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0580/41 May/Jun 2022, Question 9: Equations That Become Quadratics

Cambridge IGCSE Mathematics 0580 · Paper 41 (Extended) · Calculator · 14 marks

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The question, in short

Rectangle with sides x − 1 and 2x + 1 has area 29 cm² more than an x by x square; find k, the perimeter difference, showing working.

Summarised in our own words. Use the official paper for the exact wording and the diagram.

Part (a) · 6 marks

Using the stated area difference of 29 cm², find the value of k, the difference between the two perimeters, showing all working.

  1. 1(2x + 1)(x − 1) − x² = 29
  2. 22x² − x − 1 − x² = 29, so x² − x − 30 = 0
  3. 3(x − 6)(x + 5) = 0, x = 6 (x > 1)
  4. 4Rectangle perimeter = 2(13 + 5) = 36, square perimeter = 4 × 6 = 24
  5. 5k = 36 − 24 = 12

Answer12

Part (bi) · 4 marks

Given the larger cube's volume exceeds the smaller cube's by 5 cm³, show that 3y² + 3y − 4 = 0.

  1. 1(y + 1)³ − y³ = 5
  2. 2(y + 1)² = y² + 2y + 1
  3. 3(y + 1)³ = (y² + 2y + 1)(y + 1) = y³ + 3y² + 3y + 1
  4. 4y³ + 3y² + 3y + 1 − y³ = 5
  5. 53y² + 3y − 4 = 0

Answer3y² + 3y − 4 = 0 shown

Part (bii) · 4 marks

Hence find the volume of the smaller cube, showing working and giving the answer to 2 decimal places.

  1. 1y = (−3 ± √(3² − 4(3)(−4)))/(2 × 3) = (−3 ± √57)/6
  2. 2y = 0.7583… (y > 0)
  3. 3Volume = y³ = 0.7583…³ = 0.436…
  4. 4= 0.44 cm³

Answer0.44

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Also searched as: 0580/41/M/J/22 · 0580_s22_qp_41 · 0580 s22 41 · question 9.
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