01 · Two outside intervals
x² + 2x − 15 > 0
Hint
Factorise and check the shape.
Worked solution
(x + 5)(x − 3) > 0
x < −5 or x > 3
The roots are excluded because the inequality is strict.
Understand · explore · practise
Quadratic inequalities
Use roots and sign regions to solve quadratic inequalities, combine solution sets, handle rational inequalities and check parameter restrictions.
Before you startQuadratic equations, graphs and linear inequalities
01 / Positive or negative
Roots mark where a sign can change.
For f(x) = (x + 2)(x − 3), the roots are −2 and 3. Between them, one factor is positive and the other negative, so f(x) < 0. Outside them the factors have the same sign, so f(x) > 0.
f(x) < 0 ⇔ −2 < x < 3
f(x) > 0 ⇔ x < −2 or x > 3
The roots themselves give zero. Include them for ≤ or ≥, but exclude them for < or >.
Change the multiplier to −1 in the graph. The roots stay fixed, but every non-zero output changes sign. The highlighted vertical bands mark the x values that satisfy the selected condition.
f(x) = a(x + 2)(x − 3)Sign regions
For a = 1, f(x) < 0 between its roots: −2 < x < 3. The roots are excluded.
Watch the factors determine the sign
Pause, replay or seek freely. The notes explain the same idea and stay in view.
02 / The method
Make one side zero, then use the graph.
- Move all terms to one side.
- Solve the corresponding equation to find critical values.
- Sketch the shape or make a sign table.
- Select the intervals with the required sign.
- Include or exclude the boundary roots appropriately.
One test input in each interval is enough once the expression is a polynomial and all real roots are accounted for: it cannot change sign within an interval without crossing zero.
Finding the roots is an intermediate step. An inequality normally asks for intervals, not just those root values.
A non-unit leading coefficientWorked example
2x² + x − 6 ≥ 0
Factorise to find the critical values.
(2x − 3)(x + 2) = 0
x = −2 or x = 3/2
The upward parabola is non-negative outside the roots.
x ≤ −2 or x ≥ 3/2
Equality is allowed, so include both roots.
Test x = 0: −6 < 0
The middle interval is correctly excluded.
03 / Negative coefficient
Read the actual shape of the graph.
A downward parabola is positive between two distinct roots and negative outside them. Do not memorise “greater than means outside” without checking the leading coefficient.
You can instead multiply the whole inequality by −1, provided you reverse its sign. Either approach must give the same answer.
For 8 + 2x − x² ≥ 0, the graph opens downwards and crosses at −2 and 4. Its non-negative part lies between them.
Convert to an upward parabolaWorked example
8 + 2x − x² ≥ 0
Multiply every term by −1.
x² − 2x − 8 ≤ 0
Reverse ≥ to ≤.
(x − 4)(x + 2) ≤ 0
An upward parabola is non-positive between its roots.
−2 ≤ x ≤ 4
Both endpoints give zero and are included.
04 / Special cases
Not every quadratic changes sign twice.
At a repeated root, the graph touches the axis without changing sign. A quadratic with no real roots keeps the same sign everywhere.
A repeated root
(x − 2)² ≥ 0: all real x
(x − 2)² > 0: x ≠ 2
(x − 2)² ≤ 0: x = 2
(x − 2)² < 0: no solutions
A square is zero only when its bracket is zero. It is otherwise positive.
For x² + 2x + 3 = (x + 1)² + 2, the minimum is 2. Therefore “> 0” is true for all real x and “≤ 0” has no solutions.
With a negative multiplier, the signs reverse: −(x + 1)² − 2 is negative for every real x.
No need to force a factorisationWorked example
3x² − 6x + 8 < 0
Complete the square.
3(x − 1)² + 5 < 0
The left side is at least 5.
Solution set: ∅
No real input can satisfy the inequality.
05 / Combine sets
Solve separately, then take the overlap.
For x² − x − 6 ≤ 0 and x > 1, the first condition gives −2 ≤ x ≤ 3. Keeping only values greater than 1 gives 1 < x ≤ 3.
Two quadratic conditions work the same way. Each may already consist of more than one interval.
Test whether an endpoint satisfies both original inequalities. One strict condition can exclude it even when the other includes it.
An overlap with two piecesWorked example
x² − 9 < 0 and x² − x − 2 ≥ 0
Solve the two inequalities separately.
First: −3 < x < 3
The upward parabola is negative between −3 and 3.
Second: x ≤ −1 or x ≥ 2
Factor as (x + 1)(x − 2).
−3 < x ≤ −1 or 2 ≤ x < 3
Intersect the sets. Interval form: (−3, −1] ∪ [2, 3).
06 / Denominators
Do not multiply by an unknown sign.
For 4/x > 1, multiplying by x is unsafe without knowing whether x is positive or negative. Record x ≠ 0, then multiply by x², which is strictly positive on the allowed domain.
