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Math: Absolute value (Modulus) 15 May 2019, 16:22
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OP or Bunuel

There is a typo in the first post.

It says "(-15 is not within (-3,4) interval." for the range in (c) when it should actually say "9 is not within the -3<x<4 range"
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Math: Absolute value (Modulus) Updated on: 26 Nov 2023, 03:01
Originally posted by KarishmaB on 08 Jul 2020, 22:53.
Last edited by KarishmaB on 26 Nov 2023, 03:01, edited 1 time in total.
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Syeedarimshah
You said in each range signs of the terms will be different . I am unable to assign proper signs to terms in ranges VeritasKarishma please help

Posted from my mobile device
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The rightmost region is positive and you alternate from there (assuming the factors are in (ax + b) or (ax - b) form). Check out the logic explained in this video:

https://youtu.be/PWsUOe77__E
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Math: Absolute value (Modulus) 25 Oct 2025, 05:26
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Saun2511
When x ≥ 0, |x| = x
When x ≤ 0, |x| = -x

Here, it was x<0 right? not equal to? because then x=0 is both + and - Bunuel

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It doesn’t matter where you put the equal sign because the absolute value of zero is zero. So |x| = x when x ≥ 0 and |x| = −x when x ≤ 0. Whether the equality is with the first or the second part makes no difference at all.
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Math: Absolute value (Modulus) 15 Sep 2010, 03:58
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Can any one explain the below problem once again :(I didnt understand even thought it was explained in the above posts.. :(

Problem: 1<x<9. What inequality represents this condition?

A. |x|<3
B. |x+5|<4
C. |x-1|<9
D. |-5+x|<4
E. |3+x|<5

Stuck between C & D
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Math: Absolute value (Modulus) 01 Dec 2010, 12:19
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Sorry walker one more question on the below as I review this
Thinking of inequality with modulus as a segment at the number line.

For example,
Problem: 1<x<9. What inequality represents this condition?
Image
A. |x|<3
B. |x+5|<4
C. |x-1|<9
D. |-5+x|<4
E. |3+x|<5
Solution: 10sec. Traditional 3-steps method is too time-consume technique. First of all we find length (9-1)=8 and center (1+8/2=5) of the segment represented by 1<x<9. Now, let’s look at our options. Only B and D has 8/2=4 on the right side and D had left site 0 at x=5. Therefore, answer is D.

Why would be looking for 4 on the right side? shouldn't this be 5 as the midpoint? And why do we want the left side to be 0 at x=5? Id like to understand this a little better

II. Converting inequalities with modulus into range expression.
In many cases, especially in DS problems, it helps avoid silly mistakes.

For example,
|x|<5 is equal to x e (-5,5).
|x+3|>3 is equal to x e (-inf,-6)&(0,+inf)

How did you convert lx+3l>3 into (-inf,-6) and (0,+inf)

Sorry these maybe simple questions but I just want to grasp the concept firmly
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Math: Absolute value (Modulus) 01 Dec 2010, 19:35
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Great thanks for the explanation on that Karishma - very very very helpful.

Can you also help me with how we converted lx+3l>3 into (-inf,-6) and (0,+inf)?

Is this because x+3>=0 (non-negative) and negative x+3<0 (negative). So we get to x+3=3 so x=0 which fits condition 1 above and this will go to infinity

and then

-(x+3)=3 so you get -x=6 or x=-6 which fits condition two above so x<-6 so therefore it goes to negative infinity?

I think I am close. appreciate the help.
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Math: Absolute value (Modulus) 26 May 2011, 14:34
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walker


Let’s consider following examples,

Example #1
Q.: \(|x+3| - |4-x| = |8+x|\). How many solutions does the equation have?
Solution: There are 3 key points here: -8, -3, 4. So we have 4 conditions:

a) \(x < -8\). \(-(x+3) - (4-x) = -(8+x)\) --> \(x = -1\). We reject the solution because our condition is not satisfied (-1 is not less than -8)

b) \(-8 \leq x < -3\). \(-(x+3) - (4-x) = (8+x)\) --> \(x = -15\). We reject the solution because our condition is not satisfied (-15 is not within (-8,-3) interval.)

