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Re: Math: Absolute value (Modulus)
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26 May 2011, 14:34
walker wrote: Let’s consider following examples,
Example #1 Q.: \(x+3  4x = 8+x\). How many solutions does the equation have? Solution: There are 3 key points here: 8, 3, 4. So we have 4 conditions:
a) \(x < 8\). \((x+3)  (4x) = (8+x)\) > \(x = 1\). We reject the solution because our condition is not satisfied (1 is not less than 8)
b) \(8 \leq x < 3\). \((x+3)  (4x) = (8+x)\) > \(x = 15\). We reject the solution because our condition is not satisfied (15 is not within (8,3) interval.)
c) \(3 \leq x < 4\). \((x+3)  (4x) = (8+x)\) > \(x = 9\). We reject the solution because our condition is not satisfied (15 is not within (3,4) interval.)
d) \(x \geq 4\). \((x+3) + (4x) = (8+x)\) > \(x = 1\). We reject the solution because our condition is not satisfied (1 is not more than 4)
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 4x >= 0 and 4x<0 Could someone explain what I am doing wrong here? thanks in advance



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Re: Math: Absolute value (Modulus)
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27 May 2011, 06:44
VeritasPrepKarishma wrote: someonear wrote: walker wrote: Let’s consider following examples,
Example #1 Q.: \(x+3  4x = 8+x\). How many solutions does the equation have? Solution: There are 3 key points here: 8, 3, 4. So we have 4 conditions:
a) \(x < 8\). \((x+3)  (4x) = (8+x)\) > \(x = 1\). We reject the solution because our condition is not satisfied (1 is not less than 8)
b) \(8 \leq x < 3\). \((x+3)  (4x) = (8+x)\) > \(x = 15\). We reject the solution because our condition is not satisfied (15 is not within (8,3) interval.)
c) \(3 \leq x < 4\). \((x+3)  (4x) = (8+x)\) > \(x = 9\). We reject the solution because our condition is not satisfied (15 is not within (3,4) interval.)
d) \(x \geq 4\). \((x+3) + (4x) = (8+x)\) > \(x = 1\). We reject the solution because our condition is not satisfied (1 is not more than 4)
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 4x >= 0 and 4x<0 Could someone explain what I am doing wrong here? thanks in advance You don't need to solve anything to get these four ranges. You see that the points where the signs will vary are 8, 4 and 3. To cover all the numbers on the number line, the ranges are: x is less than 8, then between 8 and 3, then between 3 and 4 and then greater than 4. It doesn't matter where you put the '='. In each range the sign of the terms will be different. After assigning the proper signs, you will get a value for x and you have to check if the value lies in the range you were considering. If it does, it is a solution, else it is not. 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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Re: Math: Absolute value (Modulus)
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27 May 2011, 07:15
someonear wrote: 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
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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Re: Math: Absolute value (Modulus)
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21 Jun 2011, 00:01
Hi Walker,
Can you please explain this?
Example #1 Q.: x+3  4x = 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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Re: Math: Absolute value (Modulus)
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28 Oct 2012, 07:23
walker wrote: when x<8, (4x) is always positive and we don't need to modify the sign when we open it. 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 (4x) 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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Re: Math: Absolute value (Modulus)
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13 May 2013, 12:43
gettinit wrote: 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.: \(x+3  4x = 8+x\). How many solutions does the equation have? Solution: There are 3 key points here: 8, 3, 4. So we have 4 conditions:
a) \(x < 8\). \((x+3)  (4x) how did we get (x+3) here?= (8+x)\) > \(x = 1\). We reject the solution because our condition is not satisfied (1 is not less than 8)
b) \(8 \leq x < 3\). \((x+3)  (4x) = (8+x)\) > \(x = 15\). We reject the solution because our condition is not satisfied (15 is not within (8,3) interval.)
c) \(3 \leq x < 4\). \((x+3)  (4x) = (8+x)\) > \(x = 9\). We reject the solution because our condition is not satisfied (15 is not within (3,4) interval.)
d) \(x \geq 4\). \((x+3) + (4x) = (8+x)\) > \(x = 1\). 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.: \(x+3  4x = 8+x\). 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, 44=0 for second modulus sign, so key point here is 4 88 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) \(x < 8\). \((x+3)  (4x) = (8+x)\) > \(x = 1\). 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 wrote: 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 \( (4x)\)? as in this case X will turn positive after opening the bracket 2nd EQ \(8 \leq x < 3\) \((x+3)  (4x)\) = \((8+x)\) 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? Ofcourse 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) \(3 \leq x < 4\). \((x+3)  (4x) = (8+x)\) > \(x = 9\). 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) \(x \geq 4\). \((x+3) + (4x) = (8+x)\) > \(x = 1\). 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 \(+(4X)\), I think it should be \((4X)\) 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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Re: Math: Absolute value (Modulus)
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14 May 2013, 09:51
nikhil007 wrote: gettinit wrote: 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.: \(x+3  4x = 8+x\). How many solutions does the equation have? Solution: There are 3 key points here: 8, 3, 4. So we have 4 conditions:
a) \(x < 8\). \((x+3)  (4x) how did we get (x+3) here?= (8+x)\) > \(x = 1\). We reject the solution because our condition is not satisfied (1 is not less than 8)
b) \(8 \leq x < 3\). \((x+3)  (4x) = (8+x)\) > \(x = 15\). We reject the solution because our condition is not satisfied (15 is not within (8,3) interval.)
c) \(3 \leq x < 4\). \((x+3)  (4x) = (8+x)\) > \(x = 9\). We reject the solution because our condition is not satisfied (15 is not within (3,4) interval.)
d) \(x \geq 4\). \((x+3) + (4x) = (8+x)\) > \(x = 1\). 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.: \(x+3  4x = 8+x\). 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, 44=0 for second modulus sign, so key point here is 4 88 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) \(x < 8\). \((x+3)  (4x) = (8+x)\) > \(x = 1\). 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 wrote: 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 \( (4x)\)? as in this case X will turn positive after opening the bracket 2nd EQ \(8 \leq x < 3\) \((x+3)  (4x)\) = \((8+x)\) 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? Ofcourse 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) \(3 \leq x < 4\). \((x+3)  (4x) = (8+x)\) > \(x = 9\). 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) \(x \geq 4\). \((x+3) + (4x) = (8+x)\) > \(x = 1\). 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 \(+(4X)\), I think it should be \((4X)\) 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. 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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Re: Math: Absolute value (Modulus)
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14 May 2013, 15:42
VeritasPrepKarishma wrote: 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.
Hi Karishma I went through your post on the blog, but to be frank found this post of your more helpfull 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 (x2) < 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 (XK) 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) \(x < 8\). \((x+3)  (4x) = (8+x)\) > \(x = 1\). 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 (4x), because we will again get 4x 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 (xk) format? in RHS we have (8x) because again we want 8x to turn out a negative number so we put (8x) to make it always negative, let me know if I got it correctly. Quote: b) \(8 \leq x < 3\). \((x+3)  (4x) = (8+x)\) > \(x = 15\). 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 (8X) as we need to make sure that the result of this bracket is tive so 8x = (8x) Quote: c) \(3 \leq x < 4\). \((x+3)  (4x) = (8+x)\) > \(x = 9\). 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) \(x \geq 4\). \((x+3) + (4x) = (8+x)\) > \(x = 1\). 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? (4x) 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 http://www.veritasprep.com/blog/2012/07 ... nsparti/(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 = (xk) format? 2(x–0)(x–5/2)(x–6) < 0 (multiply both sides by 1)> how did you arrive at 2(x0)? i think it should be just \(2x(x\frac{5}{2})(x6) <0\)
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Re: Math: Absolute value (Modulus)
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15 May 2013, 06:30
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 x2>5 what I learnt is simply take 2 conditions, x2>5 and 2x>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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Re: Math: Absolute value (Modulus)
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04 Jun 2013, 12:35
(This question is from a GMAT club study book. It can be found here: mathabsolutevaluemodulus86462.html) x^24 = 1. What is x? Solution: There are 2 conditions: a) (x^24)\geq0 > x \leq 2 or x\geq2. x^24=1 > x^2 = 5. x e {\sqrt{5}, \sqrt{5}} and both solutions satisfy the condition. b) (x^24)<0 > 2 < x < 2. (x^24) = 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^24 = 1 x^24 = 1 ==> x^2 = 5 ==> x = \sqrt{5} OR (x^24) = 1 ==> x^2 +4 = 1 ==> x^2 = 3 ==> X^2 = 3 ==> x = \sqrt{3} Thanks!



