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0580/43 Oct/Nov 2023, Question 10: Equations That Become Quadratics

Cambridge IGCSE Mathematics 0580 · Paper 43 (Extended) · Calculator · 14 marks

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The question, in short

Expand and simplify an expression, factorise two expressions fully, and solve a speed-time problem leading to a quadratic equation and its solution.

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

Part (a) · 2 marks

Expand and simplify 4(2x − 1) − 6(3 − x).

  1. 14(2x − 1) − 6(3 − x) = 8x − 4 − 18 + 6x
  2. 2= 14x − 22

Answer14x − 22 or 2(7x − 11)

Part (bi) · 2 marks

Factorise 6x²y + 9xy completely.

  1. 1Common factor of 6x²y and 9xy is 3xy
  2. 26x²y + 9xy = 3xy(2x + 3)

Answer3xy(2x + 3)

Part (bii) · 3 marks

Factorise 4x² − y² + 8x + 4y completely.

  1. 14x² − y² = (2x + y)(2x − y) and 8x + 4y = 4(2x + y)
  2. 24x² − y² + 8x + 4y = (2x + y)(2x − y) + 4(2x + y)
  3. 3= (2x + y)(2x − y + 4)

Answer(2x + y)(2x − y + 4)

Part (ci) · 4 marks

Using 100 km at x km/h and 150 km at (x + 10) km/h over 4 h 20 min, show 13x² − 620x − 3000 = 0.

  1. 1Time = distance ÷ speed, and 4 h 20 min = 4 1/3 = 13/3 h
  2. 2100/x + 150/(x + 10) = 13/3
  3. 3Multiply by 3x(x + 10): 300(x + 10) + 450x = 13x(x + 10)
  4. 4300x + 3000 + 450x = 13x² + 130x
  5. 513x² − 620x − 3000 = 0

Answer100/x + 150/(x + 10) = 4 1/3 leading to 13x² − 620x − 3000 = 0

Part (cii) · 3 marks

Solve 13x² − 620x − 3000 = 0 for the first speed, showing working, to 1 decimal place.

  1. 1x = (620 ± √((−620)² − 4 × 13 × (−3000))) ÷ (2 × 13)
  2. 2= (620 ± √540 400) ÷ 26
  3. 3x = 52.119… or x = −4.43… (speed cannot be negative)
  4. 4Speed = 52.1 km/h

Answer52.1

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Also searched as: 0580/43/O/N/23 · 0580_w23_qp_43 · 0580 w23 43 · question 10.
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