Linear and Quadratic Inequalities
Pillar: SHAPE — "Reshape into a form where the sign is obvious."
Why Inequalities Matter
Equations ask "where does this equal zero?" Inequalities ask "where is this positive? Negative?" That second question comes up constantly in optimization, algorithm analysis, and competition math. Every time you bound a quantity, estimate a sum, or prove an algorithm's correctness, you're working with inequalities.
This lesson builds the machinery from scratch: linear inequalities, quadratic inequalities via sign charts, absolute value, and the triangle inequality. By the end, you'll have a systematic method for any inequality that involves polynomials or absolute values.
Linear Inequalities
A linear inequality looks like \(ax + b > 0\) (or \(\ge\), \(<\), \(\le\)). Solving it is almost identical to solving a linear equation, with one critical rule.
The sign-flip rule: When you multiply or divide both sides by a negative number, the inequality direction reverses.
Why? Multiplying by \(-1\) reflects the number line. Points that were to the right of \(-4\) end up to the left.
Worked Example
Solve \(5 - 2x \ge 3x + 1\).
Solution: \(x \le \frac{4}{5}\), or in interval notation \(\left(-\infty, \frac{4}{5}\right]\).
On the number line, shade everything to the left of \(\frac{4}{5}\), including the point itself (closed dot because \(\le\)).
Compound Inequalities
Two inequalities joined by AND or OR.
AND (intersection): Both must hold simultaneously.
Split into two:
- \(-3 < 2x + 1 \implies -4 < 2x \implies -2 < x\)
- \(2x + 1 \le 7 \implies 2x \le 6 \implies x \le 3\)
Solution: \(-2 < x \le 3\), which is the interval \((-2, 3]\).
OR (union): At least one must hold.
Solution: \((-\infty, -1) \cup (4, \infty)\).
| Type | Keyword | Operation | Example |
|---|---|---|---|
| AND | "and", "both", chain notation | Intersection \(\cap\) | \(-2 < x \le 3\) |
| OR | "or", "either" | Union \(\cup\) | \(x < -1\) or \(x > 4\) |
Quadratic Inequalities
To solve \(ax^2 + bx + c \le 0\) (or any variant), use the factor-then-sign-chart method.
Step 1: Factor
Get the quadratic in factored form. For \(x^2 - 5x + 6 \le 0\):
The roots are \(x = 2\) and \(x = 3\).
Step 2: Sign Chart
The roots divide the number line into intervals. Test one point in each interval to determine the sign of the product.
| Interval | Test point | \((x-2)\) | \((x-3)\) | Product |
|---|---|---|---|---|
| \((-\infty, 2)\) | \(x = 0\) | \(-\) | \(-\) | \(+\) |
| \((2, 3)\) | \(x = 2.5\) | \(+\) | \(-\) | \(-\) |
| \((3, \infty)\) | \(x = 5\) | \(+\) | \(+\) | \(+\) |
Step 3: Read the Answer
We want \((x-2)(x-3) \le 0\), i.e., the product is negative or zero. From the chart: the product is negative on \((2, 3)\) and zero at \(x = 2, 3\).
Solution: \([2, 3]\).
The Pattern
For a quadratic \(a(x - r_1)(x - r_2)\) with \(r_1 < r_2\) and \(a > 0\):
| Inequality | Solution |
|---|---|
| \(\le 0\) | \([r_1, r_2]\) |
| \(< 0\) | \((r_1, r_2)\) |
| \(\ge 0\) | \((-\infty, r_1] \cup [r_2, \infty)\) |
| \(> 0\) | \((-\infty, r_1) \cup (r_2, \infty)\) |
If \(a < 0\), the parabola opens downward, so the signs flip. The sign chart method always works regardless --- just read off the correct intervals.
What If There Are No Real Roots?
If the discriminant \(b^2 - 4ac < 0\), the quadratic never crosses zero. It's always positive (if \(a > 0\)) or always negative (if \(a < 0\)). So:
- \(x^2 + x + 1 > 0\) for all real \(x\) (discriminant \(= 1 - 4 = -3 < 0\), \(a = 1 > 0\)).
- \(-x^2 - 1 \le 0\) for all real \(x\).
Absolute Value
The absolute value \(|x|\) measures distance from zero. Its definition:
Absolute Value Inequalities
Two fundamental equivalences:
Think geometrically: \(|x| < a\) means "\(x\) is within distance \(a\) of zero." \(|x| > a\) means "\(x\) is farther than \(a\) from zero."
Worked Example
Solve \(|2x - 5| < 7\).
Using the equivalence: \(-7 < 2x - 5 < 7\).
Add 5 throughout: \(-2 < 2x < 12\).
