AQA · Mathematics · 8300
GCSE Maths: Inequalities: quadratic
Algebra · AQA GCSE Maths (8300). Cards are generated from the Corbettmaths textbook exercise and practice questions for Inequalities: quadratic (Corbettmaths Video 378), and every card links back to the exact page it came from. Practice tests generate fresh variations of these questions, so the numbers change every attempt. Source material © Corbettmaths — used with attribution.
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What is the first step in solving a quadratic inequality such as x² + 5x + 6 > 0?
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Factorise the quadratic (or otherwise find its roots) to locate where the expression equals zero, e.g. (x + 2)(x + 3) > 0 gives critical values x = −3 and x = −2.
For a < b, what is the solution set of (x − a)(x − b) < 0?
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a < x < b. The parabola has a positive x² coefficient, so it is negative only between its two roots.
For a < b, what is the solution set of (x − a)(x − b) > 0?
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x < a or x > b. An upward-opening parabola is positive outside its two roots, so the answer is two separate regions.
Solve the inequality (x − 4)(x − 1) < 0.
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1 < x < 4.
Solve the inequality x² < 36.
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−6 < x < 6. Note the answer is a single interval, not just x < 6.
Solve the inequality x² − 64 ≥ 0.
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x ≤ −8 or x ≥ 8.
How do you handle a quadratic inequality with a negative x² coefficient, e.g. −x² + 8x − 15 < 0?
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Multiply both sides by −1 and reverse the inequality sign, giving x² − 8x + 15 > 0, then solve as normal.
Solve the inequality x² + 6x + 8 < 0.
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−4 < x < −2, since x² + 6x + 8 = (x + 2)(x + 4).
Solve the inequality x² + 2x − 35 > 0.
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x < −7 or x > 5, since x² + 2x − 35 = (x + 7)(x − 5).
Write a possible quadratic inequality whose solution set is −5 < x < −2.
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(x + 5)(x + 2) < 0, i.e. x² + 7x + 10 < 0. A strict 'between the roots' answer needs a '< 0' inequality with roots −5 and −2.
Write a possible quadratic inequality whose solution set is x < −3 or x > 6.
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(x + 3)(x − 6) > 0, i.e. x² − 3x − 18 > 0. Two outer regions come from a '> 0' inequality with roots −3 and 6.
A rectangle has width x m and length (x + 30) m, perimeter under 500 m and area over 4000 m². Find the range of x.
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50 < x < 110. Perimeter: 4x + 60 < 500 gives x < 110; area: x² + 30x − 4000 > 0 gives (x + 80)(x − 50) > 0, so x > 50 for a positive width.
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