AQA · Mathematics · 8300

GCSE Maths: Quadratics: forming and solving

Algebra · AQA GCSE Maths (8300). Cards are generated from the Corbettmaths practice questions for Quadratics: forming and solving (Corbettmaths Video 267), 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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  1. What are the solutions to (x − 2)(x + 9) = 0?

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    x = 2 or x = −9. When a product equals zero, at least one factor must be zero.

  2. What are the roots of (x − 6)(x + 1) = 0?

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    x = 6 or x = −1. Set each factor equal to zero.

  3. What are the solutions to x² + 5x + 6 = 0?

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    x = −2 or x = −3. Factor as (x + 2)(x + 3) = 0.

  4. What are the solutions to x² − 49 = 0?

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    x = 7 or x = −7. This is a difference of two squares: (x − 7)(x + 7) = 0.

  5. If x² + bx + c = 0 has solutions x = −8 and x = −2, what are the values of b and c?

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    b = 10 and c = 16. The equation is (x + 8)(x + 2) = 0, which expands to x² + 10x + 16 = 0.

  6. If x² + bx + c = 0 has one solution x = −3, what are the values of b and c?

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    b = 6 and c = 9. A single solution means a repeated root: (x + 3)² = 0, which expands to x² + 6x + 9 = 0.

  7. What are the solutions to (2x − 1)(x − 3) = 0?

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    x = 1/2 or x = 3. Set each factor to zero: 2x − 1 = 0 gives x = 1/2, and x − 3 = 0 gives x = 3.

  8. What are the solutions to 4x² − 9 = 0?

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    x = 3/2 or x = −3/2. Factor as (2x − 3)(2x + 3) = 0, a difference of squares.

  9. When solving y² − 6y = 27, what is the first step?

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    Rearrange to standard form: y² − 6y − 27 = 0. Then factor and solve.

  10. In a right triangle problem where the perpendicular height is x + 9 cm, the base is x cm, and the area is 95 cm², what equation represents this?

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    x² + 9x − 190 = 0. From the area formula (1/2) × base × height = 95, we get (1/2)x(x + 9) = 95, which simplifies to this equation.

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