c) \(-3 \leq x < 4\). \((x+3) - (4-x) = (8+x)\) --> \(x = 9\). We reject the solution because our condition is not satisfied (-15 is not within (-3,4) interval.)

d) \(x \geq 4\). \((x+3) + (4-x) = (8+x)\) --> \(x = -1\). We reject the solution because our condition is not satisfied (-1 is not more than 4)
Show more

This is a great topic.However i have a little confusion on the range of numbers that are defined in the 4 scenarios.
As we are looking at x+3 >=0 we get x>=-3
Similarly when x+3 <0 we have x<-3

Based on these the ranges I have for the 4 cases
1. x<-8
2. -8<= x <-3
3. -3<= x <= 4
4. x> 4
the last two cases are whats different .
This obviously stems from solving 4-x >= 0 and 4-x<0 Could someone explain what I am doing wrong here? thanks in advance
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Math: Absolute value (Modulus) 21 Jun 2011, 00:01
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Hi Walker,

Can you please explain this?

Example #1
Q.: |x+3| - |4-x| = |8+x|. How many solutions does the equation have?
Solution: There are 3 key points here: -8, -3, 4. So we have 4 conditions:


How do we get 3 key points and 4 conditions?
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Math: Absolute value (Modulus) 28 Oct 2012, 07:23
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walker
when x<-8, (4-x) is always positive and we don't need to modify the sign when we open it.
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sorry Walker, I still don't get it :(

when we consider x<-8 then we made all the x to be negative so multiplied each with negative terms.. is that (4-x) is already having negative for x so we didn't have to multiply with negative here.

Sorry i am really novice for this kind of problem.. please also let me know if there is any post related with this :)
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Math: Absolute value (Modulus) 13 May 2013, 12:43
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Let’s consider following examples,

Example #1 I am not understanding this example and really struggling with modulus? Can someone please elaborate and explain in further detail? From this post I can't see how I would use this on every modulus problem?
Q.: \(|x+3| - |4-x| = |8+x|\). How many solutions does the equation have?
Solution: There are 3 key points here: -8, -3, 4. So we have 4 conditions:

a) \(x < -8\). \(-(x+3) - (4-x) how did we get -(x+3) here?= -(8+x)\) --> \(x = -1\). We reject the solution because our condition is not satisfied (-1 is not less than -8)

b) \(-8 \leq x < -3\). \(-(x+3) - (4-x) = (8+x)\) --> \(x = -15\). We reject the solution because our condition is not satisfied (-15 is not within (-8,-3) interval.)

c) \(-3 \leq x < 4\). \((x+3) - (4-x) = (8+x)\) --> \(x = 9\). We reject the solution because our condition is not satisfied (-15 is not within (-3,4) interval.)

d) \(x \geq 4\). \((x+3) + (4-x) = (8+x)\) --> \(x = -1\). We reject the solution because our condition is not satisfied (-1 is not more than 4)

I am totally lost with this post and also with other modulus problems I looked up in Gmat club thank you very much for your help in advance!!!!!
Show more

like the gentlemen above, I continue to be puzzled by this post. I tried searching for answer in this topic post itself, but couldn't get a compilation to all my questions, can any expert please make me understand this, here are my queries.

Q.: \(|x+3| - |4-x| = |8+x|\). How many solutions does the equation have?
Solution: There are 3 key points here: -8, -3, 4. So we have 4 conditions:

I get this part, perfectly fine, basic goal to arrive at points is to make every value within modulus "0"
so we have 3 key points and 4 solutions Ie,
-3+3= 0 for 1st modulus sign so key point here is -3,
4-4=0 for second modulus sign, so key point here is 4
8-8 in third modulus, so key point here is -8

Therefore on a number line it will be 3 points something like this ---------\((-8)\)---------\((-3)\)------------------------\((4)\)

second step:

Quote:
A. a) \(x < -8\). \(-(x+3) - (4-x) = -(8+x)\) --> \(x = -1\). We reject the solution because our condition is not satisfied (-1 is not less than -8)
Show more

I do understand in first Bracket \(-(x+3)\), since we are testing X against x < -8[/m], so we need to make \(-X\) here. as per Walkers quote

walker

if x < -8, (x + 3) is always negative. So, modulus is nonnegative and we need to change a sign: |x+3| = - (x+3) for x<-8
For example, if x = -10,
|-10+3| = |-7| = 7
-(-10+3) = -(-7) = 7

