sajibfin06 wrote:If x is an integer and (x² - 16)(x² + x - 2) < 0, what is the sum of all the possible values of x?
A. -1
B. 0
C. 1
D. 2
E. 3
Please explain in detail, if possible without using wavvy line.
(x+4)(x-4)(x+2)(x-1) < 0.
The CRITICAL POINTS are x=-4, x=-2, x=1, x=4.
These are the only values of x such that (x² -16)(x² + x - 2) = 0.
To determine the values of x such that (x² -16)(x² + x - 2) < 0, test one value to the LEFT AND RIGHT of each critical point.
In other words, test one value in each of the following ranges:
x < -4
-4 < x < -2
-2 < x < 1
1 < x < 4
x > 4.
Note the following:
If x<-4 or x>4 is a valid range, then x will have an INFINITE number of solutions, with the result that the sum of all possible values of x cannot be determined.
Since the answer choices imply that the sum must be between -1 and 3, inclusive, neither x<-4 nor x>4 can be a valid range.
Thus, we need test only one value in each of the remaining ranges:
-4 < x < -2
-2 < x < 1
1 < x < 4.
-4 < x < -2:
If we plug x=-3 into (x² - 16)(x² + x - 2) < 0, we get:
[ (-3)² - 16 ] [ (-3)² - 3 - 2 ] < 0
(-7)(4) < 0
-28 < 0.
This works.
Since -4 < x < -2 is a valid range, x=-3 is a valid integer value for x.
-2 < x < 1:
If we plug x=0 into (x² - 16)(x² + x - 2) < 0, we get:
(0² - 16 )(0² + 0 - 2) < 0
(-16)(-2) < 0
32 < 0.
Doesn't work.
Since -2 < x < 1 is not a valid range, no integer value between -2 and 1 is a valid solution for x.
Since the only valid solution thus far is x=-3, and the answer choices do not include -3 as an option for the sum of all possible values of x, the only remaining range MUST be valid.
Since 1 < x < 4 must be a valid range, x=2 and x=3 are both valid integer values for x.
Sum of all possible values of x = -3 + 2 + 3 = 2.
The correct answer is
D.
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