Inequalities — Linear, Quadratic, and Absolute Value

Inequalities: Beyond ”=” Signs

In school, we spend most time on equations — finding exact values. But most real problems are about constraints: a bus can carry at most 40 people, a student needs at least 35% marks to pass, a factory must produce at most 100 units per day to avoid overtime costs.

These are inequalities, and solving them means finding a range of values (an interval), not a single number. The techniques look similar to equations, but there are crucial differences — especially when multiplying or dividing by negative numbers, or squaring both sides.

For CBSE Class 11, inequalities (specifically linear inequalities in two variables) is a dedicated chapter. For JEE, quadratic and absolute value inequalities appear in almost every paper under “Functions,” “Sets,” and “Algebra” sections.

Key Terms & Definitions

Inequality: A mathematical statement comparing two expressions using <, >, ≤, ≥, or ≠.

Linear inequality: Inequality where the variable appears to the first power. Example: $2x - 3 > 5$ .

Quadratic inequality: Inequality involving $x^2$ . Example: $x^2 - 5x + 6 < 0$ .

Absolute value inequality: Inequality involving $|x|$ . Example: $|x - 3| \leq 2$ .

Solution set: The set of all values of the variable that satisfy the inequality. Usually an interval or union of intervals.

Open interval $(a, b)$ : All values strictly between $a$ and $b$ ; $a$ and $b$ not included.

Closed interval $[a, b]$ : All values between $a$ and $b$ , including $a$ and $b$ .

The Key Rule: Flip the Sign When Multiplying by Negative

The most important difference between equations and inequalities:

\text{If } a &lt; b \text{ and } c &lt; 0, \text{ then } ac &gt; bc

When you multiply (or divide) both sides of an inequality by a negative number, the direction of the inequality reverses.

Example: $-3x < 12 \Rightarrow x > -4$ (dividing by $-3$ flips ”<” to ”>”)

Not flipping this sign is the single most common source of errors in inequalities.

Linear Inequalities

Single Variable

Solve: $3x - 7 \geq 2x + 5$

3x - 7 \geq 2x + 5

x \geq 12

Solution: $x \in [12, \infty)$ .

Solve: $-4x + 2 > 14$

-4x &gt; 12

x &lt; -3 \quad \text{(sign flipped when dividing by -4)}

Solution: $x \in (-\infty, -3)$ .

Two-Variable Linear Inequalities (CBSE Class 11)

To solve $2x + y \leq 4$ :

Draw the line $2x + y = 4$ (use solid line for ≤ or ≥; dashed line for strict < or >)
Test a point not on the line (e.g., origin $(0,0)$ : $2(0) + 0 = 0 \leq 4$ ✓)
Shade the half-plane containing the test point (since origin satisfies the inequality)

The solution is the shaded region, including the boundary line.

System of linear inequalities: Solve each inequality graphically; the solution is the intersection of all shaded regions.

CBSE Class 11 exams frequently ask you to find the feasible region for a system of 3–4 linear inequalities. Graph each inequality, shade correctly, and mark the feasible region (the area satisfying ALL inequalities simultaneously). Corner points of this region are important — they are used in Linear Programming (Class 12).

Quadratic Inequalities

Quadratic inequalities require a different approach. The key method is the sign chart (wavy curve / number line method).

Factorisation Method

Solve: $x^2 - 5x + 6 < 0$

Step 1: Factorise: $(x-2)(x-3) < 0$

Step 2: Find the critical points (roots): $x = 2$ and $x = 3$ .

Step 3: Draw number line with critical points marked.

Step 4: Test the sign of $(x-2)(x-3)$ in each interval:

$x < 2$ : e.g., $x = 0$ : $(0-2)(0-3) = (-)(-)= (+) > 0$ ✗
$2 < x < 3$ : e.g., $x = 2.5$ : $(0.5)(-0.5) = (-) < 0$ ✓
$x > 3$ : e.g., $x = 4$ : $(2)(1) = (+) > 0$ ✗

Solution: $x \in (2, 3)$

The pattern for factored quadratics: $(x-a)(x-b) < 0$ (with $a < b$ ) has solution $x \in (a, b)$ . For > 0, the solution is $x \in (-\infty, a) \cup (b, \infty)$ .

For $(x-a)(x-b)$ where $a < b$ :

$(x-a)(x-b) < 0 \Rightarrow x \in (a, b)$

$(x-a)(x-b) > 0 \Rightarrow x \in (-\infty, a) \cup (b, \infty)$

Mnemonic: “Less than → between the roots. Greater than → outside the roots.”

The Wavy Curve Method (for Higher Degrees)

For $f(x) = \frac{(x-1)(x-3)}{(x-2)} > 0$ :

Step 1: Find all critical points (zeros of numerator and denominator): $x = 1, 2, 3$ .

Step 2: Plot on number line in order: 1, 2, 3.

Step 3: The highest power coefficient determines the rightmost sign. For large positive $x$ , the expression is positive (all three factors are positive). Start with + on the right.