In other words, |x| = x if x is positive and |x|=-x if x is negative.
Show more
but my Question is If we eventually want to see a negative X inside the bracket than why \(- (4-x)\)? as in this case X will turn positive after opening the bracket

2nd EQ------
\(-8 \leq x < -3\) \(-(x+3) - (4-x)\) = \((8+x)\)

again in 2nd equation my doubt is why do we have the \((8+X)\) as non negative, I mean it should be same as \(-(8+x)\), like in 1st test case. as X is still negative. in this test case? Of-course this is fine if I can get answer to my 1st query, if we have to make X negative than this is not ok.


in 3rd test case
Quote:
c) \(-3 \leq x < 4\). \((x+3) - (4-x) = (8+x)\) --> \(x = 9\). We reject the solution because our condition is not satisfied (-15 is not within (-3,4) interval.)
Show more

in this case X can be negative or positive, so why don't we put \(-(x+3)\) here? rather than \((X+3)\) ?

Quote:
d) \(x \geq 4\). \((x+3) + (4-x) = (8+x)\) --> \(x = -1\). We reject the solution because our condition is not satisfied (-1 is not more than 4)
Show more

Again in the equation above we are testing against positive X test point, than why \(+(4-X)\), I think it should be \(-(4-X)\) to turn X into positive after opening the brackets.?

All of the above questions may sound stupid, but I need to understand this, as inequalities is my weak topic.
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Math: Absolute value (Modulus) 14 May 2013, 09:51
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gettinit
Let’s consider following examples,

Example #1 I am not understanding this example and really struggling with modulus? Can someone please elaborate and explain in further detail? From this post I can't see how I would use this on every modulus problem?
Q.: \(|x+3| - |4-x| = |8+x|\). How many solutions does the equation have?
Solution: There are 3 key points here: -8, -3, 4. So we have 4 conditions:

a) \(x < -8\). \(-(x+3) - (4-x) how did we get -(x+3) here?= -(8+x)\) --> \(x = -1\). We reject the solution because our condition is not satisfied (-1 is not less than -8)

b) \(-8 \leq x < -3\). \(-(x+3) - (4-x) = (8+x)\) --> \(x = -15\). We reject the solution because our condition is not satisfied (-15 is not within (-8,-3) interval.)

c) \(-3 \leq x < 4\). \((x+3) - (4-x) = (8+x)\) --> \(x = 9\). We reject the solution because our condition is not satisfied (-15 is not within (-3,4) interval.)

d) \(x \geq 4\). \((x+3) + (4-x) = (8+x)\) --> \(x = -1\). We reject the solution because our condition is not satisfied (-1 is not more than 4)

I am totally lost with this post and also with other modulus problems I looked up in Gmat club thank you very much for your help in advance!!!!!

like the gentlemen above, I continue to be puzzled by this post. I tried searching for answer in this topic post itself, but couldn't get a compilation to all my questions, can any expert please make me understand this, here are my queries.

Q.: \(|x+3| - |4-x| = |8+x|\). How many solutions does the equation have?
Solution: There are 3 key points here: -8, -3, 4. So we have 4 conditions:

I get this part, perfectly fine, basic goal to arrive at points is to make every value within modulus "0"
so we have 3 key points and 4 solutions Ie,
-3+3= 0 for 1st modulus sign so key point here is -3,
4-4=0 for second modulus sign, so key point here is 4
8-8 in third modulus, so key point here is -8

Therefore on a number line it will be 3 points something like this ---------\((-8)\)---------\((-3)\)------------------------\((4)\)

second step:

Quote:
A. a) \(x < -8\). \(-(x+3) - (4-x) = -(8+x)\) --> \(x = -1\). We reject the solution because our condition is not satisfied (-1 is not less than -8)

I do understand in first Bracket \(-(x+3)\), since we are testing X against x < -8[/m], so we need to make \(-X\) here. as per Walkers quote

walker

if x < -8, (x + 3) is always negative. So, modulus is nonnegative and we need to change a sign: |x+3| = - (x+3) for x<-8
For example, if x = -10,
|-10+3| = |-7| = 7
-(-10+3) = -(-7) = 7

