For everyone interested in modulus practices, I have listed,

B. |x-3|<2x-4

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

o If x >= 3, then the inequation becomes:
x-3<2x-4
<=> x > 1
=> x >= 3

Finally, we have : x > 7/3

C. x + 2 < |x2 - 4|

|x2 - 4| = |(x+2)(x-2)|

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

If -2 < x < 2, the inequation becomes :
x + 2 < -(x^2 - 4)
<=> x^2 + x - 2 < 0 (B)

Delta = 1 + 4*2 = 9.

The roots are:
o Root1 = (-1 - 3)/2 = -2
o Root2 = (-1 + 3)/2 = 1

So, (B) is true when x is between the roots of x^2 + x - 2. That implies -2 < x < 1.

If x < -2 or x > 2, the inequation becomes :
x + 2 < x^2 - 4
<=> x^2 - x - 6 > 0 (A)

Delta = 1 + 4*6 = 25.

The roots are:
o Root1 = (1 - 5)/2 = -2
o Root2 = (1 + 5)/2 = 3

(A) : x^2 - x - 6 > 0
<=> (x+2)(x-3) > 0

One more time, a is positive here. Thus (x+2)(x-3) > 0 outside of the roots.
In other words, x > 3 or x < -2


Finally, the domain of solutions of x is : x > 3 or x < -2 or -2 < x < 1.

FIG I HAVE A QUESTION HERE ABOUT THE ANSWER

THE RANGE , 7/4 <= x <= 9/4 INCLUDES 2 , AND THE FUNCTION IS NOT DEFINED AT X=2 AM I RIGHT???

Yes .... U are right... My conclusion should clarify it .... We have to remove the x=2

Hello Fig , hope u r fine and in best health ,

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

Thanks man , this is no catch this the fruits of your help and coaching

thanks a lot

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

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.

Hi buddy

could you please explain this,
<=> x > 7/3
=> 7/3 < x < 3
<=> x > 1
=> x >= 3

i solved and got x>1 & x>7/3

in case If -2 < x < 2, shouldn't it also have one more solution to problem.i.e. x<-2 and x>1

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 \(x>3\).

Solving inequalities:
x2-4x-94661.html#p731476
inequalities-trick-91482.html
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html?hilit=extreme#p873535
everything-is-less-than-zero-108884.html?hilit=extreme#p868863

Hope it helps.

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