Algebra
Grade 8-9Quadratic Inequalities
Solving a quadratic inequality starts with the critical values found by treating it as an equation, but the shape of the parabola then decides which side of those values actually satisfies the inequality. This lesson covers finding the critical values, deciding whether the solution is a single bounded region or two separate unbounded regions, showing the solution on a number line, and combining a quadratic inequality with a linear inequality.
Written by Asad, Co-Founder of Teachably
Video walkthrough coming soon
The written lesson below covers everything you need in the meantime.
Finding the critical values
Factorise the quadratic as if solving the equation equal to zero, to find the two critical values where the expression is exactly zero.
For x² − x − 6, factorising gives (x + 2)(x − 3), so the critical values are x = −2 and x = 3.
Deciding between a bounded and unbounded region
Since the graph of a positive quadratic is a U-shape, the expression is negative (below the x-axis) between the two critical values, and positive (above the x-axis) outside them. A "less than" inequality gives the bounded region between the values, while a "greater than" inequality gives the two unbounded regions outside them.
For x² − x − 6 < 0, with critical values −2 and 3, the solution is the bounded region −2 < x < 3. For x² − x − 6 > 0, the solution is the two unbounded regions x < −2 or x > 3.
Showing the solution on a number line
Use an open circle for a strict inequality (< or >) and a closed circle for ≤ or ≥, then shade or arrow the correct region or regions based on the critical values.
For x² − x − 6 ≥ 0, draw closed circles at x = −2 and x = 3, with arrows extending outward from each circle, since the solution is x ≤ −2 or x ≥ 3.
Combining with a linear inequality
Solve the quadratic inequality to find its solution set, then find the overlap between this set and the linear inequality given.
To find the values of x that satisfy both x² − x − 6 < 0 and x > 0, the quadratic gives −2 < x < 3. Combining this with x > 0 gives the overlap 0 < x < 3.
Worked Examples
Three exam-style questions, fully solved.
Solve x² − x − 6 < 0.
Easy- 1.Factorise to find the critical values: (x + 2)(x − 3) = 0, so x = −2 or x = 3
- 2.Since this is a "less than" inequality, the solution is the bounded region between the critical values
Answer: −2 < x < 3
Solve x² − x − 6 > 0.
Medium- 1.Factorise to find the critical values: (x + 2)(x − 3) = 0, so x = −2 or x = 3
- 2.Since this is a "greater than" inequality, the solution is the two unbounded regions outside the critical values
Answer: x < −2 or x > 3
Find the set of values of x that satisfy both x² − x − 6 < 0 and x > 0.
Hard- 1.Solve the quadratic inequality: (x + 2)(x − 3) < 0, giving −2 < x < 3
- 2.Find the overlap with x > 0
Answer: 0 < x < 3
Avoid These
The most common mistakes students make.
Solving the quadratic to find the critical values but not checking which side of them actually satisfies the original inequality, instead guessing based on the inequality symbol alone.
Writing the solution to a "less than" inequality as two unbounded regions, or a "greater than" inequality as a single bounded region, the wrong way round.
Using an open circle where a closed circle is needed, or vice versa, when showing the solution on a number line for a non-strict inequality.
When combining a quadratic inequality with a linear inequality, combining both ranges instead of finding the overlap between them.
Misreading the roots from a graph, which leads to the critical values, and therefore the whole solution, being stated incorrectly.
FAQ
Questions parents and students ask.
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