In other words, |x| = x if x is positive and |x|=-x if x is negative.
but my Question is If we eventually want to see a negative X inside the bracket than why \(- (4-x)\)? as in this case X will turn positive after opening the bracket

2nd EQ------
\(-8 \leq x < -3\) \(-(x+3) - (4-x)\) = \((8+x)\)

again in 2nd equation my doubt is why do we have the \((8+X)\) as non negative, I mean it should be same as \(-(8+x)\), like in 1st test case. as X is still negative. in this test case? Of-course this is fine if I can get answer to my 1st query, if we have to make X negative than this is not ok.


in 3rd test case
Quote:
c) \(-3 \leq x < 4\). \((x+3) - (4-x) = (8+x)\) --> \(x = 9\). We reject the solution because our condition is not satisfied (-15 is not within (-3,4) interval.)

in this case X can be negative or positive, so why don't we put \(-(x+3)\) here? rather than \((X+3)\) ?

Quote:
d) \(x \geq 4\). \((x+3) + (4-x) = (8+x)\) --> \(x = -1\). We reject the solution because our condition is not satisfied (-1 is not more than 4)

Again in the equation above we are testing against positive X test point, than why \(+(4-X)\), I think it should be \(-(4-X)\) to turn X into positive after opening the brackets.?

All of the above questions may sound stupid, but I need to understand this, as inequalities is my weak topic.
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In a post above, I have given the links to 3 posts which explain the process in detail (including the reasoning behind the process). Check those out to get answers to your questions.
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Math: Absolute value (Modulus) 14 May 2013, 15:42
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VeritasPrepKarishma


In a post above, I have given the links to 3 posts which explain the process in detail (including the reasoning behind the process). Check those out to get answers to your questions.
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Hi Karishma

I went through your post on the blog, but to be frank found this post of your more helpfull

VeritasPrepKarishma
Of course it can be done using algebra as well. It doesn't matter how many mods there are. you always deal with them in the same way.
|x|= x when x is >= 0,
|x|= -x when x < 0

|x - 2|= (x - 2) when x - 2 >= 0 (or x >= 2),
|x - 2|= -(x - 2) when (x-2) < 0 (or x < 2)

Then you solve the equations using both conditions given above. That is the importance of the points.
So if you have:
|x - 2|= |x + 3|

You say, |x - 2|= (x - 2) when x >= 2.
|x - 2|= -(x - 2) when x < 2
|x + 3| = (x + 3) when x >= -3
|x + 3| = -(x + 3) when x < -3
Show more

Ok, Now after literally banging my head for 3 hrs and reading you blog articles back and forth, I get it that to make an EQ in (X-K) format we manipulate it by taking -tive sign out
but I guess in this example its this concept that we need to understand

|x|= x when x is >= 0,
|x|= -x when x < 0

ok, so based on this understanding I will take a fresh shot, please let me know what's wrong

Quote:
a) \(x < -8\). \(-(x+3) - (4-x) = -(8+x)\) --> \(x = -1\). We reject the solution because our condition is not satisfied (-1 is not less than -8)
Show more

In this test case, since we will always have |x+3| negative we put a -tive sign outside because modulus will turn it into non negative, so to do that we take multiply it by (-1), is this understanding correct?
and since we are ok with -(4-x), because we will again get |4-x| positive with a negative x, the -tive sign outside the bracket will make sure its always -tive when out of the Modulus. However to be frank, a little confusion here is, as you mentioned in the blog, why don't we try to convert it into (x-k) format?
in RHS we have -(8-x) because again we want |8-x| to turn out a negative number so we put -(8-x) to make it always negative, let me know if I got it correctly.


Quote:
b) \(-8 \leq x < -3\). \(-(x+3) - (4-x) = (8+x)\) --> \(x = -15\). We reject the solution because our condition is not satisfied (-15 is not within (-8,-3) interval.)
Show more

again, I get it why LHS is that way, however I still don't get it why we don't have -(8-X) as we need to make sure that the result of this bracket is -tive so |8-x| = -(8-x)


Quote:
c) \(-3 \leq x < 4\). \((x+3) - (4-x) = (8+x)\) --> \(x = 9\). We reject the solution because our condition is not satisfied (-15 is not within (-3,4) interval.)
Show more

I still dont get it, if we test x against both -tive and positive scenario, why is that we just have 1 equation? in my view we should split it in 2 eq. to test against both negative and positive value.

