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TMUA 2018 · Paper 1 · Question 4 of 20

TMUA 2018 Paper 1 Question 4

Algebra and functions — Simultaneous equations · discriminant condition. Try it first; the answer and a full worked solution are below.

TMUA 2018 · Paper 1Algebra and functionsSimultaneous equations · discriminant condition7 options
Consider the simultaneous equations 3x2 + 2xy = 4 x + y = a where a is a real constant.

Find the complete set of values of a for which the equations have two distinct real solutions for x.

  1. AThere are no values of a.
  2. B−2 < a < 2
  3. C−1 < a < 1
  4. Da = 0
  5. Ea < −1 or a > 1
  6. Fa < −2 or a > 2
  7. GAll real values of a
Show the answer and worked solution
answer · G
  1. AThere are no values of a.
  2. B−2 < a < 2
  3. C−1 < a < 1
  4. Da = 0
  5. Ea < −1 or a > 1
  6. Fa < −2 or a > 2
  7. GAll real values of a
Substitute y = a − x into the first equation: 3x2 + 2x(a − x) = 4, which tidies to x2 + 2ax − 4 = 0. The discriminant is 4a2 + 16, and this is positive for every real a because 4a2 ≥ 0. So there are always two distinct real values of x, and the answer is all real values of a.