Quadratic Inequalities — I can solve quadratic inequalities by analyzing a graph or sign chart and interpret the solution set. - Math 2

SECTION 1

Historical Context & Motivation

Mathematicians have grappled with equations involving squares for thousands of years, but the idea of asking when an expression is greater or less than a value — rather than exactly equal — opened an entirely new branch of algebra. A quadratic inequality replaces the equals sign of a quadratic equation with an inequality symbol such as <, >, ≤, or ≥. Instead of finding a couple of specific x-values, you find entire intervals of x-values that make the statement true.

Why does this matter? In the real world, problems rarely ask for one exact answer. Engineers need to know the range of speeds at which a bridge remains safe. Economists analyze the range of prices that keep a company profitable. These situations are modeled by inequalities, not equations. The story of how we got here stretches back millennia.

~1800 BCE • Babylonian Quadratics

Babylonian scribes solved quadratic-style area problems on clay tablets, laying the groundwork for quadratic expressions — though they worked only with positive solutions.

~300 BCE • Euclid's Geometric Algebra

Euclid described relationships between lengths and areas that correspond to quadratic expressions, reasoning about when one quantity exceeds another — an early form of inequality thinking.

1637 • Descartes & Coordinate Geometry

René Descartes introduced the coordinate plane, making it possible to visualize a parabola and see where it lies above or below the x-axis — a powerful tool for solving quadratic inequalities graphically.

1800s • Formal Inequality Notation

Mathematicians standardized the inequality symbols (<, >, ≤, ≥) and developed systematic algebraic methods, including sign analysis, for solving polynomial inequalities.

Today, you already know how to solve a quadratic equation to find the x-intercepts of a parabola. The next natural question is: for which x-values is the parabola above or below the x-axis? That is exactly what quadratic inequalities answer.

SECTION 2

Core Principles & Definitions

Before diving into techniques, you need to lock down four foundational ideas that every quadratic inequality problem depends on. These principles connect what you already know about parabolas with the new skill of identifying solution intervals.

Standard Form of a Quadratic Inequality

A quadratic inequality looks like ax² + bx + c > 0 (or <, ≤, ≥). The expression on one side must be a quadratic, and the other side is zero. Always rearrange to this form first.

Roots as Boundary Points

The solutions to ax² + bx + c = 0 are called boundary points (or critical values). They divide the number line into intervals where the expression keeps a constant sign — either always positive or always negative.

Sign of the Leading Coefficient

When a > 0 the parabola opens upward, so the expression is negative between the roots. When a < 0 it opens downward, and the expression is positive between the roots.

Open vs. Closed Intervals

Strict inequalities (< or >) use open circles and parentheses ( ). Non-strict inequalities (≤ or ≥) use closed circles and brackets [ ] because the boundary points are included.

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KEY TAKEAWAY

Think of the roots of the quadratic as the two ends of a bridge. The parabola is either above the water (positive) or below the water (negative) between those endpoints. Your job is to figure out which side of the bridge is dry — that is, where the expression has the sign you want. The shape of the parabola (opening up or down) tells you whether the "dry" section is between the roots or outside them.

SECTION 3

Graphical Approach — Reading the Parabola

The most intuitive way to solve a quadratic inequality is to graph the related quadratic function y = ax² + bx + c and then observe where the curve is above or below the x-axis. The x-axis represents y = 0, so the portions of the parabola above the axis correspond to y > 0, and the portions below correspond to y < 0.

Consider the inequality x² − 4x − 5 > 0. The related function is y = x² − 4x − 5. Factoring gives y = (x − 5)(x + 1), so the roots are x = −1 and x = 5. Because a = 1 > 0, the parabola opens upward. The diagram below shows the curve crossing the axis at those roots.

The parabola y = x² − 4x − 5 crosses the x-axis at x = −1 and x = 5. The cyan-shaded regions show where y > 0 (outside the roots), and the pink-shaded region shows where y < 0 (between the roots). Open circles at the roots indicate that the roots themselves are not included in a strict inequality.

