5.4 Quadratic Inequalities and Quadratic-Linear Systems
Key Takeaways
- Solve ax^2 + bx + c > 0 by finding the roots, then using a sign chart or the parabola’s position above or below the x-axis.
- x^2 − x − 6 < 0 factors as (x − 3)(x + 2) < 0, so the solution is −2 < x < 3, or the open interval (−2, 3).
- A line and a parabola meet at 0, 1, or 2 points. After substitution, the discriminant of the resulting quadratic counts the intersections.
- Substitution y = x^2 − 2x − 3 into y = x + 1 yields x^2 − 3x − 4 = 0, so the intersection points are (4, 5) and (−1, 0).
- Table 11 explicitly includes systems that involve a quadratic; report ordered pairs, not only the x-roots, unless the stem asks only for x.
5.4 Quadratic Inequalities and Quadratic-Linear Systems
Solving ax^2 + bx + c > 0 with a sign chart
- Solve the equation
ax^2 + bx + c = 0to find the critical points (roots). - Place those roots on a number line. They split the line into intervals.
- Test a point in each interval, or sketch the parabola (opens up vs down).
- Include or exclude the roots according to
>versus≥.
The graph picture: y = ax^2 + bx + c is positive where the parabola is above the x-axis and negative where it is below. That is the same Skills Insight 250–262 graph connection, now used for inequalities rather than for naming a vertex.
Worked example: x^2 − x − 6 < 0
Factor: (x − 3)(x + 2) < 0. Roots x = 3 and x = −2.
The parabola y = x^2 − x − 6 opens up (a = 1 > 0). It is below the x-axis between the roots.
| Interval | Test x | (x − 3)(x + 2) | Sign |
|---|---|---|---|
| x < −2 | x = −3 | (−6)(−1) = 6 | positive |
| −2 < x < 3 | x = 0 | (−3)(2) = −6 | negative |
| x > 3 | x = 4 | (1)(6) = 6 | positive |
We want negative, so −2 < x < 3. Strict inequality excludes the roots (where the product is 0). Interval notation: (−2, 3).
If the inequality had been ≤ 0, include the roots: [−2, 3]. If it had been > 0, take the outer rays: x < −2 or x > 3, written (−∞, −2) ∪ (3, ∞). Mixing those three answers is the standard CAT trap: the algebra of the factors is right, but the inequality direction or the endpoints are wrong.
Check an endpoint: at x = −2 or x = 3, y = 0, which is not less than 0. That is why the original strict inequality is open.
Opens-down inequality
−x^2 + 4x − 3 ≥ 0. Multiply by −1 and reverse: x^2 − 4x + 3 ≤ 0, or factor the original: −(x − 1)(x − 3) ≥ 0.
Roots 1 and 3. Original a = −1 < 0, so the parabola opens down and is above the x-axis between the roots. With ≥, include the roots: 1 ≤ x ≤ 3.
Never reverse the inequality when you do not multiply or divide by a negative. Expanding (x − 1)^2 ≥ 0 does not reverse anything; that inequality is true for all real x because a square is never negative.
Never test only one interval and assume. One test point per open interval is cheap insurance, and it catches a forgotten sign on a.
Quadratic inequalities that do not factor nicely
Use the formula for the roots, then the same sign chart. x^2 − 4x − 1 > 0 has roots 2 ± √5. Because a > 0, the parabola is positive outside those roots:
x < 2 − √5 or x > 2 + √5.
Do not approximate √5 ≈ 2.236 unless the options are decimals. Exact endpoints are the AAF default, calculator icon or not.
If D < 0 and a > 0, then ax^2 + bx + c > 0 for all real x (the parabola never goes below the axis). If D < 0 and a < 0, then ax^2 + bx + c < 0 for all real x. If D = 0 and the inequality is strict (> or <), the double root does not satisfy a strict inequality. For (x − 4)^2 > 0, the solution is all reals except x = 4. For (x − 4)^2 < 0, the solution is empty.
Systems: a line and a parabola
Solve
y = x^2 − 2x − 3
y = x + 1
Substitution: set the expressions for y equal.
x^2 − 2x − 3 = x + 1
x^2 − 3x − 4 = 0
(x − 4)(x + 1) = 0
x = 4 or x = −1
When x = 4, y = 4 + 1 = 5. When x = −1, y = −1 + 1 = 0.
