Events & Promotions
|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Sep 1507:00 PM EDT
-09:00 PM EST
Elite scores are possible when you enroll in a powerful GMAT course, taught live online + 6 months access to TTP OnDemand video courses included! Class starts Tues/Thurs Sept. 15, 2026 - Nov. 15, 2027, 7:00pm-9:00pm EST
Sep 2107:00 PM EDT
-10:00 PM EDT
Boost your GMAT score in less than one month in a live online class + 6 months access to TTP OnDemand video courses included!
riteshpatnaik
Let me first tell you that in this question, x < 0 has no significance.
|x| = x when x >= 0
|x| = -x when x < 0
On the same lines,
|x + 4| = x + 4 when (x + 4) >= 0
|x + 4| = -(x + 4) when (x + 4 ) < 0
In the first definition, x is just a placeholder for any expression.
|x^2 - 8| = x^2 - 8 when (x^2 - 8) >= 0
|x^2 - 8| = -(x^2 - 8) when (x^2 - 8) < 0
So, how do you get rid of |x + 3| in the original question? You take two cases: (x + 3) >= 0 or (x + 3) < 0
|x + 3| = x + 3 when (x + 3) >= 0 (i.e. when x >= -3)
|x + 3| = -(x + 3) when (x + 3) < 0 (i.e. when x < -3)
That is how you get -3 as a transition point.
Do the same for other expressions.
_________________
-
Start FREE with selected GMAT/EA Content upon registration at anaprep.com
YouTube Channel: karishma.anaprep
Start FREE with selected GMAT/EA Content upon registration at anaprep.com
YouTube Channel: karishma.anaprep
Posts: 66
I solved it very differently and got the right answer. Not sure whether it was a fluke.
I squared both sides. so the expression becomes: (x^2 +9 + 6x) - (16 +x^2 - 8x) = 64 + x^2 + 16x ---> 0 = x^2 + 2x + 71.
Now using the quadratic equation formula you find that the discriminant i.e. \sqrt{b^2 - 4ac} will be negative and hence 0 solutions.
Is this approach correct?
I squared both sides. so the expression becomes: (x^2 +9 + 6x) - (16 +x^2 - 8x) = 64 + x^2 + 16x ---> 0 = x^2 + 2x + 71.
Now using the quadratic equation formula you find that the discriminant i.e. \sqrt{b^2 - 4ac} will be negative and hence 0 solutions.
Is this approach correct?
guerrero25
ameyaprabhu
Note that \((a + b)^2 = a^2 + b^2 + 2ab\)
So \((|x+3| - |4-x|)^2 =|x + 3|^2 + |4 - x|^2 - 2*|x + 3|*|4 - x|\)
You missed out the last term. You would need to square it yet again and that will complicate the question further.
_________________
-
Start FREE with selected GMAT/EA Content upon registration at anaprep.com
YouTube Channel: karishma.anaprep
Start FREE with selected GMAT/EA Content upon registration at anaprep.com
YouTube Channel: karishma.anaprep
smritidabas
It doesn't matter whether you use |x - 4| or |4 - x|. The answer will stay the same.
I. Consider |x - 4|
When x >= 4, then (x - 4) is positive (try putting in x = 5) so
|x - 4| = x - 4
When x < 4, then (x - 4) is negative (try putting in x = 3) so
|x - 4| = -(x - 4) = 4 - x
II. Consider |4 - x|
When x >= 4, then (4 - x) is negative (try putting in x = 5 to see) so
|4 - x| = - (4 - x) = x - 4
When x< 4, then (4 - x) is positive (try putting in x = 3 to see) so
|4 - x| = 4 - x
Note that in both cases, when x >= 4, you get (x - 4) and when x < 4, you get (4 - x).
Just that case I is more intuitive so it's easier to flip first and then solve.
_________________
-
Start FREE with selected GMAT/EA Content upon registration at anaprep.com
YouTube Channel: karishma.anaprep
Start FREE with selected GMAT/EA Content upon registration at anaprep.com
YouTube Channel: karishma.anaprep
Bunuel
Hi Bunuel,
I am getting confused at the sign changes. Please explain a bit clearly.
We have three transition points for \(|x+3| - |4-x| = |8+x|\): -8, -3, and 4 (transition point is the value of x for which an expression in the modulus equals to zero). Thus we have four ranges to check:
1. \(x<-8\);
2. \(-8\leq{x}\leq{-3}\);
3. \(-3<x<4\)
4. \(x\geq{4}\)
Now we have four ranges:
As per property...................... |x| < 0 ... | some expression | = - some expression ---> Property 1
|x| > 0 ... | some expression | = some expression ----> Property 2.