From the graph you can read the solution directly. Because the inequality asks where the expression is greater than zero, you look for the parts of the parabola that are above the x-axis. Those are the cyan-shaded regions: x < −1 or x > 5. In interval notation, the solution set is (−∞, −1) ∪ (5, ∞). If the inequality were ≥ instead of >, the roots would be included, and you would use brackets: (−∞, −1] ∪ [5, ∞).

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GRAPH-READING STRATEGY

Step 1 — Set the expression equal to zero and find the roots. Step 2 — Sketch the parabola, noting whether it opens up or down. Step 3 — Shade the region above the axis for > or ≥, or below the axis for < or ≤. Step 4 — Read off the x-intervals from the shaded region.

SECTION 4

Mathematical Framework — Sign Charts

A sign chart (sometimes called a sign analysis or sign diagram) is an algebraic method that does the same work as the graph but without drawing a curve. You identify the boundary points, test a value in each interval, and record whether the expression is positive or negative there. This technique is especially useful when you need an exact algebraic answer and cannot easily sketch the parabola.

STANDARD FORM

ax² + bx + c {>, <, ≥, ≤} 0

a, b, c are real-number coefficients; a ≠ 0. The quadratic expression must be on one side with 0 on the other.

FIND BOUNDARY POINTS

ax² + bx + c = 0 → x = (−b ± √(b² − 4ac)) / (2a)

Use factoring, completing the square, or the quadratic formula to find the roots. These roots divide the number line into intervals.

TEST POINT RULE

Pick any x-value inside each interval → substitute → record + or −

The sign of the expression cannot change within an interval that contains no roots. So one test point per interval is enough.

After completing the sign chart, select the intervals whose sign matches the inequality. If the inequality is ≤ or ≥, include the boundary points (closed circles / brackets). If it is < or >, exclude them (open circles / parentheses).

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IMPORTANT NOTE

If the discriminant b² − 4ac is negative, the quadratic has no real roots. The parabola never crosses the x-axis, so the expression is always positive (if a > 0) or always negative (if a < 0). The solution set is either all real numbers or the empty set.

SECTION 5

Detailed Breakdown — Building a Sign Chart

Let's build a complete sign chart for the inequality x² − 4x − 5 < 0 (the same quadratic from Section 3, but now with a "less than" sign). The factored form is (x − 5)(x + 1) < 0, so the boundary points are x = −1 and x = 5. These two points split the number line into three intervals: (−∞, −1), (−1, 5), and (5, ∞). The diagram below visualizes the process.

The sign chart tests one x-value in each of the three intervals created by the roots x = −1 and x = 5. Each factor's sign is recorded, and the product's sign determines whether the interval belongs to the solution. Because (x − 5)(x + 1) < 0 asks for a negative product, only Interval II qualifies, giving the solution (−1, 5).

Interval I (x < −1): Both factors are negative, so the product is positive. Does NOT satisfy < 0.
Interval II (−1 < x < 5): One factor is positive and one is negative, so the product is negative. SATISFIES < 0.
Interval III (x > 5): Both factors are positive, so the product is positive. Does NOT satisfy < 0.

SECTION 6

Worked Example

Let's solve a complete problem from start to finish using both a sign chart and graphical reasoning.

Solve −2x² + 8x − 6 ≥ 0

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Step 1 — Write in Standard Form

The inequality is already in the form ax² + bx + c ≥ 0, with a = −2, b = 8, and c = −6. Because a is negative, the parabola opens downward.
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Step 2 — Find the Boundary Points

Set −2x² + 8x − 6 = 0. Divide both sides by −2 (and remember: dividing or multiplying an equation by a constant doesn't flip anything — that rule is for inequalities only when you multiply by a negative). This gives x² − 4x + 3 = 0. Factor: (x − 1)(x − 3) = 0.

