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# Math: Absolute value (Modulus)

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Manager
Joined: 19 Jul 2016
Posts: 50
Re: Math: Absolute value (Modulus) [#permalink]

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21 Jan 2017, 07:55
Bunuel wrote:
gupta87 wrote:
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
Math Expert
Joined: 02 Sep 2009
Posts: 44320
Re: Math: Absolute value (Modulus) [#permalink]

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22 Jan 2017, 03:41
1
KUDOS
Expert's post
gupta87 wrote:
Bunuel wrote:
gupta87 wrote:
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

-2x < 9

Divide by -2 and flip the sign x > -4.5,
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Re: Math: Absolute value (Modulus) [#permalink]

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06 May 2017, 17:54
Thanks for the post. For the 3 step approach for complex problems, how do you find x (E.g. x=-1, x=-15)?
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Joined: 19 Aug 2016
Posts: 73
Re: Math: Absolute value (Modulus) [#permalink]

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11 Oct 2017, 19:08
walker wrote:
ABSOLUTE VALUE
(Modulus)

This post is a part of [GMAT MATH BOOK]

created by: walker
edited by: bb, Bunuel

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Definition

The absolute value (or modulus) $$|x|$$ of a real number x is x's numerical value without regard to its sign.

For example, $$|3| = 3$$; $$|-12| = 12$$; $$|-1.3|=1.3$$

Graph:

Important properties:

$$|x|\geq0$$

$$|0|=0$$

$$|-x|=|x|$$

$$|x|+|y|\geq|x+y|$$

$$|x|\geq0$$

How to approach equations with moduli

It's not easy to manipulate with moduli in equations. There are two basic approaches that will help you out. Both of them are based on two ways of representing modulus as an algebraic expression.

1) $$|x| = \sqrt{x^2}$$. This approach might be helpful if an equation has × and /.

2) |x| equals x if x>=0 or -x if x<0. It looks a bit complicated but it's very powerful in dealing with moduli and the most popular approach too (see below).

3-steps approach:

General approach to solving equalities and inequalities with absolute value:

1. Open modulus and set conditions.
To solve/open a modulus, you need to consider 2 situations to find all roots:
• Positive (or rather non-negative)
• Negative

For example, $$|x-1|=4$$
a) Positive: if $$(x-1)\geq0$$, we can rewrite the equation as: $$x-1=4$$
b) Negative: if $$(x-1)<0$$, we can rewrite the equation as: $$-(x-1)=4$$
We can also think about conditions like graphics. $$x=1$$ is a key point in which the expression under modulus equals zero. All points right are the first condition $$(x>1)$$ and all points left are second condition $$(x<1)$$.

2. Solve new equations:
a) $$x-1=4$$ --> x=5
b) $$-x+1=4$$ --> x=-3

3. Check conditions for each solution:
a) $$x=5$$ has to satisfy initial condition $$x-1>=0$$. $$5-1=4>0$$. It satisfies. Otherwise, we would have to reject x=5.
b) $$x=-3$$ has to satisfy initial condition $$x-1<0$$. $$-3-1=-4<0$$. It satisfies. Otherwise, we would have to reject x=-3.

3-steps approach for complex problems

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)

(Optional) The following illustration may help you understand how to open modulus at different conditions.

Example #2
Q.: $$|x^2-4| = 1$$. 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.

(Optional) The following illustration may help you understand how to open modulus at different conditions.

Answer: $$-\sqrt{5}$$, $$-\sqrt{3}$$, $$\sqrt{3}$$, $$\sqrt{5}$$

Tip & Tricks

The 3-steps method works in almost all cases. At the same time, often there are shortcuts and tricks that allow you to solve absolute value problems in 10-20 sec.

I. Thinking of inequality with modulus as a segment at the number line.

For example,
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
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.

II. Converting inequalities with modulus into a 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)

III. Thinking about absolute values as the distance between points at the number line.

For example,
Problem: A<X<Y<B. Is |A-X| <|X-B|?
1) |Y-A|<|B-Y|
Solution:

We can think of absolute values here as the distance between points. Statement 1 means than the distance between Y and A is less than that between Y and B. Because X is between A and Y, |X-A| < |Y-A| and at the same time the distance between X and B will be larger than that between Y and B (|B-Y|<|B-X|). Therefore, statement 1 is sufficient.

Pitfalls

The most typical pitfall is ignoring the third step in opening modulus - always check whether your solution satisfies conditions.

Official GMAC Books:

The Official Guide, 12th Edition: PS #22; PS #50; PS #130; DS #1; DS #153;
The Official Guide, Quantitative 2th Edition: PS #152; PS #156; DS #96; DS #120;
The Official Guide, 11th Edition: DT #9; PS #20; PS #130; DS #3; DS #105; DS #128;

Generated from [GMAT ToolKit]

Resources

Absolute value DS problems: [search]
Absolute value PS problems: [search]

Fig's post with absolute value problems: [Absolute Value Problems]

A lot of questions as well as a separate topic of PrepGame on absolute value is included in GMAT ToolKit 2

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[Reveal] Spoiler: Images
Attachment:
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Attachment:
graph_modulus.png
Attachment:
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Attachment:
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Attachment:
Math_abs_example0.png

II. Converting inequalities with modulus into a 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 u get -inf and 0 could u pls explain?

Also, how did u get B and D as the last two 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.

Pls help..

