5.3 Quadratic Inequalities in One/Two Variables & Optimization

5.3 Quadratic Inequalities in One/Two Variables & Optimization

Quadratic inequalities expand quadratic algebra into relational constraints, defining continuous solution intervals on the real number line or multi-dimensional regions on the coordinate plane. On the FTCE Mathematics 6–12 examination, educators must demonstrate rigorous mastery of boundary value analysis, sign chart construction, coordinate plane region shading, and non-linear optimization constraints, while addressing persistent student notation errors.

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1. Single-Variable Quadratic Inequalities & Boundary Principles

A quadratic inequality in one variable can be expressed in one of four standard configurations:

$ax^2 + bx + c > 0, \quad ax^2 + bx + c \ge 0, \quad ax^2 + bx + c < 0, \quad ax^2 + bx + c \le 0$

where $a, b, c \in \mathbb{R}$ and $a \neq 0$.

The Boundary Value Principle

Because the polynomial function $f(x) = ax^2 + bx + c$ is continuous across all $\mathbb{R}$, by the Intermediate Value Theorem, $f(x)$ can change algebraic sign (from positive to negative or vice-versa) only at points where $f(x) = 0$. These real zeros are the critical boundary points.

Secondary educators teach two complementary analytical perspectives:

1. Method A: Test Point & Sign Chart Analysis: Partition the real number line into open intervals bounded by the real roots $r_1 < r_2$. Select an arbitrary test value within each interval and evaluate the sign of each factored binomial.
2. Method B: Geometric Parabola Analysis:
   • If $a > 0$, the parabola opens upward. The function dips below the $x$-axis between the roots ($f(x) < 0$ on $(r_1, r_2)$) and rises above the $x$-axis outside the roots ($f(x) > 0$ on $(-\infty, r_1) \cup (r_2, \infty)$).
   • If $a < 0$, the parabola opens downward. The function is positive between the roots ($f(x) > 0$ on $(r_1, r_2)$) and negative outside the roots ($f(x) < 0$ on $(-\infty, r_1) \cup (r_2, \infty)$).

                 Upward Parabola (a > 0)
         + + +                             + + + 
        ------\                           /------  x-axis
               \ - - - - - - - - - - - - / 
                r1                      r2
         f(x) > 0        f(x) < 0         f(x) > 0

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2. Step-by-Step Exemplar & Sign Chart Mechanics

Problem: Solve the quadratic inequality $-2x^2 + 5x + 12 \ge 0$ over $\mathbb{R}$ and express the solution set in interval notation.

Step 1: Standardize with a Positive Leading Coefficient

Multiply the entire inequality by $-1$. Recall the fundamental order axiom of real numbers: multiplying or dividing an inequality by a negative scalar reverses the inequality symbol:

$2x^2 - 5x - 12 \le 0$

Step 2: Find Critical Boundary Points

Solve the associated boundary equation $2x^2 - 5x - 12 = 0$. Factoring via the $ac$-method ($ac = 2(-12) = -24$, factors $-8$ and $3$):

$2x^2 - 8x + 3x - 12 = 2x(x - 4) + 3(x - 4) = (2x + 3)(x - 4) = 0$

Setting each factor to zero yields the boundary roots:

$r_1 = -\frac{3}{2}, \quad r_2 = 4$

Step 3: Construct the Sign Chart

The critical points partition the real line into three disjoint intervals: $\left(-\infty, -\frac{3}{2}\right)$, $\left(-\frac{3}{2}, 4\right)$, and $(4, \infty)$.

+---------------------------------------------------------------------------------+
| Interval          | (-inf, -3/2)    | x = -3/2 | (-3/2, 4)   | x = 4 | (4, inf) |
| Test Value        | x = -2          |          | x = 0       |       | x = 5    |
+-------------------+-----------------+----------+-------------+-------+----------+
| Factor (2x + 3)   | 2(-2)+3 = -1 (-) |    0     | 2(0)+3 = +3 |   +   | 2(5)+3=+ |
| Factor (x - 4)    | -2 - 4 = -6 (-) |    -     | 0 - 4 = -4  |   0   | 5 - 4 =+ |
+-------------------+-----------------+----------+-------------+-------+----------+
| Product P(x) Sign |    (+) * (-) = +|    0     | (+) * (-) = -|   0   | (+) * (+)=+|
+---------------------------------------------------------------------------------+

Step 4: Extract and Format the Solution Set

We seek values where $P(x) = (2x + 3)(x - 4) \le 0$ (negative or zero). The sign chart confirms the product is negative on $\left(-\frac{3}{2}, 4\right)$ and zero at the boundaries. Because the original inequality $\ge$ (and transformed $\le$) is non-strict, boundary roots are included with square brackets:

$\text{Solution Set: } \left[-\frac{3}{2}, 4\right]$

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3. Degenerate & Boundary Cases ($\Delta \le 0$)

When the discriminant $\Delta = b^2 - 4ac$ is non-positive, the quadratic function does not cross the $x$-axis, yielding distinct solution behaviors:

Case 1: Repeated Real Root ($\Delta = 0$)

The quadratic forms a perfect square $a(x - h)^2$:

• $(x - 3)^2 > 0 \implies x \in (-\infty, 3) \cup (3, \infty)$ (all reals except $x = 3$).
• $(x - 3)^2 \ge 0 \implies x \in (-\infty, \infty)$ (satisfied by all real numbers).
• $(x - 3)^2 < 0 \implies \emptyset$ (empty set; no real square is negative).
• $(x - 3)^2 \le 0 \implies {3}$ (the solitary isolated point $x = 3$).

Case 2: Strictly Complex Roots ($\Delta < 0$)

The parabola never touches the $x$-axis, remaining strictly on one side of zero for all $x \in \mathbb{R}$:

• If $a > 0$: The entire parabola lies above the $x$-axis ($f(x) > 0$ for all $x$). Therefore, $ax^2 + bx + c > 0$ has solution $(-\infty, \infty)$, while $ax^2 + bx + c \le 0$ has solution $\emptyset$.
• If $a < 0$: The entire parabola lies below the $x$-axis ($f(x) < 0$ for all $x$). Therefore, $ax^2 + bx + c < 0$ has solution $(-\infty, \infty)$, while $ax^2 + bx + c \ge 0$ has solution $\emptyset$.

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4. Two-Variable Quadratic Inequalities on the Coordinate Plane

A two-variable quadratic inequality takes the form $y > f(x)$, $y \ge f(x)$, $y < f(x)$, or $y \le f(x)$, where $f(x) = ax^2 + bx + c$. Its solution set represents a planar region bounded by a parabola.

Boundary Curve Conventions

• Strict Inequalities ($<, >$): Rendered as a dashed (broken) curve, indicating that points lying directly on the parabola $y = ax^2 + bx + c$ are excluded from the solution set.
• Non-Strict Inequalities ($\le, \ge$): Rendered as a solid curve, indicating that boundary points are included.

Region Testing & Shading Rules

To determine which side of the boundary to shade:

1. Choose an arbitrary test point $(x_0, y_0)$ not on the boundary parabola. The origin $(0, 0)$ is ideal unless the curve passes through it; otherwise, select the vertex shifted vertically $(h, k \pm 1)$.
2. Substitute $(x_0, y_0)$ into the inequality. If true, shade the region containing $(x_0, y_0)$; if false, shade the opposite region.
3. Geometric Orientation Shortcut:
   • $y > ax^2 + bx + c$: Shade the region vertically above the parabola (inside for $a > 0$; outside for $a < 0$).
   • $y < ax^2 + bx + c$: Shade the region vertically below the parabola (outside for $a > 0$; inside for $a < 0$).

Systems of Quadratic Inequalities

A system of inequalities represents the intersection (simultaneous overlap) of individual regions. For example, the system:

$\begin{cases} y \ge x^2 - 4 \\ y \le -x + 2 \end{cases}$

defines a closed planar region bounded below by the solid parabola $y = x^2 - 4$ and above by the solid line $y = -x + 2$. The intersection points are found by setting $x^2 - 4 = -x + 2 \implies x^2 + x - 6 = 0 \implies (x + 3)(x - 2) = 0$, giving boundary intersection vertices at $(-3, 5)$ and $(2, 0)$.

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5. Non-Linear Feasibility & Optimization Constraints

In business operations and engineering design, quadratic inequalities define feasible operational envelopes. Consider a firm whose profit function is $P(x) = -x^2 + 70x - 600$, where $x$ represents production volume in thousands of units. To remain profitable, the firm must maintain $P(x) \ge 0$:

$-x^2 + 70x - 600 \ge 0 \implies x^2 - 70x + 600 \le 0$ $(x - 10)(x - 60) \le 0 \implies 10 \le x \le 60$

The firm's viable production range is $10{,}000$ to $60{,}000$ units. Within this feasible constraint interval, vertex evaluation identifies the maximum profit at $x = 35$ ($35{,}000$ units).

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6. Common Student Misconceptions & Pedagogical Strategies