Boundary points: x = 1 and x = 3
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Step 3 — Set Up the Sign Chart

The boundary points divide the number line into three intervals: (−∞, 1), (1, 3), and (3, ∞). Choose test points x = 0, x = 2, and x = 4.
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Step 4 — Evaluate Test Points

Substitute into −2x² + 8x − 6. At x = 0: −2(0) + 0 − 6 = −6 (negative). At x = 2: −2(4) + 8(2) − 6 = −8 + 16 − 6 = 2 (positive). At x = 4: −2(16) + 8(4) − 6 = −32 + 32 − 6 = −6 (negative).

Signs: −, +, −
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Step 5 — Select Intervals and Include Boundaries

The inequality asks for ≥ 0, so you need the interval where the expression is positive (Interval II) and include the boundary points because of the ≥ symbol.

Solution: 1 ≤ x ≤ 3, or [1, 3]
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Step 6 — Verify with Graph Interpretation

Because a = −2 < 0, the parabola opens downward. It is above the x-axis between the roots and below it outside the roots. The portion on or above the axis corresponds exactly to [1, 3], confirming the sign chart result.

SECTION 7

Graph Method vs. Sign Chart — Strengths & Limitations

Both the graphical approach and the sign chart approach will always give the same answer. However, each has situations where it shines and situations where it is less convenient. The table below compares the two methods across several practical dimensions.

Criterion	Graphical Method	Sign Chart Method
Speed	Fast if you have a graphing calculator or can sketch quickly.	Fast once you practice — no graph needed, just arithmetic.
Visualization	Excellent — you see the parabola and physically identify the shaded region.	Limited — it's purely algebraic, no visual curve.
Precision	Sketches can be imprecise; you might misread intercepts.	Exact — every value is computed, so no reading errors.
Extends to higher degrees	Harder to sketch cubics, quartics, etc.	Easily generalizes to polynomial inequalities of any degree.
Best used when	You have access to graphing technology or want a quick conceptual check.	You need an algebraic proof of the solution or are working without technology.

Comparison of the two primary methods for solving quadratic inequalities.

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KEY TAKEAWAY

Think of the graph method as using a map and the sign chart method as using written turn-by-turn directions. Both get you to the same destination. The map gives you a big-picture view, while the directions ensure you never miss a turn. In practice, use the graph for understanding and the sign chart for precision.

SECTION 8

Connection to Higher-Degree & Rational Inequalities

The sign-chart technique you just learned is not limited to quadratics. In future courses, you will encounter polynomial inequalities of degree 3, 4, and beyond, as well as rational inequalities where the variable appears in a denominator. The core logic is the same: find boundary points, split the number line, test each interval, and read off the solution.

Feature	Quadratic Inequality (this lesson)	Higher-Degree / Rational Inequality
Degree	2 (parabola)	3 or more, or a ratio of polynomials
Max boundary points	2 real roots	n real roots (for degree n), plus vertical asymptotes for rationals
Sign chart intervals	Up to 3 intervals	Up to n + 1 or more intervals
Key difference	Only two factors to track	More factors; for rationals, boundary points from denominator = 0 are never included in the solution.

How quadratic inequalities relate to more advanced inequality types.

Mastering sign charts now gives you a transferable skill. When you move into Precalculus or Algebra 2, the problems will have more factors, but the process stays the same: find boundary points, test intervals, match signs. You're building a strategy, not just memorizing one type of problem.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

A parabola opens upward and has x-intercepts at x = 2 and x = 8. For what x-values is the quadratic expression negative? Explain your reasoning without computing anything.

Show answer

Because the parabola opens upward, it dips below the x-axis between its roots. Therefore the expression is negative when 2 < x < 8, which is the open interval (2, 8). Outside the roots — that is, for x < 2 or x > 8 — the parabola is above the axis and the expression is positive. This reasoning relies on the shape of the parabola (U-shape) and the meaning of x-intercepts as the points where the expression equals zero.

