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For everyone interested in modulus practices, I have listed, [#permalink]
03 Dec 2006, 04:22

2

This post received KUDOS

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For everyone interested in modulus practices, I have listed, at the bottom of this topic, several threads of modulus in a graduated level of difficulty.

Keep in mind that it's for trainning purposes and that the most advanced posts are not reprensative of the real GMAT.

FOR THE THIRD PROBLEM , I BELIEVE THE ANSWER MISS THE RANGE

-2<X<1

I'm ok thanks... it follows the way

Great catch .... Yes, I did a "silly" mistake 2-4 = 6 and ended up with a wrong x^2 + x + 6 .... I have corrected my post following your catch... very good :D

o If x < 3, then the inequation becomes: -(x-3)<2x-4 <=> 7 < 3x <x> 7/3 => 7/3 < x <3>= 3, then the inequation becomes:[/b] x-3<2x-4 <x> 1 => x >= 3

Finally, we have : x > 7/3

Hi Fig,

Why have you mentioned that x >=3 and not that x > 3.. I see absolute Questions which sometimes have the >= sign and sometimes the > sign only.. Might be a very mindless question but hope you would answer..

o If x < 3, then the inequation becomes: -(x-3)<2x-4 <=> 7 < 3x <x> 7/3 => 7/3 < x <3>= 3, then the inequation becomes:[/b] x-3<2x-4 <x> 1 => x >= 3

Finally, we have : x > 7/3

Hi Fig,

Why have you mentioned that x >=3 and not that x > 3.. I see absolute Questions which sometimes have the >= sign and sometimes the > sign only.. Might be a very mindless question but hope you would answer..

Not meaningless but rather precised question ...

Actually, it's just to avoid adding 1 solution alone from another interval... and so to meet a higher degree of "perfection" by making it a smoother way to the final solution and a straightforward simplicity to understand

Note that nothing is wrong if we say > or >= as soon as we do not miss to consider all values of x, especially here the ones that limit 2 intervals

As we are working positive value and as the function f(x) = 1/x decreases when x increases on the domain of x > 0, we can imply that:

|x-2| =< 1/4

can someone please explain why this happens. Why can't I just use |1/x-2| >=4. However, when x<2, I get x<=7/4. Also I agree with everyone else as the range include 2, doesn't lie outside of our limitations.

Since the sign of a*x^2+b*x+c is that of a for any values of x not between the 2 roots, solutions of the equation a*x^2+b*x+c = 0 if they exist.

Thus, o Sign((x+2)(x-2)) > 0 when x > 2 or x < -2 o Sign((x+2)(x-2)) < 0 when -2 < x < 2

Trying hard to grasp this concept, but don't think I'm understanding what you're trying to convey here. Ideally, I would try to solve this as such:

Given that |x^2 - 4| is same as |(x+2)(x-2)|

Solving for x in |x^2 - 4| > x + 2 in the following two scenarios: 1. When (x+2)(x-2) >= 0 ==> x^2 - 4 > x + 2 ==> x>-1 and x>2

2. When (x+2)(x-2) < 0 ==> -(x^2 - 4) > x + 2 ==> 4 - x^2 > x + 2 ==> x<1 and x<-2

Does the combination of these solutions for x represent the final answer? Am I missing a step?

First of all you won't see such equation on the GMAT, but anyway:

\(|x^2-4|>x+2\).

We should consider the cases when \(x^2-4<0\) and \(x^2-4\geq{0}\):

\(x^2-4<0\) for \(-2<x<2\) --> \(-(x^2-4)>x+2\) --> \(x^2+x-2<0\) --> \((x+2)(x-1)<0\)--> \(-2<x<1\). Since we are considering the range \(-2<x<2\) then the solution for this case will be intersections of these two (common part): \(-2<x<1\);

\(x^2-4\geq{0}\) for \(x\leq{-2}\) or \(x\geq{2}\) --> \((x^2-4)>x+2\) --> \(x^2-x-6>0\) --> \((x+2)(x-3)>0\)--> \(x<-2\) or \(x>3\). Since we are considering the range \(x\leq{-2}\) or \(x\geq{2}\) then the solution for this case will be intersections of these four (common parts): \(x<-2\) or \(x>3\);

Finally the ranges of \(x\) for which \(|x^2-4|>x+2\) true are: \(x<-2\), \(-2<x<1\) and \(x>3\).

Re: For everyone interested in modulus practices, I have listed, [#permalink]
27 Aug 2014, 03:40

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