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Math: Absolute value (Modulus) 05 Jun 2013, 00:27
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WholeLottaLove
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!

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Yes the first thing to do is finding the values of x that make the finction positive/negative.

So first you need to split the function into two cases: where is positive and where is negative (as you say), then you solve each case and find the result(s), then you check if the result is INSIDE the interval you are considering.
So \(x^2-4\) is positive for x<-2 and for x>2, so you solve \(x^2-4=1\)=>\(x=+-\sqrt{5}\), are those value inside the interval x<-2 and x>2? YES, they are valid results.
Then you repeat the same operation for the negative part \(-x^2+4\)
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Math: Absolute value (Modulus) Updated on: 26 Nov 2023, 02:57
Originally posted by KarishmaB on 14 Apr 2014, 20:54.
Last edited by KarishmaB on 26 Nov 2023, 02:57, edited 1 time in total.
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honey86
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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|x - a| = b
implies that x is at a distance b from a.

|x - a| < b
implies x is at a distance less than b from a.

|x - 5| < 4
implies x is at a distance less than 4 from 5. A distance of 4 from 5 gives two points: 1 and 9. Since we need 1 < x < 9, this is the answer.

|x + 5| < 4
implies x is at a distance less than 4 from -5. A distance of 4 from -5 gives points -9 and -1.

Check this video: https://youtu.be/oqVfKQBcnrs
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Math: Absolute value (Modulus) 15 Jun 2014, 22:24
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madn800
Can somebody help me solve the inequality:|x+5| > 3|x - 5|

This is what I did.....
\(\frac{|x+5|}{|x - 5|}>3\)

==>\(\frac{x+5}{x - 5}>3\) & \(\frac{x+5}{x - 5}<-3\)
==>\(x+5>3x-15\) & \(x+5<-3x+15\)
==>-2x>-20 & 4x<10
==>x<10 & x<5/2
BUT answer is 5/2<x<10.
What am I doing wrong??
In the red highlighted portion I changed the inequality sign because I multiplied the equation with minus. Is it correct??
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In an inequality, you cannot multiply/divide by a variable until and unless you know the sign of the variable.
Also, you cannot divide by |x - 5| since x could be 5. Even if we assume that x is not 5, still you cannot multiply the inequality by (x - 5) as you have done here: ==>\(x+5>3x-15\) & \(x+5<-3x+15\)

If x < 5 i.e. x - 5 is negative, when you multiply the inequality by x - 5, the sign of the inequality with flip.

If you would want to use algebra, you need to check in the regions between the transition points:

Case 1: x < -5
In this case x+5 is negative and x-5 is negative, so
-(x+5) > -3(x - 5)
2x > 20
x > 10
Since x should be less than -5, x > 10 doesn't give us any value for x

Case 2: -5 <= x < 5
In this case x+5 is positive (or 0) and x-5 is negative, so
(x+5) > -3(x - 5)
4x > 10
x > 2.5
Since x should be less than 5, the range we get is 2.5 < x< 5


Case 3: x >= 5
In this case x+5 is positive and x-5 is positive, so
(x+5) > 3(x - 5)
2x < 20
x < 10
Since x should be greater than or equal to 5, the range we get is 5 <= x < 10

Combining case 2 and case 3, we get 2.5 < x < 10
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Math: Absolute value (Modulus) 15 Jun 2014, 22:36
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madn800
Can somebody help me solve the inequality:|x+5| > 3|x - 5|

This is what I did.....
\(\frac{|x+5|}{|x - 5|}>3\)

==>\(\frac{x+5}{x - 5}>3\) & \(\frac{x+5}{x - 5}<-3\)
==>\(x+5>3x-15\) & \(x+5<-3x+15\)
==>-2x>-20 & 4x<10
==>x<10 & x<5/2
BUT answer is 5/2<x<10.
What am I doing wrong??
In the red highlighted portion I changed the inequality sign because I multiplied the equation with minus. Is it correct??
Show more

You can use the number line approach too.

|x+5| > 3|x - 5|
The distance of x from -5 is greater than 3 times the distance of x from 5.