Confirm the below:
Now case 1 : x < -8 then |x+3| = - (x+3) and |x-4| = - ( x-4) ( this has to -ve but you mentioned as positive ) and |x+8| = - (x +8 )
and confused for the other three ranges as well on how the expressions got +ve and -ve signs.
What all need to be checked to give positive and negative signs to the expressions, as you mentioned in the solution for the above four ranges.
msk0657
When x<-8:
x+3 becomes negative, thus |x+3|=-(x+3);
x+8 becomes negative, thus |x+8|=-(x+8);
4-x becomes positive, thus |4-x|=4-x. For example, 4-(-9)=13=positive.
Theory on Absolute Values: math-absolute-value-modulus-86462.html
The E-GMAT Question Series on ABSOLUTE VALUE: the-e-gmat-question-series-on-absolute-value-198503.html
Properties of Absolute Values on the GMAT: properties-of-absolute-values-on-the-gmat-191317.html
Absolute Value: Tips and hints: absolute-value-tips-and-hints-175002.html
DS Absolute Values Questions to practice: search.php?search_id=tag&tag_id=37
PS Absolute Values Questions to practice: search.php?search_id=tag&tag_id=58
Hard set on Absolute Values: inequality-and-absolute-value-questions-from-my-collection-86939.html
The E-GMAT Question Series on ABSOLUTE VALUE: the-e-gmat-question-series-on-absolute-value-198503.html
Properties of Absolute Values on the GMAT: properties-of-absolute-values-on-the-gmat-191317.html
Absolute Value: Tips and hints: absolute-value-tips-and-hints-175002.html
DS Absolute Values Questions to practice: search.php?search_id=tag&tag_id=37
PS Absolute Values Questions to practice: search.php?search_id=tag&tag_id=58
Hard set on Absolute Values: inequality-and-absolute-value-questions-from-my-collection-86939.html
_________________
New to the GMAT Club?
- • Questions: Question Bank | Bunuel's Signature Collection
• Tests: GMAT Club Tests | Forum Quiz
• Guides: Quantitative | Critical Reasoning | Reading Comprehension | Ultimate Quant Guide | All You Need for Quant
• Rules: Quantitative | Verbal
VeritasPrepKarishma
Responding to a pm:
[quote]
Request you to throw some more light on this concept
I have confusion why we have taken x<-8 and why not x>-8
[/quote
The transition points are -8, -3 and 4.
Draw them on the number line:
------------------- (-8) ----------- (-3) --------------------------- (4) -------------
In different sections of this number line, the terms are going to behave differently.
When x < -8,
|x + 8| = - (x + 8)
For the other terms too, when we remove the absolute value sign, we need a negative sign.
At x = -8, the sign for |x +8| turns.
When x >= -8 but less than -3, then
|x+8| = x + 8
For the other terms, when we remove the absolute value sign, we need a negative sign.
Hence each of the four sections of the number line are considered separately.
For more on this, check: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2014/06 ... -the-gmat/
_________________
-
Start FREE with selected GMAT/EA Content upon registration at anaprep.com
YouTube Channel: karishma.anaprep
Start FREE with selected GMAT/EA Content upon registration at anaprep.com
YouTube Channel: karishma.anaprep
Responding to a pm:
It comes down to whether you understand why we take ranges and change signs. First check this post:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2014/06 ... -the-gmat/
Now take a simpler example:
|x - 2| + |x - 5| = 4
What do you need to consider here?
For |x - 2|, you have to think about x < 2 and x >= 2
For |x - 5|, you have to think about x < 5 and x >= 5
To consider both together, you have
one range x < 2.
Another is x >= 2 and x < 5 so this is 2 <= x < 5.
Yet another is x >= 5
Now, will you worry about x < 2 and at the same time, x > 5? No. There will not be such a value of x. This is where |x - 2| will open as -(x - 2) and |x - 5| will open as (x - 5). This doesn't exist.
So in a question like this: |x - 2| + |x - 5| = 4, you don't have 4 cases ((x - 2) positive or negative and (x - 5) positive or negative). You have only 3 cases. The 4th case will not exist.
_________________
Quote:
It comes down to whether you understand why we take ranges and change signs. First check this post:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2014/06 ... -the-gmat/
Now take a simpler example:
|x - 2| + |x - 5| = 4
What do you need to consider here?