Step 4: Alternate signs across each critical point (wavy curve alternation, assuming each root is simple):

$x > 3$ : positive (+)
$2 < x < 3$ : negative (−)
$1 < x < 2$ : positive (+)
$x < 1$ : negative (−)

Step 5: We want > 0, so take regions with +:

$x \in (1, 2) \cup (3, \infty)$

Note: $x = 2$ is excluded (denominator = 0, undefined); $x = 1, 3$ give zero (not > 0, so excluded for strict inequality).

The wavy curve method is the standard JEE technique for all polynomial and rational inequalities. It works because every time we cross a simple root (odd multiplicity), the sign changes. If a root has even multiplicity (like $(x-2)^2$ ), the sign does NOT change at that root (the wavy curve “touches” the x-axis and bounces back).

Absolute Value Inequalities

The Two Key Rules

Rule 1: $|x| < a$ (where $a > 0$ ) $\Leftrightarrow -a < x < a$

Meaning: $x$ is within distance $a$ from zero → $x \in (-a, a)$ .

Rule 2: $|x| > a$ (where $a > 0$ ) $\Leftrightarrow x < -a$ or $x > a$

Meaning: $x$ is more than distance $a$ from zero → $x \in (-\infty, -a) \cup (a, \infty)$ .

For $|x - c| < a$ : this means $x$ is within distance $a$ from $c$ → $c-a < x < c+a$ .

Example 1: Solve $|x - 3| \leq 2$

By Rule 1: $-2 \leq x - 3 \leq 2 \Rightarrow 1 \leq x \leq 5$

Solution: $x \in [1, 5]$

Geometric interpretation: all points within 2 units of 3 on the number line.

Example 2: Solve $|2x + 1| > 5$

By Rule 2: $2x + 1 < -5$ or $2x + 1 > 5$

$2x < -6$ → $x < -3$ , or $2x > 4$ → $x > 2$

Solution: $x \in (-\infty, -3) \cup (2, \infty)$

Example 3: Solve $|x| \geq x + 2$ (case-based approach needed)

Case 1: $x \geq 0$ : $|x| = x$ , so $x \geq x + 2 \Rightarrow 0 \geq 2$ . Contradiction — no solution in this case.

Case 2: $x < 0$ : $|x| = -x$ , so $-x \geq x + 2 \Rightarrow -2x \geq 2 \Rightarrow x \leq -1$ .

Combined with $x < 0$ : solution is $x \leq -1$ .

Solution: $x \in (-\infty, -1]$

Inequalities with Fractions

Be careful: when solving $\frac{x+1}{x-2} > 0$ , do NOT multiply both sides by $(x-2)$ unless you know its sign.

Use the wavy curve instead: critical points at $x = -1$ (numerator = 0) and $x = 2$ (denominator = 0).

Sign analysis:

$x > 2$ : both positive → fraction positive ✓
$-1 < x < 2$ : numerator positive, denominator negative → negative ✗
$x < -1$ : both negative → fraction positive ✓

Solution: $x \in (-\infty, -1) \cup (2, \infty)$

Note: $x = 2$ excluded (denominator = 0); $x = -1$ excluded (strict inequality, fraction = 0).

Solved Examples

Easy — CBSE Level

Q: Solve $5x - 3 \geq 3x + 7$ .

Solution: $5x - 3x \geq 7 + 3$ $2x \geq 10$ $x \geq 5$

Solution: $x \in [5, \infty)$ .

Medium — JEE Main Level

Q: Find the range of $x$ such that $x^2 - x - 6 > 0$ .

Solution: Factorise: $(x-3)(x+2) > 0$ .

Roots: $x = 3, x = -2$ .

Sign: positive outside roots → $x \in (-\infty, -2) \cup (3, \infty)$ .

Hard — JEE Advanced Level

Q: Solve: $\frac{(x-1)(x+2)}{(x-3)^2(x+4)} \leq 0$ .

Solution: Critical points: $x = -4, -2, 1, 3$ .

For large positive $x$ : all factors positive, $(x-3)^2$ is always positive → expression positive.

Wavy curve sign (rightmost = positive):

$x > 3$ : positive (+)
$1 < x < 3$ : negative (note: passing through $x = 3$ is even multiplicity → sign doesn’t change) wait — actually $(x-3)^2$ has even multiplicity, so the sign does NOT change at $x = 3$ . The sign before and after $x = 3$ is the same. Let me redo:

Critical points with multiplicities: $x = -4$ (odd, sign changes), $x = -2$ (odd, changes), $x = 1$ (odd, changes), $x = 3$ (even, no change).