Draw the number line:


___________________-5______________0______________5______________

At 0, the distance of x from -5 is same as distance of x from 5. So you must move to the right to make distance from -5 greater. As you keep moving to the right, distance of x from -5 will keep increasing but distance from 5 will keep decreasing. Mid way between 0 and 5, distance of x from -5 is 7.5 and distance of x from 5 is 2.5. So x = 2.5 is a point where distance of x from -5 is 3 times distance of x from 5.

As you keep moving further to the right and reach x = 5, distance from -5 will be 10. Moving further to the right 5 steps will give you a distance of 15 from -5 and 5 from 5. So x = 10 is a point where distance of x from -5 is 3 times distance of x from 5.

In between these two points, distance of x from -5 is more than 3 times distance of x from 5.

Answer: -2.5 < x < 10
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Math: Absolute value (Modulus) 14 Jun 2015, 00:44
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pacifist85
I am not sure I undertand this completely. I have a problem with the signs we use depending on the conditions:

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)
Why are the all the signs here negative?
b) −8≤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.)
Why are the signs in the first part here negative and not the second?
c) −3≤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.)
Why is this similar to the original equation?
d) x≥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)
This makes more sense as x will be positive. But is this why all the signs outside the brackets are positive?

Thank you!
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hi pacifist85,
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)
Why are the all the signs here negative?
we have three mod here ..
1)|x+3|.. this term will be positive because of mod sign but if x<-8, say -9.. x+3=-9+3=-6... so the value of x+3 is actually negative but l-6l=6..
therefore to get the value of x we put a negative sign when we open modulus sign..
2)|4−x|.. here 4-(-9)=13 so the values do not change whether with mod sign or without.. so the existing -ive sign remains..
3)|8+x|... when x=-9, 8=(-9)=-1.. same as 1 above ..

you can similarly check for other three parts..
hope it helped
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Math: Absolute value (Modulus) 09 Feb 2011, 04:41
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i have a question related to Example 1:

|x+3| - |4-x| = |8+x|

-3 \leq x < 4
x \geq 4.
if we take /4-x/ than we have 4-x >=0 so x >=4
and 4-x<0 4<=0
i see in the solution that you switched x-4 and wrote x-4>= 0 and x-4 <o
i understand that it is perfectly ok to do so. my question: when do we have to know when to switch ? because if not switched than we shall have x >=4 if switched x<=4
i hope i made myself clear
thanks
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Math: Absolute value (Modulus) 27 May 2011, 07:15
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someonear


I am worried about values of x that are on the border of the ranges
Say hypothetically for a particular set of equations we end with the identical 4 cases we have here.
Now if say I have x=4 then the way I had come up with the ranges I will get a solution between -3 and 4 as x=4 exists in -3<= x <= 4. But if however I consider what was done in the OP then x=4 exists in the range x>=4
Granted either way I have a solution but will it be a biggie if I fail to show exactly in which range the solution, if it does,exists
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Your aim is to get the solution. You create the ranges to help yourself solve the problem. It doesn't matter at all in which range you consider the border value to lie. Say, when x = 4, (4 - x) = 0. You solve saying that in the range -3<= x<4, (4 - x) is positive and in the range x>= 4, (4 - x) is negative.
At the border value i.e. x = 4, (4 - x) = 0. There is no negative or positive at this point. Hence it doesn't matter where you include the '='. Put it wherever you like. I just like to go in a regular fashion like walker did above. Include the first point in the first range, the second one in the second range (but not the third one i.e. -3 <= x < 4) and so on.
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Math: Absolute value (Modulus) 21 Jun 2011, 03:28
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There are 3 points where one of the modules is zero:

1)x+3=0 --> x = -3
2)4-x=0 --> x = 4
3)8+x=0 --> x = -8

Those 3 points divide the number line by 4 pieces:
1) -inf, -8
2) -8,-3
3) -3, 4
4) 4, +inf

and for each condition we are solving the equation separately.
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Math: Absolute value (Modulus) 10 May 2012, 19:11
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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)

Two Questions:
Can this only be done when we all absolute values in the equation. Could we have done it for \(|x+3| - 5 = |8+x|\)?
States the conditions are -3 and -8?

