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.

wickedvikram · Answer

I took roughly 5 mins using 3 step method. This method was way better.

Adit_ · Answer

I think the explanation provided by most is the way to go.
I like to simply check values as there arent too many and since mod is involved always safer to check both positive and negative.
By direct glance and no calculations it looks like -3 to 3 are the only values possible, if you are quick with calculations one can quickly substitute and check the values that satisfy.
0,1,-1,-2 do it perfectly and hence summation = -2

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

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

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