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

 Nov 20, 2019
 #1
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+1

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).

 Nov 20, 2019
 #2
avatar+2854 
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Guest made a teeny tiny mistake.

 

The answer is actually

 \(2

 

EDIT: What happened to the LaTeX???????

CalculatorUser  Nov 20, 2019
edited by CalculatorUser  Nov 20, 2019
 #3
avatar+29881 
+4

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. 

 Nov 20, 2019

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