5.4 Inequalities, Sign Charts & Absolute Value

Inequalities follow the same collecting-and-clearing steps as equations, with one structural exception that GMAT Quant exploits constantly: multiplying or dividing by a negative number reverses the inequality. Absolute value is distance on the number line. Polynomial and rational inequalities are not solved by “moving terms the way you would in an equation”; they are solved with a sign chart. This is Official Guide Math Review 3.2 inequalities, with absolute value as in Math Review 3.1.

There is still no calculator, and you still have about 128 seconds. A sign chart with two critical points is faster than plugging five answer choices into a quadratic inequality, and it is the only reliable method when the question asks for all integers that work.

Core rules, including the reverse-the-sign rule

1. Addition and subtraction never reverse the sign. If $a < b$, then $a + c < b + c$ for every real $c$, positive or negative.
2. Multiplying or dividing by a positive number preserves the sign. If $a < b$ and $c > 0$, then $ac < bc$.
3. Multiplying or dividing by a negative number reverses the sign. If $a < b$ and $c < 0$, then $ac > bc$.
4. You may add inequalities that point the same way: $a < b$ and $c < d$ imply $a + c < b + d$. You may not subtract them and keep the same direction: $a - c$ need not be less than $b - d$.

Worked example (reverse). Solve $-2x < 6$.

Divide by $-2$ and reverse: $x > -3$. Check: $x = 0$ gives $0 < 6$, true; $x = -4$ gives $8 < 6$, false. The boundary $x = -3$ gives equality, so it is excluded by the strict inequality.

Worked example (do not reverse). Solve $4x - 7 \ge 9$.

$4x \ge 16 \implies x \ge 4$

Dividing by $+4$ keeps $\ge$.

Taking reciprocals of positive quantities also reverses: if $0 < a < b$, then $1/a > 1/b$. If the signs are mixed, reciprocals are unsafe; use a sign chart instead.

Compound inequalities

A compound AND inequality $a < x < b$ is one interval. Perform the same operation on all three parts.

Worked example. Solve $1 < 2x - 3 \le 7$.

Add $3$: $4 < 2x \le 10$. Divide by $2$: $2 < x \le 5$. The left end is open (because of $<$) and the right end is closed (because of $\le$). In interval notation: $(2, 5]$.

A compound OR inequality is two rays. Solve each piece separately and union the results. That is exactly what $|x - a| > b$ produces, below.

Absolute value as distance

$|x| = \begin{cases} x & \text{if } x \ge 0 \\ -x & \text{if } x < 0 \end{cases}$

Geometrically, $|x - a|$ is the distance between $x$ and $a$ on the number line. The number $b$ in $|x - a| , ? , b$ must be compared with $0$ before you unfold the inequality.

• If $b < 0$, then $|x - a| < b$ is empty (an absolute value cannot be negative) and $|x - a| > b$ is all real numbers.
• If $b = 0$, then $|x - a| < 0$ is empty, $|x - a| \le 0$ is just $x = a$, and $|x - a| > 0$ is $x \ne a$.
• If $b > 0$, use the translation table.

Worked example ($<$). Solve $|2x - 5| \le 9$.

Unfold: $-9 \le 2x - 5 \le 9$. Add $5$: $-4 \le 2x \le 14$. Divide by $2$: $-2 \le x \le 7$.

Worked example ($>$). Solve $|x + 2| > 4$, i.e. distance from $-2$ greater than $4$.

$x + 2 > 4 \quad \text{or} \quad x + 2 < -4$

So $x > 2$ or $x < -6$.

Equations: $|x - 5| = 3$ means $x = 8$ or $x = 2$. There is no “interval” for an equality; there are two points.

Sign charts for quadratic inequalities

To solve $(x - 3)(x + 4) \le 0$:

1. Move all terms to one side so you compare a product (or a rational expression) with $0$.
2. Factor completely. Critical points are the zeros of the numerator and, for rationals, the zeros of the denominator.
3. Plot the critical points on a number line. Zeros of the denominator are open (never included).
4. Test one convenient number in each open interval to read the sign of the product.
5. Select the intervals whose sign matches $< 0$ or $> 0$, and include or exclude endpoints according to $\le$ versus $<$.

For $(x - 3)(x + 4)$ the zeros are $x = 3$ and $x = -4$. Intervals: $(-\infty, -4)$, $(-4, 3)$, $(3, \infty)$.

• Test $x = -5$: $(-8)(-1) = +$.
• Test $x = 0$: $(-3)(4) = -$.
• Test $x = 4$: $(1)(8) = +$.

A product of two linear factors is negative between the roots (the parabola $y = x^2 + \cdots$ opens upward). For $\le 0$ the solution is $[-4, 3]$, endpoints included because equality is allowed.

Worked example (integers). How many integers satisfy $x^2 - 4x - 21 < 0$?

$(x - 7)(x + 3) < 0 \implies -3 < x < 7$

Integers strictly between $-3$ and $7$: $-2, -1, 0, 1, 2, 3, 4, 5, 6$. That is nine integers. Including the roots would incorrectly add $-3$ and $7$.

For $ax^2 + bx + c > 0$ with $a > 0$ and two real roots, the solution is the two outer rays, not the interval between the roots. Draw the parabola: positive outside, negative between.

Domain traps: do not divide by an expression that might be 0

The deadliest Quant inequality error is treating a variable factor like a known positive constant.

Illegal: from $\dfrac{x - 1}{x + 2} > 0$, “multiply both sides by $x + 2$” to conclude $x - 1 > 0$. That assumes $x + 2 > 0$. When $x + 2 < 0$ the inequality reverses, and when $x + 2 = 0$ the original is undefined.

Legal: sign chart. Critical points $x = 1$ (zero) and $x = -2$ (undefined). Intervals $(-\infty, -2)$, $(-2, 1)$, $(1, \infty)$.

• Test $x = -3$: $(-4)/(-1) = +$.
• Test $x = 0$: $(-1)/(2) = -$.
• Test $x = 2$: $(1)/(4) = +$.

The quotient is positive on $(-\infty, -2) \cup (1, \infty)$. Do not include $x = -2$. Do not include $x = 1$ because the inequality is strict.

Another illegal division. From $x(x - 3) > 0$, dividing by $x$ to get $x - 3 > 0$ drops the entire interval $x < 0$ where both factors are negative and the product is still positive. The sign chart (or “same sign”) gives $x < 0$ or $x > 3$, excluding $x = 0$ and $x = 3$.

If you insist on multiplying a rational inequality by the denominator, you must split cases $x + 2 > 0$ and $x + 2 < 0$ and reverse on the second case, then exclude $x = -2$. The sign chart is shorter and harder to botch under time pressure.

Traps

• Forgetting to reverse when dividing by $-3$, or reversing when dividing by $+3$.
• Translating $|x - a| > b$ into a single interval $a - b < x < a + b$ (that translation is for $<$, not $>$).
• Including a root in a strict quadratic inequality, or including a denominator zero in any rational inequality.
• Multiplying both sides by $x$ or by $x - c$ without a case split.
• Treating $|A| < B$ as having solutions when $B$ is negative.