; h + r),
\]
which corresponds to a single bounded segment on the number line.
2.4. “Outside” inequalities (\(>\) or \(\ge\))
For \(|a x + b| \ge c\) or \(|a x + b| > c\) with \(c \ge 0\), we again divide by
\(|a|\) to get
\[
|x - h| \ge r
\quad\text{or}\quad
|x - h| > r.
\]
This means that \(x\) is at least \(r\) units away from \(h\). The solution is the union
of two unbounded intervals:
\[
|x - h| \ge r
\;\Longleftrightarrow\;
x \le h - r \;\;\text{or}\;\; x \ge h + r,
\]
\[
|x - h| > r
\;\Longleftrightarrow\;
x < h - r \;\;\text{or}\;\; x > h + r.
\]
3. Sums of two absolute values
Now consider an inequality of the form
\[
|a_1 x + b_1| + |a_2 x + b_2| \,\square\, c.
\]
Each absolute value behaves like a “V-shaped” function. The sum is therefore a piecewise
linear function whose shape changes whenever one of the inside expressions
\(a_1 x + b_1\) or \(a_2 x + b_2\) changes sign.
3.1. Breakpoints
The expressions inside the absolute values are zero at
\[
x_1 = -\frac{b_1}{a_1} \quad (a_1 \ne 0),
\qquad
x_2 = -\frac{b_2}{a_2} \quad (a_2 \ne 0).
\]
These points split the real line into intervals. On each interval the signs of
\(a_1 x + b_1\) and \(a_2 x + b_2\) are fixed, so we can remove the absolute values
by replacing each expression with either itself or its negative. For example, on an
interval where both expressions are non-negative we have
\[
|a_1 x + b_1| + |a_2 x + b_2|
= (a_1 x + b_1) + (a_2 x + b_2).
\]
3.2. Solving the inequality
- Find all breakpoints where \(a_1 x + b_1 = 0\) or \(a_2 x + b_2 = 0\).
- For each interval between successive breakpoints, determine the signs of the
expressions and rewrite the left-hand side as a simple linear function of \(x\).
- Solve the resulting linear inequality on that interval.
- Take the union of all intervals where the inequality is satisfied.
The calculator automates these steps and shows the final solution as a union of
intervals on the number line.
4. Absolute value of a quadratic: \(\bigl|a x^2 + b x + c_0\bigr| \,\square\, k\)
Let
\[
Q(x) = a x^2 + b x + c_0.
\]
We are interested in inequalities of the form
\[
|Q(x)| \,\square\, k,
\qquad k \ge 0.
\]
4.1. Inside vs. outside a band
The inequality \(|Q(x)| \le k\) means that the values of the quadratic stay within
the horizontal band between \(-k\) and \(k\):
\[
|Q(x)| \le k
\;\Longleftrightarrow\;
-k \le Q(x) \le k.
\]
Similarly,
\[
|Q(x)| \ge k
\;\Longleftrightarrow\;
Q(x) \le -k
\;\text{ or }\;
Q(x) \ge k.
\]
So problems with \(|Q(x)|\) can always be rewritten as one or two quadratic inequalities.
4.2. Solving quadratic inequalities
To solve an inequality such as \(Q(x) \ge k\) or \(Q(x) \le k\), we move all terms to
one side and study the sign of a quadratic expression:
\[
Q(x) - k \ge 0
\quad\text{or}\quad
Q(x) + k \le 0.
\]
For a quadratic \(P(x) = a x^2 + b x + c\) with \(a \ne 0\), the key information is
contained in its discriminant
\[
\Delta = b^2 - 4 a c.
\]
- \(\Delta > 0\): two distinct real roots; the parabola crosses the \(x\)-axis twice.
- \(\Delta = 0\): one double root; the parabola touches the \(x\)-axis.
- \(\Delta < 0\): no real roots; the parabola lies entirely above or below the axis.
Combining the sign of \(a\) (parabola opening up or down) with the roots allows us
to describe precisely where \(P(x) \ge 0\), \(P(x) > 0\), \(P(x) \le 0\), or
\(P(x) < 0\). The calculator performs this analysis and then intersects or unions
the resulting intervals, depending on the original absolute value inequality.
5. Reading the calculator output
