What is the sum of all integers which satisfy the inequality above?

Question

|x + 5|x|| < 12 
What is the sum of all integers which satisfy the inequality above?

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

M37-66

Bunuel · Accepted Answer

Official Solution: |x + 5|x|| < 12 What is the sum of all integers which satisfy the inequality above? A. -2 B. -1 C. 0 D. 1 E. 2 |x + 5|x|| < 12 -12 < x + 5|x| < 12 CASE 1: x < 0 . -12 < x + 5(-x) < 12 -12 < -4x < 12 Multiply by -\frac{1}{4} (and don't forget to flip the signs since we multiply by a negative number): 3 > x > -3 Negative integers in this range (don't forget that we consider x < 0 for this case) are -1, and -2. CASE 2: x \geq 0 . -12 < x + 5x < 12 -12 < 6x < 12 Divide by 6 : -2 < x < 2 Non-negative integers in this range (don't forget that we consider x \geq 0 for this case) are 0, and 1. The sum of all integers which satisfy the inequality is therefore: -1+(-2)+0+1=-2 Answer: A

nick1816 · Answer

|x + 5|x|| < 12

-12 < x + 5|x| < 12

Case 1 - x<0

-12 < x-5x < 12
-12< -4x < 12
-3< x < 3

Only -2 and -1 lie in the range

Case 2- x>= 0

-12 < x+5x < 12
-12< 6x < 12
-2 < x < 2

Only 0 and 1 lie in the range

Sum of all integers which satisfy the inequality = -2+-1+0+1 = -2

preetamsaha · Answer

|x+5|x||<12
case I : x ≥ 0 , IxI =x
Ix+5xI <12
or, I6xI <12
or, 0≤6x<12  
or, 0≤x<2
so, x=0, 1

case I : x <0 , IxI =-x
Ix-5xI <12
or, I-4xI <12
or,I-4IIxI<12  
or,4IxI<12
or,IxI<3
or, -3<x<0  
so, x= -2, -1

so, Sum of all integers which satisfy the inequality = 0+1+(-2)+(-1) = -2

correct answer will be A

Fdambro294 · Answer

Rule: given the Absolute Value Expression and the "less than" Inequality:

[X] < K

----where K = some Number Value

-(K) < X < +(K)

(1st) Opening up the Outer Modulus using the above Rule:

-(12) < X + 5[X] < +(12)

(2nd) Opening up the Inner Modulus making Assumptions:

Case 1: If X >/= 0 --------> [X] = X

-12 < X + 5X < +12

-12 < 6X < +12

-2 < X < +2

however, we assumed that X is greater than or equal to 0, so the only range that works is:

0 </= X < +2

Possible Integer Values for X are:

0 and +1

Case 2: If X < 0 ------> [X] = -(X)

-12 < X + 5 * (-X) < +12

-12 < X - 5X < +12

-12 < -4X < +12

----dividing by (-)4, we need to REVERSE the Inequality Signs------

3 > X > -3

however, in this case we Assumed that X must be LESS THAN < 0

thus the only Range that Satisfies this condition:

-3 < X < 0

Possible Integers that Satisfy this range:

-2 and -1

SUM up Every possible Integer Solution from both Cases:

0 + 1 - 2 - 1 = -2

-A-

NG24umich · Answer

In the case where x < 0 why is it converted to −12<x+5(−x)<12 as opposed to −12<-x+5(−x)<12? Sorry if this is naive question.

Bunuel · Answer

When x < 0, then |x| = -x. Hence, for this case we substitute |x| with -x in -12 < x + 5|x| < 12. However, x itself does not need to be changed. So, we get -12 < x + 5(-x) < 12.

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What is the sum of all integers which satisfy the inequality above?

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