Two solutions: (4, 5) and (−1, 0).
Check in the parabola: 4^2 − 2·4 − 3 = 16 − 8 − 3 = 5. (−1)^2 − 2(−1) − 3 = 1 + 2 − 3 = 0. Both points lie on both graphs. Reporting only x = 4 and x = −1 is incomplete unless the stem asked only for x-values.
Zero, one, or two intersections
After substitution you get a quadratic in one variable. Its discriminant tells you how many intersection points:
| Result after substitution | Geometric meaning |
|---|---|
| Two distinct real x | Line cuts the parabola twice |
| One real x (double) | Line is tangent |
| No real x | Line misses the parabola |
Tangent example. y = x^2 and y = 2x − 1.
x^2 = 2x − 1 → x^2 − 2x + 1 = 0 → (x − 1)^2 = 0 → x = 1, y = 1. One solution (1, 1). The line just touches the parabola.
Miss example. y = x^2 + 2 and y = x.
x^2 + 2 = x → x^2 − x + 2 = 0 → D = 1 − 8 = −7 < 0. No real intersection. The line sits below the raised parabola.
A horizontal line y = k intersects y = ax^2 + bx + c at the solutions of ax^2 + bx + (c − k) = 0. A vertical line x = h intersects a function-parabola at exactly one point (h, a h^2 + b h + c), because a function has one y per x.
A second numeric system
Solve
x^2 + y = 7
y = 3x − 3
Substitute: x^2 + (3x − 3) = 7
x^2 + 3x − 10 = 0
(x + 5)(x − 2) = 0
x = −5 or x = 2
x = −5 ⇒ y = 3(−5) − 3 = −18. Point (−5, −18).
x = 2 ⇒ y = 6 − 3 = 3. Point (2, 3).
Check the first equation: 25 + (−18) = 7 and 4 + 3 = 7.
If the line is given as 2x + y = 5, solve for y = 5 − 2x and substitute the same way. Elimination is less natural because of the x^2 term; substitution is the AAF-default method for a quadratic-linear pair. Table 11 names this skill explicitly: systems that involve a quadratic, not only 2×2 linear systems.
Creating a system from a sentence
The point (x, y) lies on the parabola y = x^2 − 4 and on the line through (0, 5) with slope 1. Create:
y = x^2 − 4
y = x + 5
Then x^2 − 4 = x + 5 → x^2 − x − 9 = 0 → x = (1 ± √37)/2. Two real intersection points with exact coordinates. This is Table 11 creating quadratic equations in two variables meeting systems that involve a quadratic.
A motion example: height h = −16t^2 + 48t + 4 and a linear drone path h = 20 − 4t. Setting them equal is a quadratic-linear system in t and h. Solve for t first, then back-substitute for h if the item wants the meeting point, not only the time.
Simplifying before you solve
Quadratic inequalities and systems both punish unsimplified equations. Expand first:
(x − 1)^2 < 2x + 7
x^2 − 2x + 1 < 2x + 7
x^2 − 4x − 6 < 0
Roots 2 ± √10. Opens up, so 2 − √10 < x < 2 + √10.
For a system, expand (x + 1)^2 + y = 10 to x^2 + 2x + 1 + y = 10 before substituting y. Leaving the square packed makes substitution look like a cubic and wastes CAT time.
Calculator and exact form again
A system that yields x = (3 ± √17)/2 should stay in that form. The on-screen calculator can check that (3 + √17)/2 ≈ 3.56, but the matching option is usually the exact pair. Interval answers for inequalities should use the same exact roots. College Board’s calculator policy still forbids handheld calculators except with an approved accommodation.
Graphically, Skills Insight 250–262 still applies: the inequality solution is a set of x-values matching where the parabola is above or below the axis, and the system solution is a set of points matching intersections. Do not report a system’s x-roots without the corresponding y-values unless the question asks only for x.
Solve x^2 − x − 6 < 0.
Find the solution set of the system y = x^2 − 2x − 3 and y = x + 1.
How many real solutions does the system y = x^2 and y = 2x − 1 have?