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TMUA 2020 · Paper 1 · Question 19 of 20

TMUA 2020 Paper 1 Question 19

Algebra and functions — Quadratic inequality · least integer solution. Try it first; the answer and a full worked solution are below.

TMUA 2020 · Paper 1Algebra and functionsQuadratic inequality · least integer solution6 options
Find the lowest positive integer for which x2 − 52x − 52 is positive.
  1. A26
  2. B27
  3. C51
  4. D52
  5. E53
  6. F54
Show the answer and worked solution
answer · E
  1. A26
  2. B27
  3. C51
  4. D52
  5. E53
  6. F54
The parabola opens upwards, so it is positive to the right of its larger root, at x = 52 + √522 + 4 × 522 = 26 + √262 + 52 = 26 + √728. Since √728 lies between 26 and 27, the root is between 52 and 53, so the answer is one of those two. Testing directly settles it: at x = 52 the value is 522 − 522 − 52 = −52, still negative, while at x = 53 it is 2809 − 2756 − 52 = 1, positive. The answer is 53.