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IGCSE Maths · Edexcel 4XMA1 · Unit 2 Higher · Algebra

Quadratic Inequalities — Worked Solutions

Full worked solutions to all 12 questions (37 marks), set out the way an Edexcel examiner wants to see them.

Section A — Warm-up (grades 5–6)
  1. (2 marks)Solve x² − 9 < 0.
    Solution
    1. (x − 3)(x + 3) < 0, so the critical values are −3 and 3.
    2. “< 0” → between.
    −3 < x < 3
  2. (3 marks)Solve x² − 7x + 10 ≤ 0.
    Solution
    1. (x − 2)(x − 5) ≤ 0, so the critical values are 2 and 5.
    2. “≤ 0” → between, with the end points included.
    2 ≤ x ≤ 5
  3. (3 marks)Solve x² + 2x − 15 > 0.
    Solution
    1. (x + 5)(x − 3) > 0, so the critical values are −5 and 3.
    2. “> 0” → outside the roots.
    x < −5 or x > 3
  4. (3 marks)List all the integers that satisfy x² − 4x − 12 < 0.
    Solution
    1. (x − 6)(x + 2) < 0, so −2 < x < 6.
    2. Integers strictly between −2 and 6.
    −1, 0, 1, 2, 3, 4, 5
Section B — Core (grades 6–7)
  1. (3 marks)Solve 2x² − 5x − 3 ≥ 0.
    Solution
    1. (2x + 1)(x − 3) ≥ 0, so the critical values are −½ and 3.
    2. “≥ 0” → outside, including the end points.
    x ≤ −½ or x ≥ 3
  2. (3 marks)Solve x² < 3x + 10.
    Solution
    1. x² − 3x − 10 < 0 → (x − 5)(x + 2) < 0.
    2. Critical values −2 and 5; “< 0” → between.
    −2 < x < 5
  3. (3 marks)Solve 12 + x − x² > 0.
    Solution
    1. Multiply by −1 and reverse: x² − x − 12 < 0.
    2. (x − 4)(x + 3) < 0 → between −3 and 4.
    −3 < x < 4
  4. (3 marks)Solve x(x − 6) ≥ 16. Show your answer on a number line.
    Solution
    1. x² − 6x − 16 ≥ 0 → (x − 8)(x + 2) ≥ 0.
    2. Critical values −2 and 8; “≥ 0” → outside, included.
    -4-3-2-1012345678910
    x ≤ −2 or x ≥ 8
Section C — Stretch (grades 8–9)
  1. (3 marks)Solve x² + 2x − 4 < 0. Give your answer in surd form.
    Solution
    1. x² + 2x − 4 = (x + 1)² − 5.
    2. (x + 1)² − 5 = 0 → x = −1 ± √5.
    3. “< 0” → between the roots.
    −1 − √5 < x < −1 + √5
  2. (4 marks)Find all the integers that satisfy both x² < 20 and 3x − 2 > 1.
    Solution
    1. x² < 20 → −4.47… < x < 4.47…, so integers −4 to 4.
    2. 3x − 2 > 1 → x > 1.
    3. Common integers: 2, 3, 4.
    2, 3, 4
  3. (4 marks)A rectangle has width x cm and length (x + 4) cm. Its area is less than 60 cm². Find the range of possible values of x.
    Solution
    1. x(x + 4) < 60 → x² + 4x − 60 < 0.
    2. (x + 10)(x − 6) < 0 → −10 < x < 6.
    3. Width must be positive: x > 0.
    0 < x < 6
  4. (3 marks)Solve 3x² ≤ 7x + 6. Write your answer using set notation.
    Solution
    1. 3x² − 7x − 6 ≤ 0 → (3x + 2)(x − 3) ≤ 0.
    2. Critical values −⅔ and 3; “≤ 0” → between, included.
    {x : −⅔ ≤ x ≤ 3}
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