The real number x satisfies x^2-5x+6<0. Find all possible values of x^2+5x+6
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).
Guest made a teeny tiny mistake.
The answer is actually
EDIT: What happened to the LaTeX???????
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.