SOLUTION: If a and b are real numbers such that a < b, then (x-a)(x-b) is positive over the interval: A) (a, b) B) (-&#8734;, b) C) (-&#8734;, a)U(b, &#8734;) D) (-&#8734;, 0)U(b, &#8

Algebra ->  Functions -> SOLUTION: If a and b are real numbers such that a < b, then (x-a)(x-b) is positive over the interval: A) (a, b) B) (-&#8734;, b) C) (-&#8734;, a)U(b, &#8734;) D) (-&#8734;, 0)U(b, &#8      Log On


   

Question 1006088: If a and b are real numbers such that a < b, then (x-a)(x-b) is positive over the interval:
A) (a, b)
B) (-∞, b)
C) (-∞, a)U(b, ∞)
D) (-∞, 0)U(b, ∞)
E) (-∞, a]U[b,∞)
Answer by josgarithmetic(39618) About Me  (Show Source):
You can put this solution on YOUR website!
Making a number line will make this easier to analyze. The critical values of x are a and b.

%28x-a%29%28x-b%29%3E0

Look at each interval that the critical values make.

(-infinity,a)
(-)(-) positive.
TRUE

(a,b)
(+)(-) negative.
FALSE

(b,infinity)
(+)(+) positive.
TRUE

Choices C and E are exactly the same set and that is the solution set.