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WMA11/01 Jan 2019, Question 11: The Quadratic Formula and Completing the Square

Pearson Edexcel International AS Level Mathematics · Pure Mathematics 1 (WMA11) · Calculator · 12 marks

The question, in short

Sketch two cubic-related curves on blank axes with axis intercepts marked, prove an intersection equation, then find exact first-quadrant intersection coordinates.

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

Part (ai) · 2 marks

Sketch y = x(3 − x) on Diagram 1, marking where the curve crosses the coordinate axes.

  1. 1x(3 − x) = −x² + 3x has a negative x² coefficient, so it is n-shaped
  2. 2It crosses the x-axis at x = 0 and x = 3 (and the y-axis at O)

Answern-shaped parabola crossing the x-axis at (0, 0) and (3, 0)

Part (aii) · 2 marks

Sketch y = x(x − 2)(5 − x) on Diagram 1, marking axis crossing coordinates clearly.

  1. 1Expanding gives −x³ + …, so it is a negative cubic: high on the left, low on the right
  2. 2It crosses the x-axis at x = 0, 2 and 5, with a minimum between 0 and 2 and a maximum between 2 and 5

AnswerNegative cubic: comes down from the top left through (0, 0), minimum between 0 and 2, up through (2, 0) to a maximum between 2 and 5, then down through (5, 0)

Part (b) · 3 marks

Show that intersection x-coordinates of the two curves satisfy x(x² − 8x + 13) = 0.

  1. 1Set equal: x(x − 2)(5 − x) = x(3 − x)
  2. 2Expand: −x³ + 7x² − 10x = 3x − x²
  3. 3Collect terms: x³ − 8x² + 13x = 0, so x(x² − 8x + 13) = 0

AnswerShown: x(x − 2)(5 − x) = x(3 − x) leads to x(x² − 8x + 13) = 0

Part (c) · 5 marks

For first-quadrant point P on both curves, use algebra to find its exact coordinates, showing working.

  1. 1Solve x² − 8x + 13 = 0: x = (8 ± √12)/2 = 4 ± √3
  2. 2P is in the first quadrant, so it is between 0 and 3 where y = x(3 − x) > 0: x = 4 − √3
  3. 3y = (4 − √3)(3 − 4 + √3) = (4 − √3)(√3 − 1)
  4. 4= 4√3 − 4 − 3 + √3 = −7 + 5√3

AnswerP(4 − √3, −7 + 5√3)

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