Quote:
d) \(x \geq 4\). \((x+3) + (4-x) = (8+x)\) --> \(x = -1\). We reject the solution because our condition is not satisfied (-1 is not more than 4)
Show more

(x+3) since we don't have a negative value of X this bracket will always be positive, we don't need a -tive sign outside, is this the reason?
(4-x) again since a >4 will always make it positive we don't need a -tive sign outside the bracket, is this the reason?
(8+x) again same reason as above for this?


I also had one doubt in your blog question.

Complication No 3: on this post https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2012/07 ... ns-part-i/

(-2x^3 + 17x^2 – 30x) > 0

This is how I understand it,

\(x(-2x^2 + 17x - 30) > 0\) (just took out x common) ok
x(2x – 5)(6 – x) > 0(factoring the quadratic) ok
2x(x – 5/2)(-1)(x – 6) > 0 (take 2 common) ---------> I think in this you took out -1 common to make the second bracket = (x-k) format?
2(x–0)(x–5/2)(x–6) < 0 (multiply both sides by -1)-------> how did you arrive at 2(x-0)? i think it should be just \(2x(x-\frac{5}{2})(x-6) <0\)
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Math: Absolute value (Modulus) 15 May 2013, 06:30
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Karishma,

Kudos given for the post, thanks for explaining in detail,
I agree that we would be better off plugging number on such a ques, but things get tricky when it comes to DS

I basically covered this from MGmat guides and I can handle a simple Mod like |x-2|>5
what I learnt is simply take 2 conditions, x-2>5 and 2-x>5 and solve for 2 set of x, however the book never taught me this 3 step method. so I have to dig it in here.
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Math: Absolute value (Modulus) 04 Jun 2013, 12:35
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(This question is from a GMAT club study book. It can be found here: math-absolute-value-modulus-86462.html)

|x^2-4| = 1. What is x?

Solution: There are 2 conditions:

a) (x^2-4)\geq0 --> x \leq -2 or x\geq2. x^2-4=1 --> x^2 = 5. x e {-\sqrt{5}, \sqrt{5}} and both solutions satisfy the condition.

b) (x^2-4)<0 --> -2 < x < 2. -(x^2-4) = 1 --> x^2 = 3. x e {-\sqrt{3}, \sqrt{3}} and both solutions satisfy the condition.



Why do we set these problems up as >= or <= 1? I would solve this problem as follows:

|x^2-4| = 1

x^2-4 = 1 ==> x^2 = 5 ==> x = \sqrt{5}

OR

-(x^2-4) = 1 ==> -x^2 +4 = 1 ==> -x^2 = -3 ==> X^2 = 3 ==> x = \sqrt{3}

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WholeLottaLove
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Math: Absolute value (Modulus) 04 Jun 2013, 15:39
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Hmmmm...I'm not sure I follow.

This is how I solved the problem.

|x^2-4|=1

x^2 - 4 =1
OR
-x^2 + 4 = 1

SO

x^2=5 ==> x=+/- √5
OR
-x^2=-3 ==> x^2=3 ==> x=+/- √3

So, I believe that is the right answer but I'm still not sure about how the greater than/less than signs come into play. Sorry for the thick head!



Zarrolou
The first method is the correct one and will always give you the correct results.

Consider however the following case

\(|x+5|=-4\), at glance this equation has no solution because \(|x+5|\) cannot be less than 0.
But I wanna take it as example:

With the first method you'll find
if \(x>-5\)
\(x+5=-4\), \(x=-9\), out of the interval => it's not a solution

if \(x<-5\)
\(-x-5=-4\), \(x=-1\) out of the interval => it's not a solution

With the second method
\(x+5=-4\), \(x=-9\)
\(-(x+5)=-4\), or \(x=-1\)
those seem valid... but the equation we know that has no solution.

Main point: the first method works always, do not rely on the other one.
The second one does not take into consideration the intervals, so it might not work

Hope it's clear
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WholeLottaLove
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Math: Absolute value (Modulus) 04 Jun 2013, 16:57
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So, in other words, if I were to plug in -/+ √5 into x^2 = 5 it will yield me a result between -2<x<2? And where does -2<x<2 come from???

