To show x^2 - 6x + 10 0 for all x, we complete the square:

  1. (x - 3)^2 + 1
  2. (x + 3)^2 + 1
  3. (x - 3)^2 - 1
  4. (x - 6)^2 + 10
Answer

A) (x - 3)^2 + 1

(x - 3)^2 ge 0, so (x - 3)^2 + 1 ge 1 0. SLO M-09-B-15 · Knowledge

Mathematics MCQ · More Logic MCQs · Practise