A quadratic inequality asks where a quadratic expression is positive, negative, nonnegative or nonpositive. The answer is usually an interval, not a single number.
The safest method connects algebra with graph structure: find the boundary roots, divide the number line into intervals and determine the sign of the quadratic in each region.
1. Begin by moving everything to one side
To solve x²−x<6, write x²−x−6<0. Factorise: (x−3)(x+2)<0. The boundary values are x=−2 and x=3.
2. Roots divide the number line
The roots split the real line into x<−2, −2<x<3 and x>3. Test one convenient value in each interval, or use the graph’s sign pattern.
Because the leading coefficient is positive, the parabola opens upward. It is below the horizontal axis between its two distinct roots. Therefore −2<x<3.
3. Inclusive inequalities include valid roots
For (x−1)(x−5)≤0, the expression equals zero at x=1 and x=5, so both boundary values are included. The answer is 1≤x≤5.
For a strict inequality such as <0, roots are excluded because zero is not negative.
4. Negative leading coefficient reverses the visual pattern
Solve −x²+4x+5>0. The roots of −x²+4x+5=0 are x=−1 and x=5. The parabola opens downward, so it lies above the horizontal axis between its roots. Hence −1<x<5.
5. Repeated roots behave differently
Consider (x−2)²≥0. A square is nonnegative for every real x, so the inequality holds for all real numbers.
For (x−2)²>0, every real x except x=2 works. The graph touches the axis at the repeated root but does not change sign there.
6. No real roots can make the sign constant
For x²+1>0, the leading coefficient is positive and the expression has no real roots. It is positive for every real x. The graph remains above the horizontal axis.
Similarly, −x²−1<0 for every real x.
7. Product sign reasoning
For (x−4)(x+1)>0, the product is positive when both factors have the same sign. This occurs for x>4 or x<−1.
The factor-sign method and graph method agree because they describe the same quadratic.
8. Do not divide by an expression of unknown sign
In an inequality, dividing by a negative quantity reverses the inequality sign. If the sign of an algebraic expression is unknown, dividing by it without cases can invalidate the solution.
Factorisation and interval analysis avoid this trap by keeping the sign changes visible.
9. Worked comparison with a line
Find where x²−2x−3>x+1. Move all terms to one side: x²−3x−4>0. Factorise: (x−4)(x+1)>0.
Therefore x<−1 or x>4. Graphically, these are the x-values where the parabola y=x²−2x−3 lies above the line y=x+1.
10. Common mistakes
- Solving the corresponding equation and reporting only the roots.
- Including roots in a strict inequality.
- Assuming every quadratic is negative between its roots without checking the leading coefficient.
- Dividing by a factor whose sign is unknown.
11. Practice
- Solve x²−5x+6<0.
- Solve x²−9≥0.
- Solve −x²+2x+3≤0.
- Solve (x+4)²>0.
- Find where x²>3x+4.
12. Answers
- 2<x<3.
- x≤−3 or x≥3.
- x≤−1 or x≥3.
- All real x except −4.
- x²−3x−4>0=(x−4)(x+1)>0, so x<−1 or x>4.
Continue with Discriminants in Additional Mathematics or return to the Additional Mathematics Hub.