Zarrolou
WholeLottaLove
Hmmmm...I'm not sure I follow.

This is how I solved the problem.

|x^2-4|=1

x^2 - 4 =1
OR
-x^2 + 4 = 1

SO

x^2=5 ==> x=+/- √5
OR
-x^2=-3 ==> x^2=3 ==> x=+/- √3

So, I believe that is the right answer but I'm still not sure about how the greater than/less than signs come into play. Sorry for the thick head!


The original solution used \(\geq{}\) and \(\leq{}\) to define the intervals. It solved the cases:
\(x^2-4=1\) so \(x=+-\sqrt{5}\), and then it check weather those numbers are in the interval \(-2<x<2\). Both are inside so both are valid solutions
Then the other case \(-x^2+4=1\) so \(x=+-\sqrt{3}\), and then check the interval it is considering in this scenario \(x<-2\) and \(x>2\), both are inside the intervals so both are valid solutions

As I said before the correct method to solve abs values always checks if the result obtained is inside the interval is considering at that moment.
If you do not double check weather the solution you find is inside the interval, you are likely to commit errors in the problem. What I am trying to say is that your method is incomplete: it lacks the last passage - you find the solutions, but you do not check if they are possible or not.

In this example all 4 solutions are possible, so the last step does nothing; but if one of the solution were not valid, you "incomplete" method would not be albe to detect it. ( as I showed you in the \(|x+5|=-4\) example)
Hope that what I mean is clear
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Math: Absolute value (Modulus) 04 Jun 2013, 17:37
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So, I need to find the positive and negative values of |x^2-4|=1 (i.e. x^2-4=1 and -x^2+4=1) and which x values make x^2-4 positive and -x^2+4 negative?

Thanks for putting up with my slowness in picking up these concepts!


Zarrolou
WholeLottaLove
So, in other words, if I were to plug in -/+ √5 into x^2 = 5 it will yield me a result between -2<x<2? And where does -2<x<2 come from???

-2<x<2 is the interval in which the function is negative, bare with me:

Take the function |x^2-4|=1
1) Define where it is positive and where is negative => \(x^2-4>0\) if x<-2 and x>2
So if \(x<-2\) or \(x>2\) is positive, if \(-2<x<2\) is negative

2)Study each case on its own:
\(x^2-4=1\) \(x=+-\sqrt{5}\), are those results valid? Are they in the interval we are considering? Are they in the x<-2 or x>2 interval?
\(-\sqrt{5}\) is less than -2, and \(+\sqrt{5}\) is more than 2. So they are valid solutions because they are in the intervals we are considering
\(-x^2+4=1\) \(x=+-\sqrt{3}\), are those results valid? same as above
Yes they are valid because they are numbers between \(-2\) and \(2\)(the interval we are considering now, in which |abs| is negative => -x^2+4)
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Math: Absolute value (Modulus) 06 Aug 2013, 09:07
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hi,

i have a doubt in the example

Q.: |x^2-4| = 1. What is x?
Solution: There are 2 conditions:

a) (x^2-4)\geq0

how the range for this example is (X^2-4)>=0 and not (x^2-4)>0?


regards,
RRSNATHAN
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Math: Absolute value (Modulus) 11 Aug 2013, 21:30
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rrsnathan
hi,

i have a doubt in the example

Q.: |x^2-4| = 1. What is x?
Solution: There are 2 conditions:

a) (x^2-4)\geq0

how the range for this example is (X^2-4)>=0 and not (x^2-4)>0?


regards,
RRSNATHAN
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You can ignore "=0" part. (x^2 - 4) cannot be 0 since its mod is 1. So you can certainly write the range as (x^2 - 4) > 0. In the overall scheme, it doesn't change anything. People usually write the ranges are ">=0" and "<0" to cover everything. Here "=0" is not needed.
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Math: Absolute value (Modulus) 14 Apr 2014, 00:42
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I have a doubt.

In this problem
Problem: 1<x<9. What inequality represents this condition?
A. |x|<3
B. |x+5|<4
C. |x-1|<9
D. |-5+x|<4
E. |3+x|<5

can anyone please explain how can I determine in 10-20 sec that B and D are contenders among the options.
please put some light on this - D had left site 0 at x=5.

Thanks
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