4x > x²
x(x − 4) < 0
0 < x < 4
A squared denominator gives an equivalent polynomial inequality only on the original domain. Excluded denominator values must remain excluded, even if a new polynomial allows them.
A denominator already squared
6/x² + 1/x ≤ 1, x ≠ 0
6 + x ≤ x²
(x − 3)(x + 2) ≥ 0
x ≤ −2 or x ≥ 3
Multiplication by x² preserves the direction. Zero is already outside the resulting intervals.
The sign-table alternative
Bring everything into one fraction and identify every numerator zero and denominator zero. These split the number line into sign intervals. Test each interval, include allowed numerator zeros for inclusive inequalities, and always exclude poles where the denominator vanishes.
Keep a forbidden boundary outWorked example
3/(x − 1) ≤ 2, x ≠ 1
Multiply by (x − 1)² > 0.
3(x − 1) ≤ 2(x − 1)²
All terms are multiplied by the same positive expression.
(x − 1)(2x − 5) ≥ 0
The polynomial permits x ≤ 1 or x ≥ 5/2.
x < 1 or x ≥ 5/2
Remove x = 1 because the original fraction is undefined there.
07 / Parameters
An inequality can be about a coefficient.
Conditions on real roots often turn into quadratic inequalities in a parameter. Form the discriminant, solve the inequality in that parameter, then check exceptional values separately.
Real roots, allowing a repeated root
For kx² + kx − 3 = 0 and k ≠ 0:
D = k² + 12k = k(k + 12)
D ≥ 0 ⇒ k ≤ −12 or k ≥ 0
At k = 0 the equation is −3 = 0, with no solutions. Therefore the original equation has real roots exactly when k ≤ −12 or k > 0.
A parameter can make the equation linear or even remove x altogether. The discriminant alone cannot classify those cases.
No real roots, including a constant equationWorked example
kx² − 4kx + 5 = 0
For k ≠ 0, the discriminant is 16k² − 20k.
4k(4k − 5) < 0
Solve the quadratic inequality in k.
0 < k < 5/4
This covers the genuine quadratics with no real roots.
k = 0 gives 5 = 0
There are no solutions here either. For the equation as stated, the full answer is 0 ≤ k < 5/4.
08 / Your turn
Use signs, not a memorised interval pattern.
For rational inequalities, write the forbidden values first. For parameter questions, check any value that removes the quadratic term.
02 · Include the roots
2x² − 7x + 3 ≤ 0
Hint
Factorise as (2x − 1)(x − 3).
Worked solution
1/2 ≤ x ≤ 3
The upward parabola is below or on the axis between the roots.
03 · A downward parabola
12 − x − x² > 0
Hint
Multiplying by −1 reverses the direction.
Worked solution
x² + x − 12 < 0
(x + 4)(x − 3) < 0
−4 < x < 3
04 · One allowed value
(x + 1)² ≤ 0
Hint
A square cannot be negative.
Worked solution
It must be zero, so x = −1. The solution set is the singleton {−1}, not an interval of positive length.
05 · Combine two quadratics
x² − 4 < 0 and x² − x − 2 ≥ 0
Hint
Intersect (−2, 2) with the two outside intervals from the second condition.
Worked solution
First: −2 < x < 2
Second: x ≤ −1 or x ≥ 2
Combined: −2 < x ≤ −1
x = 2 fails the first strict inequality.
06 · A shifted denominator
5/(x + 2) > 1
Hint
Exclude x = −2, then multiply by (x + 2)².
Worked solution
5(x + 2) > (x + 2)²
(x + 2)(x − 3) < 0
−2 < x < 3
Both boundaries are excluded; −2 is also undefined in the original.
07 · Two fractions
4/x² − 3/x < 1
Hint
x ≠ 0. Multiply by positive x².
Worked solution
4 − 3x < x²
(x + 4)(x − 1) > 0
x < −4 or x > 1
The excluded value zero is not in either interval.
08 · An exceptional parameter
Find all p for which the equation px² + 2px + 2 = 0 has no real solution.
Hint
Use D < 0 for p ≠ 0, then inspect p = 0 directly.
Worked solution
D = 4p² − 8p = 4p(p − 2)
D < 0 ⇒ 0 < p < 2
At p = 0 the equation is 2 = 0, also with no solution. Full answer: 0 ≤ p < 2.
09 / Recap
Keep the sign and the domain together.
- Roots divide a polynomial into sign intervals.
- Check the leading coefficient and the inequality direction.
- Repeated roots need not change the sign; rootless quadratics keep one sign.
- Combine solution sets only after solving each condition.
- Clear an unknown-sign denominator using a valid case split or its positive square, retaining exclusions.
- Check parameter values that change the degree.
Section 1 of 9 · Positive or negative