AQA · Mathematics · 8300
GCSE Maths: Forming quadratic inequalities
Algebra · AQA GCSE Maths (8300). Cards are generated from the Corbettmaths practice questions for Forming quadratic inequalities (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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Solve the inequality x² + 6x + 8 < 0.
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−4 < x < −2. Factorise to (x + 2)(x + 4) < 0; the curve is below the x-axis between the roots −4 and −2.
Solve the inequality x² + 2x − 35 > 0.
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x < −7 or x > 5. Factorise to (x + 7)(x − 5) > 0; the curve is above the x-axis outside the roots.
Solve the inequality x² − 9x + 14 ≤ 0.
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2 ≤ x ≤ 7. Factorise to (x − 2)(x − 7) ≤ 0; the ≤ sign means the roots themselves are included.
Solve the inequality x² − 121 > 0.
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x < −11 or x > 11. Difference of two squares: (x − 11)(x + 11) > 0.
Solve the inequality x² − x − 30 ≥ 0.
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x ≤ −5 or x ≥ 6. Factorise to (x − 6)(x + 5) ≥ 0; the solution lies outside (and on) the roots.
Solve the inequality 7x² − 22x + 16 ≤ 0.
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8/7 ≤ x ≤ 2. Factorise to (7x − 8)(x − 2) ≤ 0; the roots are x = 8/7 and x = 2.
Find the set of values of x satisfying both x² − 2x − 24 < 0 and 12 − 5x ≥ x + 9.
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−4 < x ≤ ½. The quadratic gives −4 < x < 6; the linear inequality gives x ≤ ½; take the overlap.
For a positive quadratic, where is ax² + bx + c < 0 satisfied?
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Between the two roots. A U-shaped parabola is below the x-axis only between where it crosses the axis.
For a positive quadratic, where is ax² + bx + c > 0 satisfied?
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Outside the two roots: x less than the smaller root or x greater than the larger root.
A rectangular field has width x m and length x + 30 m. Which inequality expresses a perimeter less than 500 m?
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4x + 60 < 500, which simplifies to x < 110.
A field with width x m and length x + 30 m has perimeter under 500 m and area over 4000 m². What are the possible values of x?
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50 < x < 110. Area gives x² + 30x − 4000 > 0, i.e. (x + 80)(x − 50) > 0 so x > 50 (x is a length); perimeter gives x < 110.
If a length x satisfies 2x² − x − 171 > 0, what is the range of x?
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x > 9.5. Factorise to (2x − 19)(x + 9) > 0; the negative branch x < −9 is rejected because x is a length.
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