2.3 Linear & Quadratic Inequalities and Sign Charts

2.3 Linear & Quadratic Inequalities and Sign Charts

Inequalities test a candidate's mastery of order axioms, algebraic signs, and interval topology. Unlike equations that yield discrete points, inequalities produce continuous sets of real numbers. On the Bocconi Admission Test, questions frequently test pitfalls: cross-multiplying across variables without knowing their sign, failing to invert inequality directions upon negative multiplication, or erroneously including denominator poles in solution intervals. Mastering the Sign Chart provides a rigorous method for solving rational and quadratic inequalities.

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First-Degree Linear Inequalities and Order Axioms

A linear inequality in $x$ takes the standard form $ax + b \gtrless 0$. Manipulations follow strict order axioms:

• Adding or subtracting any real value preserves direction: $A < B \iff A + c < B + c$.
• Multiplying or dividing by a positive number ($c > 0$) preserves direction: $A < B \iff Ac < Bc$.
• Multiplying or dividing by a negative number ($c < 0$) reverses direction: $A < B \iff Ac > Bc$.

For example, solving $-3x + 6 \le 15$: $-3x \le 9 \implies x \ge \frac{9}{-3} \implies x \ge -3 \implies x \in [-3, \infty)$

Compound (Double) Inequalities

Inequalities of the form $c < ax + b \le d$ represent simultaneous conjunctions (AND). Operations apply concurrently across all three members: $-1 \le 2x + 5 < 11 \implies -6 \le 2x < 6 \implies -3 \le x < 3$, yielding $[-3, 3)$.

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Quadratic Inequalities: Parabola Concavity and Roots

Normalize any quadratic inequality so the leading coefficient is positive ($a > 0$). The function $y = ax^2 + bx + c$ represents an upward-opening parabola ($\cup$). The solution set depends on the discriminant $\Delta = b^2 - 4ac$:

Parabola Sign Rule

For $a > 0$ with two distinct roots $x_1 < x_2$:

• The expression is positive ($> 0$) on the external intervals: $x < x_1$ or $x > x_2$.
• The expression is negative ($< 0$) on the internal interval: $x_1 < x < x_2$.

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The Sign Chart (Table of Signs) Method

For products and quotients of polynomials, construct a sign chart via this 4-step process:

1. Factor completely: Decompose expressions into linear factors $(x - r)$ and irreducible quadratics.
2. Locate critical points: Identify all roots where any factor equals zero. Order them on the number line: $r_1 < r_2 < \dots < r_k$.
3. Partition into intervals: Divide $\mathbb{R}$ into $k + 1$ sub-intervals: $(-\infty, r_1), (r_1, r_2), \dots, (r_k, \infty)$.
4. Evaluate signs: Determine the sign of each factor in each column. Multiply signs: an odd count of negative factors yields a negative product; an even count yields a positive product.

Factor Multiplicity Rule

• Odd Multiplicity ($(x - r)^1, (x - r)^3$): The factor changes sign across $r$.
• Even Multiplicity ($(x - r)^2, (x - r)^4$): The factor vanishes at $r$ but maintains its sign on both sides of $r$.

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Rational Inequalities and Denominator Traps

A rational inequality has the form $\frac{P(x)}{Q(x)} \gtrless 0$.

The Variable Cross-Multiplication Trap

Never cross-multiply by an algebraic denominator containing variables. In $\frac{1}{x} < 2$, multiplying by $x$ assumes $x > 0$. If $x < 0$, multiplication flips the inequality! The correct method is moving all terms to one side: $\frac{1}{x} - 2 < 0 \implies \frac{1 - 2x}{x} < 0$ Critical points are $x = 0$ (pole) and $x = 1/2$ (zero), giving solution $(-\infty, 0) \cup (1/2, \infty)$.

Zero vs. Pole Boundaries

In inclusive inequalities ($\ge 0$ or $\le 0$):

• Numerator Zeros ($P(x) = 0$): Included in the solution set with square brackets $[;]$.
• Denominator Poles ($Q(x) = 0$): Division by zero is undefined. Denominator roots are strictly excluded with round parentheses $(;)$.

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Systems of Simultaneous Inequalities

A system of inequalities connects multiple conditions with the conjunction AND: $\begin{cases} f(x) \ge 0 \\ g(x) < 0 \end{cases}$ The solution set is the set intersection: $S_{\text{total}} = S_1 \cap S_2$.

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Worked Step-by-Step Examples

Example 1: Rational Inequality with Non-Zero Constant

Solve $\frac{2x - 5}{x + 1} \le 1$ for all real $x$:

1. Move constant to left: $\frac{2x - 5}{x + 1} - 1 \le 0$.
2. Common denominator: $\frac{2x - 5 - (x + 1)}{x + 1} \le 0 \implies \frac{x - 6}{x + 1} \le 0$.
3. Critical points: $x = 6$ (zero, included $[$), $x = -1$ (pole, excluded $($).
4. Interval signs:
   • $x < -1$: $(-)/(-) = (+)$
   • $-1 < x < 6$: $(-) /(+) = (-)$
   • $x > 6$: $(+) /(+) = (+)$
5. Select negative intervals: $S = (-1, 6]$.

Example 2: Parameter Inequality Valid for All Reals

Find all $m$ such that $(m - 1)x^2 - 2(m - 1)x + 3 > 0$ holds for all $x \in \mathbb{R}$:

1. Degenerate case $m = 1$: $0x^2 - 0x + 3 > 0 \implies 3 > 0$ (true for all $x$). So $m = 1$ is valid.
2. Quadratic case $m \neq 1$: Requires $a > 0 \implies m > 1$, and $\Delta/4 < 0$: $\frac{\Delta}{4} = (m - 1)^2 - 3(m - 1) = (m - 1)(m - 4) < 0 \implies 1 < m < 4$
3. Combine cases: $m \in [1, 4)$.

Example 3: Simultaneous System

Find all integer solutions to $\begin{cases} x^2 - 4x - 5 \le 0 \ \frac{x - 1}{x - 3} > 0 \end{cases}$:

1. Solve quadratic: $(x - 5)(x + 1) \le 0 \implies S_1 = [-1, 5]$.
2. Solve rational: Critical points $1, 3$. External positive regions: $S_2 = (-\infty, 1) \cup (3, \infty)$.
3. Intersect sets: $S = [-1, 1) \cup (3, 5]$.
4. Integer solutions: in $[-1, 1)$, integers are ${-1, 0}$; in $(3, 5]$, integers are ${4, 5}$. Total integer count $= 4$.