Intern
Joined: 02 Nov 2017
Posts: 1
Re: Math: Absolute value (Modulus) [#permalink]

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02 Nov 2017, 03:30
Hi,
In example #1, I am sorry I don't understand the logic to put either + or - after each group of the equation. How do you choose to put + or - before -/+(x+3)-/+(4−x)=+/-(8+x) for each scenario ?

Thanks
Intern
Joined: 26 Jul 2017
Posts: 6

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26 Nov 2017, 07:23
I am Able to solve most of the absolute value problems using the 3 steps approach. But every time I am taking more than 2 minutes to solve them. Is it possible to apply number line method to solve all the modulus problems? Can anyone please explain with some examples how to apply number line method.
Senior Manager
Joined: 09 Mar 2016
Posts: 327
Re: Math: Absolute value (Modulus) [#permalink]

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09 Feb 2018, 00:30
VeritasPrepKarishma wrote:
nikhil007 wrote:
VeritasPrepKarishma wrote:
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

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)

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.)

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.)

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)

(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?

Complication No 3: on this post http://www.veritasprep.com/blog/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?

First of all, if you do get a question with multiple mods and if you want to be prepared for it, using algebra will be far more time consuming than the approaches discussed in my blog. But nevertheless, you should understand it properly.

When you have an equation with x in it, you solve by taking x to one side and everything else to the other. What happens when you have mods in it?
Say |x| = 4, you still haven't got the value of x. You have the value of |x| only. So you need to remove the mod. Now there are rules to remove the mod.

|x|= x (mod removed) when x is >= 0,
|x|= -x (mod removed) when x < 0

So |x| = 4 to remove the mod, I need to know whether x is positive or negative.
If x >= 0, |x| = x so |x| = 4 = x
We get that x is 4

If x < 0, |x| = -x so |x| = 4 = -x
hence x = -4

So if we are looking for a positive value, then it is 4 and if we are looking for a negative value, it is -4.

Similarly, when you have |x+4| + |x - 3| = 10 (just an example), you need to remove the mods to solve for x. But to remove mods (which are around the entire factors x-4 and x-3 and not just around x), you need to know whether (x+ 4) and (x - 3) (the thing inside the mod) are positive/negative.

So you split it into ranges:

x > 3
Put any value greater than 3 in (x+4), (x+4) will remain positive. Put any value greater than 3 in (x - 3), (x - 3) will remain positive.

So when x > 3, we can remove the mods without any modification:
(x + 4) + (x-3) = 10
x = 9/2
Since 9/2 is greater than 3, this value of x is acceptable.

-4 < x< 3
For these values of x, (x+4) will always be positive but (x-3) will be negative. So |x - 3| = -(x-3)
(x + 4) - (x-3) = 10
You don't have any such value for x

x < -4
For these values of x, (x+4) and (x-3) will be negative. So |x - 3| = -(x-3) and |x+4| = -(x+4)
-(x + 4) - (x-3) = 10
x = -11/2
Since -11/2 is less than -4, this value of x is also acceptable.

I have discussed how to deal with such questions logically here: http://www.veritasprep.com/blog/2011/01 ... s-part-ii/

As for question with factors that are multiplied (discussed in the 3 links given above),
We know how to deal with (x-a)(x-b)(x-c) > 0 type of questions so we try to bring it that form.

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 $$2x(x-\frac{5}{2})(x-6) <0$$

(x-0) is nothing but x. I put as (x-0) to make it consistent to the (x-a)(x-b).... form to help you remember that you have to take 0 as a transition point too.

Hi VeritasPrepKarishma,

you say ' Similarly, when you have |x+4| + |x - 3| = 10 (just an example), you need to remove the mods to solve for x. But to remove mods (which are around the entire factors x-4 and x-3 and not just around x), you need to know whether (x+ 4) and (x - 3) (the thing inside the mod) are positive/negative.

So you split it into ranges:

x > 3
Put any value greater than 3 in (x+4), (x+4) will remain positive. Put any value greater than 3 in (x - 3), (x - 3) will remain positive"

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
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7992
Location: Pune, India
Re: Math: Absolute value (Modulus) [#permalink]

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09 Feb 2018, 01:04
1
KUDOS
Expert's post
dave13 wrote:

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,

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.veritasprep.com/blog/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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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Senior Manager Joined: 09 Mar 2016 Posts: 327 Re: Math: Absolute value (Modulus) [#permalink] ### Show Tags 09 Feb 2018, 12:32 VeritasPrepKarishma wrote: dave13 wrote: 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.veritasprep.com/blog/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? 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) Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7992 Location: Pune, India Re: Math: Absolute value (Modulus) [#permalink] ### Show Tags 12 Feb 2018, 05:48 1 This post received KUDOS Expert's post 1 This post was BOOKMARKED dave13 wrote: VeritasPrepKarishma wrote: dave13 wrote: 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.veritasprep.com/blog/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? 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. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: Math: Absolute value (Modulus) [#permalink]

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18 Feb 2018, 08:26
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 ?

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 ---> $$x^2 - 9x +9 = 0$$

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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Re: Math: Absolute value (Modulus) [#permalink]

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19 Feb 2018, 03:29
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dave13 wrote:
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 ?

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 wrote:
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 ---> $$x^2 - 9x +9 = 0$$

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 wrote:
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:)

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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Re: Math: Absolute value (Modulus)   [#permalink] 19 Feb 2018, 03:29

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