Learning Objectives
Use roots and factor signs to solve quadratic inequalities.
Handle negative leading coefficients and repeated roots.
Solve an inequality when the quadratic has no real roots.
Combine the solution with an additional restriction.
Prerequisites: quadratic factorisation, discriminants and linear inequalities. All practice questions in this lesson are original, not official TMUA questions.
What This Lesson Is About
A quadratic equation finds where an expression is zero; a quadratic inequality asks where it is positive, negative, or allowed to equal zero. Its roots help divide the number line into regions with a consistent sign.
We first analyse quadratics with two distinct roots, then account for a negative leading coefficient, repeated roots and no real roots. Finally, we apply domain restrictions and conditions that must hold for every real input.
1. Using Roots to Find Sign Intervals
To compare a quadratic with zero, find the points at which its sign could change. Factoring lets you determine the sign between those points from the signs of the individual factors.
1.1 Rearranging to zero and identifying the boundaries
To solve a quadratic inequality, put all terms on one side so the other side is zero. Find its real roots. Between consecutive roots, each nonzero linear factor has a fixed sign, so their product has a fixed sign.
1.2 Explaining the sign pattern for two distinct roots
For (x − r)(x − s) with r < s, both factors are negative to the left of r, opposite in sign between r and s, and positive to the right of s. Thus the product is positive outside the roots and negative between them. This pattern assumes a positive leading coefficient.
Interval | x + 2 | x − 3 | Product |
|---|---|---|---|
x < −2 | − | − | + |
−2 < x < 3 | + | − | − |
x > 3 | + | + | + |
Worked Example: A Product Below Zero
Question: Solve x² − x − 6 < 0.
Method: Factor the expression, order its two roots, and choose the interval where the factors have opposite signs.
Factor: (x − 3)(x + 2) < 0. The roots are −2 and 3. Between them x − 3 is negative and x + 2 is positive, making the product negative.
Therefore −2 < x < 3. Both roots are excluded because they make the expression zero. The test value 0 gives −6, while −3 and 4 each give positive values.
2. Accounting for the Leading Sign and Equality
The familiar sign pattern depends on a positive leading coefficient. Check that coefficient first, then decide separately whether each root belongs to the solution.
2.1 Reversing the pattern for a negative leading coefficient
If the leading coefficient is negative, an extra negative factor reverses every sign. Either keep that factor in the sign analysis or multiply the inequality by −1 and reverse the inequality symbol. Do not use an outside/inside shortcut before checking the leading sign.
2.2 Including roots only when equality is allowed
Include a root when equality is allowed; exclude it for a strict inequality. A root of a polynomial is an allowed input, unlike a zero of a denominator.
Worked Example: An Included Boundary
Question: Solve −2x² + 6x + 8 ≥ 0.
Method: Divide by the negative coefficient with a reversed inequality, then include the roots because equality is allowed.
Divide by −2 and reverse the inequality: x² − 3x − 4 ≤ 0. Factor as (x − 4)(x + 1) ≤ 0.
The product is negative between −1 and 4 and zero at the roots. Hence −1 ≤ x ≤ 4. At x = 0 the original is 8 ≥ 0, confirming the middle interval.
3. Handling Repeated Roots and No Real Roots
Not every quadratic has two distinct boundaries. A completed square shows what happens when the roots coincide or when the expression never reaches zero.
3.1 Why the sign does not switch at a repeated root
A repeated linear factor appears squared and is nonnegative on both sides of its zero. Its sign does not switch there. For example, (x − 2)² > 0 holds for every x except 2, while (x − 2)² < 0 has no real solutions.
3.2 Determining the sign when no real roots exist
A genuine quadratic with no real roots never reaches zero and has the sign of its leading coefficient everywhere. A completed square proves this directly. For instance, x² + 2x + 5 = (x + 1)² + 4 > 0 for every real x.
Worked Example: Only One Point Works
Question: Solve x² − 4x + 4 ≤ 0.
Method: Recognise a perfect square and use its nonnegativity to identify when it can also be nonpositive.
This is (x − 2)² ≤ 0. A square is nonnegative, so the inequality can hold only when the square is zero.
The solution is x = 2 alone. Treating it as an interval “between two roots” without recognising that they coincide can hide this point.
Worked Example: No Boundary Points
Question: Solve −x² − 2x − 3 < 0.
Method: Complete the square to show the expression remains strictly negative for every real input.
Rewrite the expression as −(x + 1)² − 2, which is at most −2.
It is strictly negative for every real x, so every real x is a solution.
4. Applying Domain Restrictions and All-Input Conditions
The interval found from sign analysis must still answer the original question. Some problems restrict the input set; others require the inequality to hold for every real input.
4.1 Intersecting the solution with the stated domain
An inequality may be one part of a larger problem. Solve it first, then intersect with all stated restrictions. If x must be an integer, list the integers in the interval rather than keeping a continuous range.
4.2 Requiring strict positivity for every real input
If a quadratic is required to be strictly positive for every real x, its leading coefficient must be positive and its discriminant negative. A zero discriminant would allow a zero value and fail strict positivity.
Worked Example: A Parameter for Positivity
Question: Find k such that x² − 6x + k > 0 for every real x.
Method: Complete the square, locate the minimum, and require that minimum to be strictly positive.
Complete the square: (x − 3)² + k − 9. Its minimum is k − 9, attained at x = 3.
Strict positivity for all real x therefore requires k > 9. For nonnegativity instead, the condition would be k ≥ 9.
Common Mistakes
Solve the inequality on intervals, not just at its roots.
Check the leading sign before using an inside/outside pattern.
A squared factor does not change sign at its zero.
Strict positivity for all x excludes any root, including a repeated root.
5. Practice
For each inequality, determine the sign on the relevant intervals and inspect the roots separately. In the final questions, adapt the result to integer inputs or to a condition that must hold for all real inputs.
1. Solve x² − 5x + 6 ≥ 0.
2. Solve 2x² + x − 3 < 0.
3. Solve −x² + 4x − 4 ≥ 0.
4. Solve x² + 4x + 7 ≤ 0.
5. Find integer x satisfying x² − x − 6 < 0.
6. Find k such that 2x² + 8x + k ≥ 0 for all real x.
Worked Solutions
1. Selecting the outside intervals with included roots
(x − 2)(x − 3) ≥ 0 gives x ≤ 2 or x ≥ 3. The roots are included; x = 0 and x = 4 give positive values, while 5/2 gives a negative value.
2. Selecting the negative interval for a non-monic quadratic
(2x + 3)(x − 1) < 0 gives −3/2 < x < 1. At x = 0 the value is −3, and both roots are excluded.
3. Recognising a single-point solution
−(x − 2)² ≥ 0 holds only at x = 2, where it equals zero.
4. Proving that the solution set is empty
(x + 2)² + 3 is strictly positive, so there are no solutions.
5. Restricting a real interval to integer inputs
The real interval is −2 < x < 3. Its integers are −1, 0, 1 and 2, all of which give negative values.
6. Requiring a nonnegative minimum
2(x + 2)² + k − 8 has minimum k − 8. Hence k ≥ 8. At equality the expression is zero only at x = −2.
6. Summary
- Factor or complete the square to determine signs.
- Use the leading coefficient and multiplicity of each root.
- Treat endpoint equality explicitly.
- Apply domain or integer restrictions after finding the real solution set.