For |x - 2|, you have to think about x < 2 and x >= 2
For |x - 5|, you have to think about x < 5 and x >= 5
To consider both together, you have
one range x < 2.
Another is x >= 2 and x < 5 so this is 2 <= x < 5.
Yet another is x >= 5
Now, will you worry about x < 2 and at the same time, x > 5? No. There will not be such a value of x. This is where |x - 2| will open as -(x - 2) and |x - 5| will open as (x - 5). This doesn't exist.
So in a question like this: |x - 2| + |x - 5| = 4, you don't have 4 cases ((x - 2) positive or negative and (x - 5) positive or negative). You have only 3 cases. The 4th case will not exist.
_________________
-
Start FREE with selected GMAT/EA Content upon registration at anaprep.com
YouTube Channel: karishma.anaprep
Start FREE with selected GMAT/EA Content upon registration at anaprep.com
YouTube Channel: karishma.anaprep
johnwesley
--------------------------------------------------------------------------------
I find the above technique easier for these problems.
Can someone please confirm if my below approach is correct :
The roots/criticial values for the given equations are : -8,-3,4.
Now placing these values on the no. line we have four ranges & checking for each ranges :
1. For x greater than 4
Randomly selecting any value greater than 4 , lets say x = 10
|x+3| - |4-x| = |8+x|
|10+3| - |4-10| is not equal to |8+10|
So any value greater than 4 doesnot satisfy the equation.
2. For x between -3 & 4
Let x = 2
|x+3| - |4-x| = |8+x|
|2+3| - |4-2| is not equal to |8+2|
So any value between -3 & 4 doesnot satisfy the equation.
3. For x between -8 & -3,
Let x = -2
|x+3| - |4-x| = |8+x|
|-2+3| - |4-(-2)| is not equal to |8+(-2)|
So any value between -8 & -3 doesnot satisfy the equation.
4. For x lesser than -8,
Let x = -9
|x+3| - |4-x| = |8+x|
|-9+3| - |4-(-9)| is not equal to |8+(-9)|
So any value less than -8 doesnot satisfy the equation.
So the total solution is 0.
Please confirm if my above approach is correct .
Thanks
Kshitij
kshitij89
Note that in this case, all values in the range will not satisfy the equation.
Say when you take x < -8, you could have got x = -9. But what if you had actually tried x = -10 only and decided that no value in the range satisfies?
_________________
-
Start FREE with selected GMAT/EA Content upon registration at anaprep.com
YouTube Channel: karishma.anaprep
Start FREE with selected GMAT/EA Content upon registration at anaprep.com
YouTube Channel: karishma.anaprep
Posts: 198
VeritasPrepKarishma
hi mam
in "a" when assumed that x is less than -8, the RHS modulus (8+x) is correctly negated. I am okay with this operation. Side by side, however, the LHS modulus (x + 3) is also negated...
what is the main logic underlying this negation ..? is this such that to make the LHS negative, some number should be subtracted from some bigger negatives..? if so, how can we conclude that (x+3) is bigger in value than is (4-x) ..?
thanks in advance ...
ssislam
Consider this:
If x < -8 (i.e. x can be -9, -10, -11, -11.4, -20 and so on...),
what can you say about the sign of (x + 3) ?
Can we say that when x is less than -3, (x + 3) is negative? Sure. So if we know that x is less than -8, then obviously, it is less than -3 too. So in this case, (x+3) will certainly be negative.
Hence |x + 3| will also translate into -(x+3).
To see it easily, drawing a number line helps.
......................... (-8) ...................... (-3) ..................................................... (4) ...................................
For (x + 3), all values of x to the left of -3 will give |x + 3| = -(x + 3). So if x < -8, it is to the left of -3 and hence will lead to |x+3| = -(x + 3).
To the right of -3, all values of x will give |x + 3| = (x + 3).
For more on this, check: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2014/0 ... -the-gmat/
_________________
-
Start FREE with selected GMAT/EA Content upon registration at anaprep.com
YouTube Channel: karishma.anaprep
Start FREE with selected GMAT/EA Content upon registration at anaprep.com
YouTube Channel: karishma.anaprep
Hello Bunuel, Can you explain me the following.
When x >=4 ,
Then when we take x=4, we get
(x+3)-(4-x)=8+x i.e x=9 ,i am considering |x|=x,when x=0.
But when we take x=5, we get
(x+3)+(4-x)=x+5 i.e x=-1
How we decide now?
Posted from my mobile device
When x >=4 ,
Then when we take x=4, we get
(x+3)-(4-x)=8+x i.e x=9 ,i am considering |x|=x,when x=0.