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Re: Math: Absolute value (Modulus)
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04 Jun 2013, 15:39
Hmmmm...I'm not sure I follow. This is how I solved the problem. x^24=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 wrote: 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 \(x<5\) \(x5=4\), \(x=1\) 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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Re: Math: Absolute value (Modulus)
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04 Jun 2013, 16:57
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 wrote: WholeLottaLove wrote: Hmmmm...I'm not sure I follow.
This is how I solved the problem.
x^24=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: \(x^24=1\) 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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Re: Math: Absolute value (Modulus)
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04 Jun 2013, 17:37
So, I need to find the positive and negative values of x^24=1 (i.e. x^24=1 and x^2+4=1) and which x values make x^24 positive and x^2+4 negative? Thanks for putting up with my slowness in picking up these concepts! Zarrolou wrote: WholeLottaLove wrote: 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^24=1 1) Define where it is positive and where is negative => \(x^24>0\) 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: \(x^24=1\) \(x=+\sqrt{5}\), 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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Re: Math: Absolute value (Modulus)
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11 Jul 2013, 00:06
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Re: Math: Absolute value (Modulus)
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06 Aug 2013, 09:07
hi,
i have a doubt in the example
Q.: x^24 = 1. What is x? Solution: There are 2 conditions:
a) (x^24)\geq0
how the range for this example is (X^24)>=0 and not (x^24)>0?
regards, RRSNATHAN



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Re: Math: Absolute value (Modulus)
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11 Aug 2013, 21:30
rrsnathan wrote: hi,
i have a doubt in the example
Q.: x^24 = 1. What is x? Solution: There are 2 conditions:
a) (x^24)\geq0
how the range for this example is (X^24)>=0 and not (x^24)>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)
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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. x1<9 D. 5+x<4 E. 3+x<5
can anyone please explain how can I determine in 1020 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)
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14 Apr 2014, 21:18
Thank you Karishma. That was helpful



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Re: Math: Absolute value (Modulus)
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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.
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Re: Math: Absolute value (Modulus)
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14 Jun 2014, 19:05
Can somebody help me solve the inequality:x+5 > 3x  5
This is what I did..... \(\frac{x+5}{x  5}>3\)
==>\(\frac{x+5}{x  5}>3\) & \(\frac{x+5}{x  5}<3\) ==>\(x+5>3x15\) & \(x+5<3x+15\) ==>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??




Re: Math: Absolute value (Modulus)
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14 Jun 2014, 19:05



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