Quadratic Inequality: Graphical Approach

Why the Graphical Approach Works Best

Quadratic inequalities such as x² − 5x + 6 < 0 ask you to find the range of x-values where a quadratic expression is positive or negative. While you can try to solve these algebraically using sign analysis, the graphical approach is more intuitive and less error-prone — especially under exam pressure.

The method is simple: sketch the quadratic curve, find where it crosses the x-axis, and then read off the regions where the curve is above or below the x-axis. This topic appears on IGCSE Extended Paper 4 and is worth three to five marks.

The Method in Five Steps

Rearrange the inequality so that one side is zero
Factorise (or use the quadratic formula to find the roots)
Sketch the parabola, marking the x-intercepts
Identify the region that satisfies the inequality
Write the solution using inequality notation

Worked Example 1: x² − 5x + 6 < 0

Step 1: The inequality is already in the correct form (right side is zero).

Step 2: Factorise the left side.

x² − 5x + 6 = (x − 2)(x − 3)

The roots are x = 2 and x = 3.

Step 3: Sketch the parabola y = x² − 5x + 6.

Since the coefficient of x² is positive (it is +1), the parabola opens upward — it is U-shaped. It crosses the x-axis at x = 2 and x = 3.

Step 4: We need x² − 5x + 6 < 0, meaning the curve is below the x-axis.

Looking at the U-shaped curve, it is below the x-axis between the two roots.

Step 5: The solution is 2 < x < 3.

Worked Example 2: x² − 2x − 8 ≥ 0

Step 1: Right side is already zero.

Step 2: Factorise.

x² − 2x − 8 = (x − 4)(x + 2)

Roots are x = 4 and x = −2.

Step 3: Sketch the U-shaped parabola crossing the x-axis at x = −2 and x = 4.

Step 4: We need x² − 2x − 8 ≥ 0, meaning the curve is on or above the x-axis.

The U-shaped curve is above the x-axis to the left of the left root and to the right of the right root. It equals zero at the roots themselves.

Step 5: The solution is x ≤ −2 or x ≥ 4.

Note the “or” — the solution is in two separate regions, not one continuous interval. This is a key difference from Example 1.

The Big Rule

For a U-shaped parabola (positive coefficient of x²):

Less than zero (< 0): the solution is between the roots
Greater than zero (> 0): the solution is outside the roots (two separate regions)

For an inverted parabola (negative coefficient of x²):

Less than zero (< 0): the solution is outside the roots
Greater than zero (> 0): the solution is between the roots

If the inequality includes “equals” (≤ or ≥), include the roots themselves in the solution.

Worked Example 3: Negative Leading Coefficient

Solve −x² + 4x − 3 > 0.

Step 1: The expression has a negative leading coefficient. You can either work with it directly or multiply through by −1 (which flips the inequality sign).

Let us multiply by −1:

x² − 4x + 3 < 0

Step 2: Factorise.

x² − 4x + 3 = (x − 1)(x − 3)

Roots are x = 1 and x = 3.

Step 3: Sketch the U-shaped parabola y = x² − 4x + 3 (after multiplying by −1, the parabola opens upward).

Step 4: We need x² − 4x + 3 < 0, so the curve is below the x-axis — between the roots.

Step 5: The solution is 1 < x < 3.

Alternatively, if you worked with the original expression −x² + 4x − 3 > 0 without multiplying by −1, you would sketch an inverted parabola. The curve is above the x-axis between the roots, giving the same answer.

Worked Example 4: When the Quadratic Does Not Factorise Neatly

Solve 2x² + 3x − 7 ≤ 0.

Step 2: This does not factorise with integer values. Use the quadratic formula to find the roots:

x = (−3 ± √(9 + 56)) / 4
x = (−3 ± √65) / 4
x = (−3 + 8.062) / 4 = 1.266 or x = (−3 − 8.062) / 4 = −2.766

Step 3: The parabola opens upward (coefficient of x² is positive) and crosses the x-axis at approximately x = −2.77 and x = 1.27.

Step 4: For ≤ 0, we want the curve on or below the x-axis — between the roots.

Step 5: The solution is −2.77 ≤ x ≤ 1.27 (to 3 significant figures).

Writing the Answer Correctly

IGCSE mark schemes are specific about notation:

For a single interval: 2 < x < 3 (not “x > 2 and x < 3”)
For two separate intervals: x ≤ −2 or x ≥ 4 (the word “or” is essential)
Never write −2 ≥ x ≥ 4 — this implies x is simultaneously less than −2 and greater than 4, which is impossible

If the question asks you to “list the integer values” in a solution set, enumerate them. For example, if 2 < x < 7 and x is an integer, the values are 3, 4, 5, 6.

Connection to Simultaneous Equations

Quadratic inequalities sometimes arise from problems involving simultaneous equations. For example, finding where a line lies below a curve:

“For what values of x is the line y = x + 1 below the curve y = x² − 3?”

Set up the inequality: x + 1 < x² − 3, which gives 0 < x² − x − 4, or x² − x − 4 > 0.

Solve this quadratic inequality using the method above. The graphical interpretation is clear: you are finding the x-values where the parabola is above the line.

Common Mistakes

Forgetting to flip the inequality when multiplying by −1. If you multiply both sides by a negative number, the inequality sign reverses.
Writing a single interval for a “greater than” inequality. If x² − 2x − 8 > 0, the answer is two separate regions: x < −2 or x > 4, not −2 < x < 4.
Not sketching the curve. Even a rough sketch prevents you from picking the wrong region. Always draw it.
Confusing strict and non-strict inequalities. For < and >, the roots are not included (open circles on a number line). For ≤ and ≥, the roots are included (filled circles).
Not finding the roots accurately. If the quadratic does not factorise, you must use the formula. An incorrect root leads to an incorrect solution interval.

Practice Question

Solve x² − x − 12 > 0 and list all integer values of x that satisfy the inequality and also satisfy −5 ≤ x ≤ 8.

Hint: Factorise first, sketch the parabola, identify the regions where the curve is above the x-axis, then list integers in those regions within the given range.

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