But when we take x=5, we get
(x+3)+(4-x)=x+5 i.e x=-1
How we decide now?
Posted from my mobile device
parthabar
When \(x\geq{4}\), then \(x+3\) is positive, \(4-x\) is negative and \(8+x\) is positive. Thus \(|x+3|=x+3\), \(|4-x|=-(4-x)=x-4\) and \(|8+x|=8+x\).
Therefore for this range \(|x+3| - |4-x| = |8+x|\): transforms to \(x+3 - (x-4) =8+x\): --> \(x=-1\). This solution is NOT OK, since \(x=-1\) is NOT in the range we consider (\(x\geq{4}\)).
Thus no value of x satisfies \(|x+3| - |4-x| = |8+x|\).
Check step-by-step solution here: https://gmatclub.com/forum/x-3-4-x-8-x- ... l#p1241355
Hope it helps.
_________________
New to the GMAT Club?
- • Questions: Question Bank | Bunuel's Signature Collection
• Tests: GMAT Club Tests | Forum Quiz
• Guides: Quantitative | Critical Reasoning | Reading Comprehension | Ultimate Quant Guide | All You Need for Quant
• Rules: Quantitative | Verbal
parthabar wrote:
Hello Bunuel, Can you explain me the following.
When x >=4 ,
Then when we take x=4, we get
(x+3)-(4-x)=8+x i.e x=9 ,i am considering |x|=x,when x=0.
But when we take x=5, we get
(x+3)+(4-x)=x+5 i.e x=-1
How we decide now?
Posted from my mobile device
When x≥4
x≥4
, then x+3
x+3
is positive, 4−x
4−x
is negative and 8+x
8+x
is positive. Thus |x+3|=x+3
|x+3|=x+3
, |4−x|=−(4−x)=x−4
|4−x|=−(4−x)=x−4
and |8+x|=8+x
|8+x|=8+x
.
Therefore for this range |x+3|−|4−x|=|8+x|
|x+3|−|4−x|=|8+x|
: transforms to x+3−(x−4)=8+x
x+3−(x−4)=8+x
: --> x=−1
x=−1
. This solution is NOT OK, since x=−1
x=−1
is NOT in the range we consider (x≥4
x≥4
).
Thus no value of x satisfies |x+3|−|4−x|=|8+x|
|x+3|−|4−x|=|8+x|
.
Hello Bunuel, Thanks for your reply.
I can understand the solution for x>4 , which turns out to be x=-1 and not in the range we considered.
But I am not able to understand the part when we take boundary value say x=4 , then 4-x=4-4=0 , and we know |x|=x ,x>=0, so why are we not considering this point in solution. So why are we considering |4-x| =-(4-x) only for x>4 ?
I think I am missing something here .Please explain.
Hello Bunuel, Can you explain me the following.
When x >=4 ,
Then when we take x=4, we get
(x+3)-(4-x)=8+x i.e x=9 ,i am considering |x|=x,when x=0.
But when we take x=5, we get
(x+3)+(4-x)=x+5 i.e x=-1
How we decide now?
Posted from my mobile device
When x≥4
x≥4
, then x+3
x+3
is positive, 4−x
4−x
is negative and 8+x
8+x
is positive. Thus |x+3|=x+3
|x+3|=x+3
, |4−x|=−(4−x)=x−4
|4−x|=−(4−x)=x−4
and |8+x|=8+x
|8+x|=8+x
.
Therefore for this range |x+3|−|4−x|=|8+x|
|x+3|−|4−x|=|8+x|
: transforms to x+3−(x−4)=8+x
x+3−(x−4)=8+x
: --> x=−1
x=−1
. This solution is NOT OK, since x=−1
x=−1
is NOT in the range we consider (x≥4
x≥4
).
Thus no value of x satisfies |x+3|−|4−x|=|8+x|
|x+3|−|4−x|=|8+x|
.
Hello Bunuel, Thanks for your reply.
I can understand the solution for x>4 , which turns out to be x=-1 and not in the range we considered.
But I am not able to understand the part when we take boundary value say x=4 , then 4-x=4-4=0 , and we know |x|=x ,x>=0, so why are we not considering this point in solution. So why are we considering |4-x| =-(4-x) only for x>4 ?
I think I am missing something here .Please explain.
parthabar
The solution considers the range when x >= 4, so it must be true for x = 4 too bit let's still consider x = 4 case separately.