Starting with + for $x > 3$ :

$1 < x < 3$ : + (no change at $x = 3$ )
$-2 < x < 1$ : − (change at $x = 1$ )
$-4 < x < -2$ : + (change at $x = -2$ )
$x < -4$ : − (change at $x = -4$ )

We want ≤ 0, so negative regions or where expression = 0:

$-4 < x < -2$ : positive, no ✗
$-2 < x < 1$ : negative ✓
$x = -2$ : expression = 0 ✓ (include)
$x = 1$ : expression = 0 ✓ (include)
$x = -4$ : denominator defined ≠ 0, but wait: $x=-4$ gives denominator 0 → excluded
$x = 3$ : denominator = 0 → excluded

Solution: $x \in [-2, 1]$ , excluding $x = -4$ (undefined) — but $-4$ isn’t in this range anyway.

Negative regions also include $x < -4$ , so: $x \in (-\infty, -4) \cup [-2, 1]$ (excluding $x = -4$ , so the first part is $(-\infty, -4)$ ).

Final answer: $x \in (-\infty, -4) \cup [-2, 1]$

Exam-Specific Tips

CBSE Class 11: Linear inequalities in one and two variables. Know how to graph two-variable inequalities and find feasible regions. Interval notation is required.

JEE Main: Wavy curve method for polynomial and rational inequalities. Absolute value inequalities using the two standard rules. Combined inequalities (find x satisfying both conditions simultaneously).

JEE Advanced: Complex rational inequalities with multiple factors, including even-multiplicity roots. Finding domains of functions (which often reduces to solving inequalities).

Common Mistakes to Avoid

Mistake 1: Not flipping the inequality when multiplying/dividing by a negative. $-3x > 9$ → $x < -3$ (not $x > -3$ ). This is the most tested error.

Mistake 2: Squaring both sides of an inequality without checking signs. $x > 2$ does NOT imply $x^2 > 4$ for negative $x$ values. Squaring is safe only when both sides are non-negative. Use the wavy curve instead of squaring.

Mistake 3: Including the point where the denominator is zero. In $\frac{1}{x-2} > 0$ , the solution is $x > 2$ (strictly, since $x = 2$ makes the expression undefined).

Mistake 4: In absolute value inequalities, solving $|x - 3| < -2$ without realising there’s no solution. An absolute value is always ≥ 0, so it can never be less than a negative number. The solution set is the empty set.

Mistake 5: In the wavy curve method, not accounting for even-multiplicity roots. If a factor appears as $(x-a)^2$ , the sign does NOT change at $x = a$ . Draw the curve “touching” the axis at $a$ rather than crossing it.

Practice Questions

1. Solve $2x - 5 < 3$ and represent on a number line.

$2x < 8 \Rightarrow x < 4$ . Solution: $x \in (-\infty, 4)$ . On the number line: open circle at 4, arrow pointing left.

2. Solve $|3x - 6| \leq 9$ .

$|3x - 6| \leq 9 \Rightarrow -9 \leq 3x - 6 \leq 9 \Rightarrow -3 \leq 3x \leq 15 \Rightarrow -1 \leq x \leq 5$ . Solution: $x \in [-1, 5]$ .

3. For what values of $x$ is $x^2 - 4x + 3 \leq 0$ ?

Factorise: $(x-1)(x-3) \leq 0$ . Roots at $x = 1, 3$ . Between the roots, expression is negative (or zero at the roots). Solution: $x \in [1, 3]$ .

4. Solve $\frac{x-2}{x+3} > 0$ .

Critical points: $x = -3$ (denominator zero), $x = 2$ (numerator zero). Sign analysis: for $x > 2$ : positive ✓; for $-3 < x < 2$ : negative ✗; for $x < -3$ : positive ✓. Solution: $x \in (-\infty, -3) \cup (2, \infty)$ , excluding $x = -3$ .

5. Find all real $x$ such that $x^2 - 7|x| + 10 < 0$ .

Let $u = |x| \geq 0$ . Inequality becomes $u^2 - 7u + 10 < 0 \Rightarrow (u-2)(u-5) < 0 \Rightarrow 2 < u < 5$ . So $2 < |x| < 5 \Rightarrow x \in (-5, -2) \cup (2, 5)$ .

FAQs

Q: What is the difference between open and closed intervals in solutions? Use open parentheses ( ) for strict inequalities (< or >) — endpoints NOT included. Use closed brackets [ ] for non-strict inequalities (≤ or ≥) — endpoints included. For infinity, always use open brackets: $(-\infty, 5]$ is correct; $[-\infty, 5]$ is incorrect since infinity is not a real number.

Q: Can I always use the wavy curve method? Yes, for any polynomial or rational inequality that can be fully factored. For absolute value inequalities, use the modulus rules (|x| < a → between, |x| > a → outside). For transcendental inequalities (like $\sin x > 0$ ), you need a different approach based on the function’s graph.

Q: Why does the sign alternate in the wavy curve method? Each time we cross a simple root $x = r$ (where the factor $(x - r)$ appears once), the sign of $(x - r)$ changes from negative to positive (or vice versa). All other factors’ signs don’t change at $x = r$ . So the product changes sign exactly once, causing the alternation. For even-multiplicity roots, $(x - r)^2$ doesn’t change sign (it’s always ≥ 0), so the overall product’s sign doesn’t change.