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TMUA 2022 · Paper 1 · Question 16 of 20

TMUA 2022 Paper 1 Question 16

Algebra and functions — Roots of a quartic · transforming sum and product. Try it first; the answer and a full worked solution are below.

TMUA 2022 · Paper 1Algebra and functionsRoots of a quartic · transforming sum and product6 options
The solutions to 7x4 − 6x2 + 1 = 0 are ±cosθ and ±cosβ.

Which one of the following equations has solutions ±sinθ and ±sinβ?

  1. A7x4 − 8x2 − 5 = 0
  2. B7x4 − 8x2 + 2 = 0
  3. C7x4 − 6x2 − 2 = 0
  4. D7x4 − 6x2 + 1 = 0
  5. E7x4 + 6x2 − 1 = 0
  6. F7x4 + 6x2 + 5 = 0
Show the answer and worked solution
answer · B
  1. A7x4 − 8x2 − 5 = 0
  2. B7x4 − 8x2 + 2 = 0
  3. C7x4 − 6x2 − 2 = 0
  4. D7x4 − 6x2 + 1 = 0
  5. E7x4 + 6x2 − 1 = 0
  6. F7x4 + 6x2 + 5 = 0
Put u = x2, so 7u2 − 6u + 1 = 0 has roots cos2θ and cos2β, with sum 67 and product 17. The new roots are sin2θ = 1 − cos2θ and sin2β = 1 − cos2β. Their sum is 2 − 67 = 87 and their product is 1 − 67 + 17 = 27. So they satisfy v2 − 87 v + 27 = 0, that is 7x4 − 8x2 + 2 = 0.