Divide by 2: \(-1 < x < 6\).
Solution: \((-1, 6)\).
Absolute Value Equations
Solve \(|3x + 1| = 5\).
Case split: \(3x + 1 = 5\) or \(3x + 1 = -5\).
- Case 1: \(3x = 4 \implies x = \frac{4}{3}\)
- Case 2: \(3x = -6 \implies x = -2\)
Always verify both solutions in the original equation. Here both check out.
Nested Absolute Values
Solve \(||x - 1| - 3| = 2\).
Let \(u = |x - 1|\). Then \(|u - 3| = 2\), so \(u - 3 = 2\) or \(u - 3 = -2\).
- \(u = 5\): \(|x - 1| = 5 \implies x = 6\) or \(x = -4\)
- \(u = 1\): \(|x - 1| = 1 \implies x = 2\) or \(x = 0\)
Four solutions: \(x \in \{-4, 0, 2, 6\}\).
The Triangle Inequality
For any real numbers \(a\) and \(b\):
Equality holds when \(a\) and \(b\) have the same sign (or one is zero).
Why it's called "triangle": In geometry, any side of a triangle is at most the sum of the other two sides. The algebraic version is the same idea on the number line.
Reverse Triangle Inequality
This says the absolute difference of magnitudes is at most the distance between the values. It's useful for lower-bounding expressions.
Application: Bounding Sums
Given \(|a_i| \le M\) for \(i = 1, \ldots, n\), the triangle inequality gives:
This style of argument appears constantly in analysis and algorithm proofs: bound each term, then sum the bounds.
Graphing in the Coordinate Plane
Linear inequalities in two variables define half-planes.
\(2x + 3y \le 6\): First graph the boundary line \(2x + 3y = 6\). Then test a point (say \((0,0)\)): \(0 \le 6\) is true, so shade the side containing the origin.
A system of linear inequalities defines a feasible region --- the intersection of half-planes. This is the foundation of linear programming, which you'll see in optimization contexts.
For quadratic inequalities in two variables, the boundary is a parabola (or other conic). The method is the same: graph the boundary, test a point, shade the correct region.
Common Mistakes
| Mistake | Correction |
|---|---|
| Forgetting to flip sign when multiplying by negative | Always check: did I multiply/divide by something negative? |
| Writing \(\|x\| < -3\) and trying to solve | \(\|x\|\) is always \(\ge 0\), so \(\|x\| < -3\) has no solution |
| Squaring both sides of an inequality without checking signs | Only valid when both sides are non-negative |
| Confusing open/closed endpoints | \(<\) and \(>\) give open; \(\le\) and \(\ge\) give closed |
Practice Problems
Problem 1. Solve \(3(2x - 1) > 4x + 7\) and express the solution in interval notation.
Problem 2. Solve \(x^2 - 4x - 5 \ge 0\).
Problem 3. Solve \(|4x - 3| \le 9\).
Problem 4. Find all real \(x\) satisfying \(|x - 2| + |x + 3| \le 8\).
Problem 5. Prove that for all real numbers \(a, b, c\): \(|a - c| \le |a - b| + |b - c|\).
Solution 1
\(6x - 3 > 4x + 7 \implies 2x > 10 \implies x > 5\). Solution: \((5, \infty)\).
Solution 2
Factor: \((x - 5)(x + 1) \ge 0\). Roots at \(x = -1\) and \(x = 5\). Sign chart shows positive on \((-\infty, -1]\) and \([5, \infty)\). Solution: \((-\infty, -1] \cup [5, \infty)\).
Solution 3
\(|4x - 3| \le 9 \iff -9 \le 4x - 3 \le 9 \iff -6 \le 4x \le 12 \iff -\frac{3}{2} \le x \le 3\). Solution: \(\left[-\frac{3}{2}, 3\right]\).
Solution 4
Consider cases based on where \(x\) falls relative to \(-3\) and \(2\).
- If \(x \ge 2\): \((x - 2) + (x + 3) = 2x + 1 \le 8 \implies x \le 3.5\). So \(2 \le x \le 3.5\).
- If \(-3 \le x < 2\): \((2 - x) + (x + 3) = 5 \le 8\). Always true. So \(-3 \le x < 2\).
- If \(x < -3\): \((2 - x) + (-x - 3) = -2x - 1 \le 8 \implies -2x \le 9 \implies x \ge -4.5\). So \(-4.5 \le x < -3\).
Union: \([-4.5, 3.5]\).
Solution 5
Apply the triangle inequality to \(a - c = (a - b) + (b - c)\):
This is exactly the triangle inequality with the substitution \(u = a - b\), \(v = b - c\).