Second, How did you know whether to put a negative or make positive the terms in the conditions?

For example in (a) you made \((x+3) and (8+x)\) negative
in (b) you made \((x+3)\) negative and everything positive.

What gives?

Thank you!


How do you know
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Math: Absolute value (Modulus) 04 Jun 2013, 12:46
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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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Math: Absolute value (Modulus) 04 Jun 2013, 16:20
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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!

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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, 17:16
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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???
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-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) 11 Jul 2013, 00:06
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Bumping for review*.

*New project from GMAT Club!!! Check HERE

All Theory Topics: search.php?search_id=tag&tag_id=351
MATH BOOK: gmat-math-book-87417.html
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Math: Absolute value (Modulus) 14 Jun 2015, 22:14
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pacifist85
I am not sure I undertand this completely. I have a problem with the signs we use depending on the conditions:

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)
Why are the all the signs here negative?
b) −8≤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.)
Why are the signs in the first part here negative and not the second?
c) −3≤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.)
Why is this similar to the original equation?
d) x≥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)
This makes more sense as x will be positive. But is this why all the signs outside the brackets are positive?

Thank you!
Show more
IT is derives from the definition of absolute value:

|x| = x if x is positive and -x if x is negative.

You have three transition points: -8, -3 and 4.

When x < -8, all three expressions (x + 8), (x + 3) and (x - 4) will be negative (check for say x = -10 to understand). So when you remove the absolute value from |x + 8|, you will get -(x + 8) (because x + 8 is negative).Similarly, when you remove absolute value from |x + 3|, you will get - (x + 3) because x+3 is negative and so on...

Similarly, when -8 < x < -3, (x is to the right of -8)
(x+8) will be positive but other two expressions will be negative (check for say x = -5 to understand).

and so on...
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Math: Absolute value (Modulus) 22 Jan 2017, 03:41
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If x is an integer, what is the value of x?
(1) –2(x + 5) < –1
(2) –3x > 9

from S1 it should be x>4.5...........but in explanation its given x>-4.5.
pls explain the logic

–2(x + 5) < –1

Divide by -2 and flip the sign: x + 5 > 0.5;

Deduct 5: x > 0.5 - 5

x > -4.5

Discussed here: if-x-is-an-integer-what-is-the-value-of-x-109983.html

Hope it helps.

understood.but please correct me where am i going wrong??
-2x-10<-1 => -2x<9 =>2x>9 => x>4.5
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-2x < 9

Divide by -2 and flip the sign x > -4.5,
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Math: Absolute value (Modulus) Updated on: 26 Nov 2023, 02:59
Originally posted by KarishmaB on 09 Feb 2018, 01:04.
Last edited by KarishmaB on 26 Nov 2023, 02:59, edited 1 time in total.
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dave13


regarding this |x - 3| well \(x\) could be any value be negative or positive ? why are saying to put value greater than 3? :? \(x\) could be -5 no :? ?

can you advice in which case can i square both sides od modulus ? can i do it |x+4| + |x - 3| = 10


How about this question \(|x-2| + |x+5| - |x-3| > |x+13|\) woukd i be able to square both sides ? if not please explain why ? :? :)


thank you in advance for your kind explanation and have an awesome day :)
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Hey Dave,

I am not sure I understand your question but here is something that might help you:

Why do we take the range as 'x - 3 > 0' ? or 'x + 5 > 0' ?

The definition of absolute value will help you understand this. It is discussed here: https://youtu.be/oqVfKQBcnrs

Can you square |x+4| + |x - 3| = 10?
Yes you can. It is an equation. You can always square an equation. Even if it were an inequality such as |x+4| + |x - 3| > 10, you can still square it since both sides are definitely non negative. But what will we achieve by squaring it? It doesn't remove the absolute value sign.

