Solving a quadratic inequality means sorting out where the parabola lies below the x-axis and where it lies above.
The sign of the discriminant D tells you first whether the parabola meets the x-axis at all, and if it does, the answer depends on whether you need the part between the two roots or the parts outside.
The absolute-value inequality |x| < a describes the points on the number line less than a from the origin, and a system of inequalities asks where the solutions of the separate inequalities overlap.
In the first picture, the green interval of solutions grows and shrinks with the coefficients and vanishes when D turns negative; in the second, the width of the interval on the number line changes with a.
Quadratic inequalities: where the parabola dips below
The solution of ax² + bx + c ≤ 0 (a > 0) is the interval where the parabola is below or on the x-axis. Use the discriminant D to see first whether such an interval exists.
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The part of the parabola y = x² + px + q below the x-axis, with the solution interval of x² + px + q ≤ 0 in green
💡 Reading off the solution
①The solution of x² + px + q ≤ 0 is between the two roots α and β: α ≤ x ≤ β
②The solution of x² + px + q ≥ 0 is outside the roots: x ≤ α or x ≥ β
③If D = 0 the parabola touches the x-axis, so x² + px + q ≤ 0 has the single solution x = −p/2
④If D < 0 the parabola stays above the x-axis, so x² + px + q ≤ 0 has no solution and x² + px + q > 0 holds for every real x
⑤In the picture the coefficient of x² is fixed at 1
How to solve quadratic inequalities
Solving by factorising
a(x − α)(x − β) ≤ 0, a > 0 → α ≤ x ≤ β
Factorise, then work out the sign in each interval
When D < 0
if a > 0, ax² + bx + c > 0; if a < 0, ax² + bx + c < 0 (for every real x)
With a negative discriminant, the quadratic never changes sign
📐 Using a sign table
①Factorise f(x) as a(x − α)(x − β)
②Mark the roots α and β on a number line
③Decide the sign of f(x) in each of the three intervals
④The intervals with the sign the inequality asks for are the solution
Inequalities with absolute values
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The solution of |x| < a shown as an interval centred on the origin of a number line
Absolute-value inequality
|x| < a ⟺ −a < x < a
Less than a from the origin (a > 0)
Absolute-value inequality (other way)
|x| > a ⟺ x < −a or x > a
More than a from the origin (a > 0)
Systems of inequalities
Solution of a system
(solution of the first) ∩ (solution of the second)
Solve each inequality and find where the solutions overlap
🔑 Order of steps
①Solve the inequalities one at a time
②Draw all the solutions on one number line
③The interval where they all overlap is the solution of the system
④If nothing overlaps, there is no solution
Worked examples and unit test
Example 1
Solve x² − x − 6 < 0.
1
Factorise the left-hand side.
x² − x − 6 = (x − 3)(x + 2) < 0
2
A parabola opening upwards is below the x-axis between its roots.
−2 < x < 3
▸ −2 < x < 3
The inequality is strict, so the roots −2 and 3 themselves are not included.
Example 2
Solve the system 2x − 1 > 3, x² − 5x + 4 ≤ 0.
1
Solve the two inequalities separately.
x > 2 and (x − 1)(x − 4) ≤ 0 ⇒ 1 ≤ x ≤ 4
2
Find where the two solutions overlap.
2 < x ≤ 4
▸ 2 < x ≤ 4
Drawing x > 2 and 1 ≤ x ≤ 4 on one number line, they overlap only between 2 and 4; 2 itself fails the first inequality, so it is left out.
Summary
Quadratic inequalities
a(x − α)(x − β) ≤ 0, a > 0 → α ≤ x ≤ β
Read from where the parabola sits relative to the x-axis
Absolute-value inequalities
|f(x)| < a ⟺ −a < f(x) < a
Removing the absolute value gives an inequality bounded on both sides
Grade-10 education-office assessment type
For what range of the constant k does x² − 2x + k > 0 hold for all real x?
①k < 1
②k > 1
③k ≤ 1
④k ≥ 1
⑤k > 2
▸ ② k > 1
1
It opens upward (leading coefficient 1 > 0), so always positive requires discriminant D < 0.
D = (−2)² − 4·1·k < 0
2
Solve the inequality.
4 − 4k < 0 ⇒ k > 1
🎯 Points that often come up
①For a quadratic inequality, check the sign of D first
②If a > 0 and D < 0, ax² + bx + c > 0 holds for every real number
③|x − a| < b means a − b < x < a + b, an interval of half-width b centred on a
④For a system, draw a number line and check where the solutions overlap
⑤A condition 'true for every real x' is turned into D < 0 or D ≤ 0