Question 2: BASIC CALCULATION

Solve x² − 9 > 0. Express the solution in interval notation.

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Factor: (x − 3)(x + 3) > 0. The boundary points are x = −3 and x = 3. The leading coefficient is positive, so the parabola opens upward and the expression is positive outside the roots. Test: at x = 0, (0 − 3)(0 + 3) = −9 < 0 (between roots, negative — confirms the pattern). The solution is x < −3 or x > 3, which in interval notation is (−∞, −3) ∪ (3, ∞). The strict inequality means the boundary points are excluded.

Question 3: INTERMEDIATE

Solve 2x² − 7x + 3 ≤ 0. Show a complete sign chart and give the answer in interval notation.

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Set 2x² − 7x + 3 = 0. Using the quadratic formula: x = (7 ± √(49 − 24)) / 4 = (7 ± √25) / 4 = (7 ± 5) / 4. So x = 3 or x = 1/2. The three intervals are (−∞, 1/2), (1/2, 3), and (3, ∞). Test x = 0: 2(0) − 0 + 3 = 3 > 0 (positive). Test x = 1: 2(1) − 7 + 3 = −2 < 0 (negative). Test x = 4: 2(16) − 28 + 3 = 7 > 0 (positive). The expression is ≤ 0 in Interval II, and the ≤ symbol includes the endpoints. Solution: [1/2, 3].

Question 4: APPLIED

A toy rocket's height in meters is modeled by h(t) = −5t² + 30t, where t is time in seconds. During what time interval is the rocket at or above 25 meters?

Show answer

Set up the inequality: −5t² + 30t ≥ 25. Rearrange: −5t² + 30t − 25 ≥ 0. Divide by −5 (and flip the inequality): t² − 6t + 5 ≤ 0. Factor: (t − 1)(t − 5) ≤ 0. Boundary points are t = 1 and t = 5. Since the coefficient of t² is positive and the inequality is ≤ 0, the solution is between the roots (inclusive). Test t = 3: (3 − 1)(3 − 5) = 2 × (−2) = −4 ≤ 0 ✓. The rocket is at or above 25 meters for 1 ≤ t ≤ 5, or [1, 5] seconds.

Question 5: CRITICAL THINKING

Consider the inequality x² + 4x + k < 0. For what values of the constant k does this inequality have a non-empty solution set? Explain your reasoning using the discriminant and the shape of the parabola.

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The parabola y = x² + 4x + k opens upward (a = 1 > 0). For the expression to be negative somewhere, the parabola must dip below the x-axis, which means it must have two distinct real roots. This requires the discriminant to be positive: b² − 4ac > 0 → 16 − 4(1)(k) > 0 → 16 − 4k > 0 → k < 4. If k = 4, the parabola touches the x-axis at one point but never goes below it, so x² + 4x + 4 = (x + 2)² ≥ 0 always. If k > 4, the parabola is entirely above the axis, so x² + 4x + k < 0 has no solution. Therefore the inequality has a non-empty solution set precisely when k < 4.

SUMMARY

Lesson Summary

Summary

A quadratic inequality asks where a quadratic expression is positive, negative, or zero. To solve one, first move everything to one side so the other side is zero. Then find the boundary points by setting the expression equal to zero. These roots divide the number line into intervals. Use either a graph (look at where the parabola is above or below the x-axis) or a sign chart (test one point per interval) to determine the sign of the expression in each interval. Select the intervals that satisfy the inequality.

Remember that the leading coefficient determines whether the parabola opens up or down, which controls whether the expression is negative between or outside the roots. Use open circles and parentheses for strict inequalities (< or >) and closed circles and brackets for non-strict inequalities (≤ or ≥). This same sign-chart strategy extends naturally to higher-degree polynomial and rational inequalities.