helppp unavoidable review problem

Question

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The real number x satisfies x^2-5x+6<0. Find all possible values of x^2+5x+6

Guest Nov 20, 2019

Guest · Answer 1 · 2019-11-20T01:54:18

The solutoin so x^2 - 5x - 6 < 0 is -1 < x < 6, so the set of all possible values of x^2 + 5x + 6 is (2,72).

Alan · Answer 2 · 2019-11-20T10:53:30

The question specifies that the real number x satisfies x^2 - 5x + 6<0 (not x^2 - 5x - 6 < 0).

This means x is in the range from 2 to 3. Substitute these values into x^2 + 5x + 6 to find the end points of the required set of values.