\(|x+4| + |x - 3| = 10\)

\(|x + 4|^2 + |x - 3|^2 + 2*|x+4|*|x - 3| = 10^2\)

The term 2*|x+4|*|x - 3| will continue to have absolute value sign. How will you take care of it?
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Math: Absolute value (Modulus) 12 Feb 2018, 05:48
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VeritasPrepKarishma
dave13


regarding this |x - 3| well \(x\) could be any value be negative or positive ? why are saying to put value greater than 3? :? \(x\) could be -5 no :? ?

can you advice in which case can i square both sides od modulus ? can i do it |x+4| + |x - 3| = 10


How about this question \(|x-2| + |x+5| - |x-3| > |x+13|\) woukd i be able to square both sides ? if not please explain why ? :? :)


thank you in advance for your kind explanation and have an awesome day :)

Hey Dave,

I am not sure I understand your question but here is something that might help you:

Why do we take the range as 'x - 3 > 0' ? or 'x + 5 > 0' ?

The definition of absolute value will help you understand this. It is discussed here:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2014/0 ... -the-gmat/

Can you square |x+4| + |x - 3| = 10?
Yes you can. It is an equation. You can always square an equation. Even if it were an inequality such as |x+4| + |x - 3| > 10, you can still square it since both sides are definitely non negative. But what will we achieve by squaring it? It doesn't remove the absolute value sign.

\(|x+4| + |x - 3| = 10\)

\(|x + 4|^2 + |x - 3|^2 + 2*|x+4|*|x - 3| = 10^2\)

The term 2*|x+4|*|x - 3| will continue to have absolute value sign. How will you take care of it?
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Hi VeritasPrepKarishma, many thanks for you reply and the article :)

i wish i could answer your question, :) i dont know how i would take care off it but there is one math wizzard called pushpitkc (perhaps he learnt from you :) ) who actually applied techique of squaring modulus in this question :) https://gmatclub.com/forum/how-many-val ... l#p2011543

i extracted his solution here (in wine red :-) with question stem and answer choices ), so you dont overload your laptop with many internet browser tabs opened :) otherwise your laptop risks to be frozen like mine :)

so see below this magician squares modulus :) so my question when can i apply squaring to modulus ? in which cases can I apply and which cases it wont work ?

How many values of x satisfy the equation ||x-7|-9|=11?

A. 1
B. 2
C. 3
D. 4
E. 5


If |x-7| = a, the equation will become |a - 9| = 11 ( how did he break this equation into two btw ? :)

Squaring on both sides,
\(a^2\)−18a+81=121
\(a^2\)−18a+81=121

\(a^2\)−18a−40=0
\(a^2\)−18a−40=0

Solving for a, a = 20 or -2

If |x-7| = 20
x could take the value 27 or -13
However, it is not possible to get a value of x where |x-7| = -2.

Therefore, there are 2 values of x which satisfy the equation(Option B)


Note here that in this case squaring will take care of the absolute value sign since the whole expression on the left is within the sign.

\(|x|^2 = (x)^2 = (-x)^2\)

\(|x + 2|^2 = (x + 2)^2 = x^2 + 2^2 + 2*x*2\) (No absolute value sign left)

Look at the example you gave:
\(||x-7|-9|=11\)

Just denote |x - 7| as a. You get

\(|a - 9| = 11\)
Square it and get 'a'.

But you don't need a. You need the value of x. For that, we know
\(a = 20 = |x - 7|\) (as assumed before)
Squaring again will get the answer.


How about an expression such as
\(|2x + 3| = |x - 4|\)
Again squaring will help since absolute value sign from both sides will go away.

