4PM1/02R Jun 2023, Question 1: The Discriminant and Nature of Roots
Pearson Edexcel International GCSE Further Pure Mathematics 4PM1 · Paper 2R · Calculator · 4 marks
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The question, in short
Given f(x) = 2x² + (k + 8)x + k, prove that f(x) = 0 always has two different real solutions, whatever the value of k.
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Solution · 4 marks
- 1b² − 4ac = (k + 8)² − 4 × 2 × k
- 2= k² + 16k + 64 − 8k = k² + 8k + 64
- 3= (k + 4)² − 16 + 64 = (k + 4)² + 48
- 4(k + 4)² ≥ 0, so b² − 4ac ≥ 48 > 0
- 5So f(x) = 0 has distinct real roots for all k
Answerb² − 4ac = (k + 4)² + 48 > 0 for all k, so f(x) = 0 always has distinct real roots — shown
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