If x is an integer and (x^2 - 16)(x^2 + x - 2)

Question

If x is an integer and (x^2 - 16)(x^2 + x - 2) < 0, what is the sum of all the possible values of x?

A. -1
B. 0
C. 1
D. 2
E. 3

SaquibHGMATWhiz · Answer

Solution: x is an integer It is given (x^2 - 16)(x^2 + x - 2) < 0 ⇒(x^2-4^2)(x^2+2x-x-2)<0 ⇒(x-4)(x+4) <0 ⇒(x-4)(x+4)(x+2)(x-1)<0 So, the zero points are 4, -4, -2 and 1 range66.png So, the range of x is −4

Sanil98 · Answer

Hi can you please provide me with a better explanation with respect to the signs? How do we get the sum as 2

LetsDoIt2022 · Answer

Both functions are parabolas, opened upwards. Means going to  +inf  from the greater root and to  -inf  from the lower root, the function's value will be positive.

Once you find the roots of the functions by factorizing, you'll notice that the roots of the second function,  x^2+x-2  lie between those of the other one.

Hence, multiplying both the functions will only yield a negative value in the corridor "inside" of the outer parabola ( x^2-16 ), where it is negative, and "outside" of the other, where it is positive. See image for clarification.

As x can only be of an integer value, the only values it can take is {-3,2,3}.

Sum of these values then equals  2 .

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If x is an integer and (x^2 - 16)(x^2 + x - 2) < 0, what is the sum of

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