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Math: Absolute value (Modulus)

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

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New post 05 Jun 2013, 00:27
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WholeLottaLove wrote:
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!



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

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

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New post 06 Aug 2013, 09:07
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?


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

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New post 11 Aug 2013, 21:30
rrsnathan wrote:
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


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

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New post 14 Apr 2014, 00:42
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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Re: Math: Absolute value (Modulus)  [#permalink]

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New post 14 Apr 2014, 20:54
1
honey86 wrote:
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


|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.

For more on absolute values, check: http://www.veritasprep.com/blog/2011/01 ... edore-did/
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Re: Math: Absolute value (Modulus)  [#permalink]

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New post 14 Apr 2014, 21:18
Thank you Karishma. That was helpful
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Re: Math: Absolute value (Modulus)  [#permalink]

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New post 20 May 2014, 14:04
Dear Honey

Here's a quick shortcut to solving this problem:

You know that 1<x<9. So, pick x=6 and substitute in each of the options. Only options C and D remain.

But, if you analyze option C, you'll find that it'll also hold true for x=0 or x=-1, which are outside the given domain of x. So, the given domain cannot be for option C.

So, the solution must be the only option remaining- D. :-D
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Re: Math: Absolute value (Modulus)  [#permalink]

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New post 14 Jun 2014, 19:05
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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New post 15 Jun 2014, 22:24
2
madn800 wrote:
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??


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

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New post 15 Jun 2014, 22:36
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madn800 wrote:
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??


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

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New post 17 Jun 2014, 10:07
In the absolute value question, I don't get why the distance between A and X is not explained. There's discussion about Y and A, Y and B, X between A and Y, distance between X and B and Y and B but not A and X , I need a reexplanation, maybe with numbers
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New post 17 Jun 2014, 20:27
sagnik242 wrote:
My questions: if we have 3 points why do we have 4 conditions? Next, for each of the letters (a,b,c,d) how do we know which equation to use and why is the inequality used? For example, for a) we do x = -8, but then how to we know the equation to use is -(x+3) - (4-x) = -(8+x), and how did the negative sign come in front of the -x+3 and why? I have the same type of questions for b,c,d so maybe an answer to a will help


Plot the 3 points: -8, -3 and 4 on the number line. Now there are 4 distinct regions on the number line:
The part that lies to the left of -8,
the part that lies between -8 and -3,
the part that lies between -3 and 4 and
the part that lies to the right of 4.

In each of these regions, each of the absolute value expressions will behave differently.

You should know the definition of absolute value:

|x| = x if x >= 0
|x| = -x if x < 0

Consider the equation:
|x+3| - |4-x| = |8+x|

What happens in the region which is at the left of -8 say x = -10
x+3 = -10+3 = -7
Since x+3 is negative, |x+3| = -(x + 3)

4-x = 4 - (-10) = 14
Since 4-x is positive, |4 - x| = 4 - x

8 + x = 8 - 10 = -2
Since x+8 is negative, |8+x| = -(8+x)

This is the logic used. Now repeat it for each region.
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Re: Math: Absolute value (Modulus)  [#permalink]

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New post 15 Oct 2014, 23:25
I dont seem to understand that in the last option (d), why the sign of (4-x) is reversed to positive?
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New post 08 Dec 2014, 09:31
VeritasPrepKarishma wrote:

If |x- 5| < 4, now we are looking for points at a distance less than 4 away from 5.
Attachment:
Ques2.jpg


VeritasPrepKarishma wrote:
OR consider that lx+3l>3 means distance of x from -3 is more than 3.
If you go to 3 steps to right from -3, you reach 0. Anything after than is ok.
If you go 3 steps to left from -3, you reach -6. Anything to its left is ok.
Attachment:
Ques2.jpg


I didn't understand the highlighted portion, Could you please explain:
1. How we are getting to positive 5 from -5 inside of the absolute value; then similarly negative 3 from +3 inside of the absolute value?
2. I can work with the more than or less than from the greater or less than sign but how can I get the value next to "from" , inside of the absolute value, to make a number line?

TIA!
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New post 08 Dec 2014, 21:09
appleid wrote:
VeritasPrepKarishma wrote:

If |x- 5| < 4, now we are looking for points at a distance less than 4 away from 5.
Attachment:
Ques2.jpg


VeritasPrepKarishma wrote:
OR consider that lx+3l>3 means distance of x from -3 is more than 3.
If you go to 3 steps to right from -3, you reach 0. Anything after than is ok.
If you go 3 steps to left from -3, you reach -6. Anything to its left is ok.
Attachment:
Ques2.jpg


I didn't understand the highlighted portion, Could you please explain:
1. How we are getting to positive 5 from -5 inside of the absolute value; then similarly negative 3 from +3 inside of the absolute value?
2. I can work with the more than or less than from the greater or less than sign but how can I get the value next to "from" , inside of the absolute value, to make a number line?

TIA!


Here is the post that explains this logic: http://www.veritasprep.com/blog/2011/01 ... edore-did/
It shows you how to deal with absolute values using the number line.
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Re: Math: Absolute value (Modulus)  [#permalink]

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New post 22 Jan 2015, 08:24
Hello,

There are a few things I don't find clear in the theory about absolute value, and particularly the 3 step method.

1) How do we decide which are the points we are interested in?
In this case for example, |x+3| - |4-x|=|8+x|, why are we interested in -8, -3, 4 and not -4? Is it because if we replace -x with 4 the expression becomes zero? So, in this case we are not saying that we need -4, becuase the minus sign is already there in -x?
2) Why are these values ordered like -8, -3, 4? Are they going from the least to the highest?
3) How are we choosing our conditions? So, why are the conditions these: x<-8, -8<x<=-3, -3<=x<=4 and x>=4? Does this have to do with the order of the values, as mentioned in 3 above? Does a negative sign go with a x< and a positive with a x>=?
4) How do we decide how to change the sign while we compute the inequality? E.g, when x<-8, why is it -(x+3) - (4-x) = - (8+x) or when -8<x<=-3 it is -(x+3) - (4-x) = (8+x)?

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

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New post 29 Jan 2015, 10:32
pacifist85 wrote:
1) How do we decide which are the points we are interested in?
In this case for example, |x+3| - |4-x|=|8+x|, why are we interested in -8, -3, 4 and not -4? Is it because if we replace -x with 4 the expression becomes zero? So, in this case we are not saying that we need -4, becuase the minus sign is already there in -x?
2) Why are these values ordered like -8, -3, 4? Are they going from the least to the highest?
3) How are we choosing our conditions? So, why are the conditions these: x<-8, -8<x<=-3, -3<=x<=4 and x>=4? Does this have to do with the order of the values, as mentioned in 3 above? Does a negative sign go with a x< and a positive with a x>=?
4) How do we decide how to change the sign while we compute the inequality? E.g, when x<-8, why is it -(x+3) - (4-x) = - (8+x) or when -8<x<=-3 it is -(x+3) - (4-x) = (8+x)?


1) When the expression under modulus equals 0. The expression may or may not change its sign in those special points.
2) Yes, because the number line goes from -infinity all the way to +infinity. -4 is not a special point (see #1; 4 - (-4) = 8 ≠ 0)
3) All conditions cover the number line and break at the special points.
4) Using the definition of modulus: for y = |x|, y = x if x >=0 and y = -x if x < 0.
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Re: Math: Absolute value (Modulus)  [#permalink]

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New post 12 Jun 2015, 02:29
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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