Albegra: Inequality

[sajibfin06]

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.

[Jim@StratusPrep]

First, factor the equation --> (x + 4)(x + 2)(x - 1)(x - 4) < 0 --> I put these in order to simplify the process

If x is greater than 4 all are positive
If x is 2 or 3, only the last term is negtive [Add 2 and 3 to the sum]
If x is -1 or 0, then we have 2 positive terms and entire product is positive
If x is -3, 3 terms are negative and the product is negative. [Add -3 to the sum]
All other cases are positive


2 + 3 + -3 = 2

[Brent@GMATPrepNow]

Quadratic inequality questions are pretty rare on the GMAT, and I think you need to be scoring in the top 20% to actually see one. That said, if you want to learn more about how to solve quadratic inequalities, we have a free video on the topic:
- https://www.gmatprepnow.com/module/gmat- ... ing?id=986

Afterwards, here's a follow-up practice question to try: https://www.gmatprepnow.com/module/gmat- ... ing?id=987

Cheers,
Brent

[GMATGuruNY]

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.

[[email protected]]

Hi sajibfin06,

Certain questions on Test Day can be solved with 'brute force' - you don't have to do any complex math, you just have to 'crunch' enough tiny calculations to prove what the answer is.

Here, the answers are relatively small (-1 to +3, inclusive); since we're asked for the SUM of the POSSIBLE values of X, chances are pretty good that each of those values is relatively small.

We're told that X is an INTEGER and (X^2 - 16)(X^2 + X -2) < 0. This information is based on Number Property rules. For THAT product to be LESS than 0, we need either....

(Postive)(Negative)

or

(Negative)(Positive)

Starting with (X^2 - 16), there are only a few integers that will make this 'piece' NEGATIVE....3, 2, 1, 0, -1, -2 and -3....so we should 'brute force' those first...

IF....
X = 3 --> (neg)(pos), so X could = 3
X = 2 --> (neg)(pos), so X could = 2
X = 1 --> (neg)(0), so X CANNOT = 1
X = 0 --> (neg)(neg), so X CANNOT = 0
X = -1 --> (neg)(neg), so X CANNOT= -1
X = -2 --> (neg)(neg), so X CANNOT = -2
X = -3 --> (neg)(pos), so X could = -3

So far, we have 3 possible values: X = 3, 2 and -3

Now we have to consider what it would take for (X^2 + X - 2) to be negative...there are NOT that many options though... 0 and -1.

From our prior work (above), we already know what happens when X=0 and X=-1 (and we have proof that those values are NOT possible solutions, so we're done....

The sum is 3 + 2 + (-3) = 2

Final Answer: D

GMAT assassins aren't born, they're made,
Rich