If x = 4, then |x+3|−|4−x|=|8+x| will be |4+3|−|4−4|=|8+4| --> |7|−|0|=|12| --> 7 = 12, which is NOT true, so x = 4 is not a solution for given equation. This is exactly what we got in the solution, when we considered x > = 4 range: NO solution in this range.
_________________
New to the GMAT Club?
- • Questions: Question Bank | Bunuel's Signature Collection
• Tests: GMAT Club Tests | Forum Quiz
• Guides: Quantitative | Critical Reasoning | Reading Comprehension | Ultimate Quant Guide | All You Need for Quant
• Rules: Quantitative | Verbal
Bhu750
Joined: 19 Nov 2020
Last visit: 12 Dec 2023
Posts: 85
Location: India
Schools: Fuqua '23 MIT MFin "22 Kellogg '23 Johnson '24 McDonough '24 Kellogg (S) LBS '25 (D) Emory 1YR '24 (WL)
GMAT 1: 700 Q48 V38
GPA: 3.6
Schools: Fuqua '23 MIT MFin "22 Kellogg '23 Johnson '24 McDonough '24 Kellogg (S) LBS '25 (D) Emory 1YR '24 (WL)
GMAT 1: 700 Q48 V38
Posts: 85
please advice.. I knew how to solve this question.. despite that I took 4 mins.. what do we do when such questions come up.. is there any faster way to do or do we solve or skip?
shubhamjha22
Current Student
Joined: 11 Dec 2016
Last visit: 11 May 2025
Posts: 27
Location: India
Schools: ISB '23 (A) IIMA PGPX'23 (I)
GMAT 1: 740 Q50 V41
GPA: 3
Hey Everyone,
I had a query.
|x+3| – |4-x| = |8+x|==> -|4-x|=|x+8|-|x+3|
Now the RHS of the equation is always 5 for all values of x
therefore,
-|4-x|=5
Solving for x we get x=-1 or x= 9
if x= -1
Equation==> |-1+3|-|4-(-1)|=|8-1|==> |2|-|5|=|7| LHS=/=RHS so not applicable
if x=9
Equation==> |9+3|-|4-9|=|8+9|==> |12|-|-5|=|17| LHS=/=RHS so not
There fore Option A is correct.
What is the flaw in this solution. and if there is no flaw then can I replicate this for other such questions? In what situations would this logic not work and why? Can anyone help?
Many thanks in advance
I had a query.
|x+3| – |4-x| = |8+x|==> -|4-x|=|x+8|-|x+3|
Now the RHS of the equation is always 5 for all values of x
therefore,
-|4-x|=5
Solving for x we get x=-1 or x= 9
if x= -1
Equation==> |-1+3|-|4-(-1)|=|8-1|==> |2|-|5|=|7| LHS=/=RHS so not applicable
if x=9
Equation==> |9+3|-|4-9|=|8+9|==> |12|-|-5|=|17| LHS=/=RHS so not
There fore Option A is correct.
What is the flaw in this solution. and if there is no flaw then can I replicate this for other such questions? In what situations would this logic not work and why? Can anyone help?
Many thanks in advance
Posts: 152
For the people who are confused about where to use equal sign, I have a possible solution. Always stick to the standard definition of mod.
|x| = x when x>=0 OR -x when x<0
Please see attached the snippet for reference
|x| = x when x>=0 OR -x when x<0
Please see attached the snippet for reference
Attachments
Mod Soln.png [ 149.5 KiB | Viewed 2082 times ]
File comment: Way to assign equal sign in mod
Mod Solution.pdf [367.63 KiB]
Downloaded 85 times
| To download please login or register as a user |
Posts: 152
shubhamjha22
Hello Shubham,
Your reasoning got flawed when you took RHS as always 5.
It depends on the range of x in which you're opening the mod.
You need to open the mod in all the 4 number line ranges I have drawn on the snippet.
Attachments
mod soln 2.png [ 40.74 KiB | Viewed 2110 times ]
Moderator:
Announcements
Monday, Sep 14, 2026
11:30am NY / 3:30pm London / 9pm Mumbai
Subscribe to GMAT Club for New Podcast Episodes, Debriefs, & Insightful Sessions on GMAT Preparation
Tuck at Dartmouth
AI at Dartmouth Tuck: How the School Is Preparing Leaders for the Age of Artificial Intelligence
Inside the courses, research, and hands-on learning defining wise, decisive leadership in an AI-driven world.
Read More
Inside the courses, research, and hands-on learning defining wise, decisive leadership in an AI-driven world.
Read More
Top 5 GMAT Debrief Videos
Success stories and strategies from high-scoring candidates.