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

Adit_ · Answer

Best way to solve such problems is to do trial and error:

Take a rough number for x where 5x is close to 12:
Take x=3. It is clear that it exceeds comfortably.
But wait what if you have -3 now?
It becomes 12 and is really close to 12.
So now it is obvious that -2,-1,0 are values less than or equal to 0 that satisfies.

Taking the positive side:
Since we tried 3 was quite high, try 2, it is 12.
Thus 1 has to be valid.

We have no converged with our solutions and have thus covered all possibilities.

Final sum = -2-1+0+1 = -2.

Answer:  Option A

__________________________________________________________________________________

This method will be useful in time crunch situations IMO, as one would need a sense of guarantee that they have covered all possible values in a complex mod function.
Doing this way we know for sure these are the values due to trial and error of taking extremes and thus there is a guarantee that those are the only values.

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?

Prep Toolkit

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