Quadratic Inequalities: Sketch and Sign Table Methods

From Equations to Inequalities

Solving x² − 5x + 6 = 0 gives two values, x = 2 and x = 3. But what about x² − 5x + 6 > 0, or x² − 5x + 6 ≤ 0? Now the answer is a whole range of values. Quadratic inequalities like these appear in WAEC Further Mathematics, JAMB and first-year university courses, and in practical problems such as finding when a projectile is above a certain height or when a business makes a profit. This tutorial shows two reliable methods, the graph sketch and the sign table, explains how to write the answer correctly, and works through the special cases where the quadratic has no real roots.

The Key Idea: Where Is the Quadratic Positive?

A quadratic expression ax² + bx + c can only change sign where it equals zero, at its roots. Between and outside the roots its sign stays the same. So the method is always:

Rearrange so that one side is 0.
Find the critical values (the roots of the quadratic).
Decide the sign of the quadratic in each region, using a sketch or a sign table.
Choose the regions that satisfy the inequality and write the answer.

Never "solve" a quadratic inequality by taking square roots of both sides or by treating it like an equation. For example, x² > 9 does not mean x > 3 only; x = −4 also works because (−4)² = 16 > 9.

Method 1: Sketching the Graph

The graph of y = ax² + bx + c is a parabola. When a > 0 it opens upwards (a U shape), so it is below the x-axis between the roots and above it outside the roots.

Example 1. Solve x² − 5x + 6 > 0.

Factorise: (x − 2)(x − 3) > 0. The roots are 2 and 3. The parabola opens upwards, so it is above the axis (positive) to the left of 2 and to the right of 3.

Answer: x < 2 or x > 3.

Example 2. Solve x² − 5x + 6 ≤ 0.

The same parabola is below or on the axis between the roots, including the roots themselves because of the "or equal to".

Answer: 2 ≤ x ≤ 3.

This pattern is worth remembering for any quadratic with a > 0 and two roots α < β:

(x − α)(x − β) > 0 gives x < α or x > β (outside the roots).
(x − α)(x − β) < 0 gives α < x < β (between the roots).

Method 2: The Sign Table

The sign table is useful when you prefer not to draw, and it extends to products of more than two factors. List the critical values in order, then find the sign of each factor in each interval.

Example 3. Solve (x + 1)(x − 4) < 0.

The critical values are −1 and 4, which split the number line into three intervals.

x < −1 (test x = −2): (x + 1) is negative, (x − 4) is negative, product positive.
−1 < x < 4 (test x = 0): (x + 1) is positive, (x − 4) is negative, product negative.
x > 4 (test x = 5): both positive, product positive.

We want the product to be negative, so the answer is −1 < x < 4.

Solving Quadratic Inequalities That Need Rearranging

Example 4. Solve 2x² + x ≥ 6.

Move everything to one side: 2x² + x − 6 ≥ 0. Factorise: (2x − 3)(x + 2) ≥ 0. Check: 2x² + 4x − 3x − 6 = 2x² + x − 6 ✓.

The roots are x = 3/2 and x = −2. The coefficient of x² is positive, so the quadratic is positive outside the roots and zero at them.

Answer: x ≤ −2 or x ≥ 3/2.

Example 5. Solve x² < 4x.

Do not divide by x: you don't know whether x is positive or negative, and dividing by a negative number would reverse the inequality. Instead rearrange: x² − 4x < 0, so x(x − 4) < 0. The roots are 0 and 4.

Answer: 0 < x < 4. (Dividing by x would have given x < 4, wrongly including negative numbers like x = −1, where 1 < −4 is false.)

Example 6. Solve 6 + x − x² > 0.

Here the coefficient of x² is negative. Multiply by −1 and reverse the sign: x² − x − 6 < 0, so (x − 3)(x + 2) < 0.

Answer: −2 < x < 3. (Alternatively, sketch y = 6 + x − x², an upside-down parabola, which is above the axis between its roots.)

When the Roots Are Not Whole Numbers

Example 7. Solve x² − 2x − 4 ≤ 0.

This doesn't factorise neatly, so use the quadratic formula: x = [2 ± √(4 + 16)] / 2 = (2 ± √20)/2 = 1 ± √5.

So the roots are 1 − √5 ≈ −1.24 and 1 + √5 ≈ 3.24. The quadratic is at or below zero between them.

Answer: 1 − √5 ≤ x ≤ 1 + √5, or approximately −1.24 ≤ x ≤ 3.24.

Special Cases: No Real Roots or a Repeated Root

Use the discriminant, b² − 4ac.

If b² − 4ac < 0 and a > 0, the parabola never touches the x-axis and lies entirely above it. Then ax² + bx + c > 0 for all real x, and ax² + bx + c < 0 has no solution.
If b² − 4ac = 0, the parabola just touches the axis at one point.

Example 8. Solve x² + 2x + 5 > 0.

The discriminant is 4 − 20 = −16 < 0 and a = 1 > 0, so the expression is always positive. Completing the square confirms it: x² + 2x + 5 = (x + 1)² + 4, which is at least 4. The answer is all real numbers x.

Example 9. Solve x² − 6x + 9 ≤ 0.

x² − 6x + 9 = (x − 3)², which is never negative and equals 0 only at x = 3. So the only solution is x = 3. (The strict inequality (x − 3)² < 0 would have no solution, and (x − 3)² > 0 holds for every x except 3.)

Applications of Quadratic Inequalities

Range of k for real roots. A classic Further Mathematics question: for what values of k does x² + kx + 9 = 0 have real roots? Real roots need b² − 4ac ≥ 0, so k² − 36 ≥ 0, giving (k − 6)(k + 6) ≥ 0, so k ≤ −6 or k ≥ 6.

Projectile height. A ball's height in metres after t seconds is h = 20t − 5t². When is it above 15 m? Solve 20t − 5t² > 15, so 5t² − 20t + 15 < 0, so t² − 4t + 3 < 0, so (t − 1)(t − 3) < 0. The ball is above 15 m for 1 < t < 3 seconds.

Profit. A trader's daily profit is P = −x² + 40x − 300 naira (in hundreds) when selling x items. P > 0 when x² − 40x + 300 < 0, that is (x − 10)(x − 30) < 0, so the trader makes a profit when selling between 10 and 30 items.

Writing the Answer Correctly

"Between the roots" is written as one combined inequality: −2 < x < 3.
"Outside the roots" needs two parts joined by "or": x < −2 or x > 3. Never write 3 < x < −2, which is impossible.
In set or interval notation: −2 < x < 3 is (−2, 3), and x ≤ −2 or x ≥ 3/2 is (−∞, −2] ∪ [3/2, ∞). Square brackets include the endpoint.
On a number line, use a filled dot for ≤ or ≥ and an open circle for < or >.

Summary

Rearrange to compare with 0, find the roots, then find where the quadratic is positive or negative.
For a > 0: "> 0" means outside the roots; "< 0" means between them.
Multiplying by a negative number reverses the inequality; never divide by an unknown whose sign you don't know.
If the discriminant is negative, the quadratic keeps one sign for every x.
Quadratic inequalities give ranges for parameters, times, quantities and prices in real problems.