How about \(|x+4| + |x - 3| = 10\) When you square the left hand side, you use \((a + b)^2 = a^2 + b^2 + 2ab\) where \(a = |x + 4|\) and \(b = |x - 3|\)

So \((|x+4| + |x - 3| )^2 = |x + 4|^2 + |x - 3|^2 + 2*|x + 4|*|x - 3|\)

\(= (x + 4)^2 + (x - 3)^2 + 2*|x + 4|*|x - 3|\)

Here is the problem. 2*|x + 4|*|x - 3| still has absolute value signs.
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Math: Absolute value (Modulus) 19 Feb 2018, 03:29
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dave13
Hi VeritasPrepKarishma

many thanks for taking to explain ! :)

i still have some questions :-)

you say: "
Note here that in this case squaring will take care of the absolute value sign since the whole expression on the left is within the sign."

what does "within the sign" mean ?
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It means the entire left side is within the absolute value sign.

Note that these 2 cases are different:

\(|x + y|^2 = (x + y)^2 = x^2 + y^2 + 2xy\)

\((|x| + |y|)^2 = |x|^2 + |y|^2 + 2*|x|*|y| = x^2 + y^2 + 2*|x|*|y|\)

So if x were say -4 and y were 2, in the first case, we would have \((-4)^2 + 2^2 +2*(-4)*2 = 4^2 + 2^2 - 2*4*2\)
and in the second case we would have \((-4)^2 + 2^2 + 2*|-4|*|2| = 4^2 + 2^2 + 2*4*2\)

So to solve the second case, we need to know the signs of x and y because even after squaring, we still have absolute value signs.


dave13

Another question: ok so we have \(|x+4| + |x - 3| = 10\)

tha formula we use: \((a + b)^2 = a^2 + b^2 + 2ab\)

Do you mean this formula wont work because

\(a = |x + 4|\) and \(b = |x - 3|\) whereas is in other equations we only one modulus on each side like this one \(|2x + 3| = |x - 4|\) or this one \(|a - 9| = 11\)

but on the other hand \(a = |x + 4|\) and \(b = |x - 3|\) i could write / divide left side into two modulus |x + 4| and |x - 3|

so \(|x + 4|\) --> following this formula \((a + b)^2 = a^2 + b^2 + 2ab\) i get ---> \(x^2+8x+16 = 0\)

\(|x - 3|\)---> following this formula \((a + b)^2 = a^2 + b^2 + 2ab\) i get ---> \(x^2 - 9x +9 = 0\)
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So the same case is here. When you have \(a = |x + 4|\) and \(b = |x - 3|\), the 2ab term retains the absolute signs and hence you cannot solve it. To solve an equation, you need to get rid of all absolute value signs.

dave13

Another point: you write this -->

"So\((|x+4| + |x - 3| )^2\)= \(|x + 4|^2 + |x - 3|^2 + 2*|x + 4|*|x - 3|\)

\(= (x + 4)^2 + (x - 3)^2 + 2*|x + 4|*|x - 3|\)"

i dont understand why after this (|x+4| + |x - 3| )^2 you still keep modulus sign :? we square it right

for example here everything is simple and clear ---> \(|x + 2|^2 = (x + 2)^2 = x^2 + 2^2 + 2*x*2\) (No absolute value sign left) <--- you square and modulus sign is gone:)
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Yes, the absolute value sign is gone from \(|x + 2|^2\) and from \(|x - 3|^2\) but what about from \(2*|x + 2|*|x - 3|\) ? (this is the 2ab term).
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Math: Absolute value (Modulus) 23 Aug 2018, 03:12
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How frequent or how many absolute value questions can we approximately expect on the GMAT? I understand there may not be a sure way to know or guess but just looking for an estimate based on past and present experiences.
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I'd say 1 or 2.
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Math: Absolute value (Modulus) 23 Aug 2018, 05:02
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How frequent or how many absolute value questions can we approximately expect on the GMAT? I understand there may not be a sure way to know or guess but just looking for an estimate based on past and present experiences.
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Not more than a couple. The problem is that often GMAT likes to combine topics. So, you might have a geometry question which will make you use some basic absolute value concept such as area bounded by the graph of |x| + |y| = 4. So a basic understanding of all topics is a good idea even if you plan to ignore some topics.
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