inequality having tough problem solving..

[a_sin]

if |x|>|x+1| what is x?

1) 1/2

2) 1

3)undefined

4)-1/2

5)-1

Here is another one

if x

1)x

2)undefined

3)-x

4)1

5)0

[hitzs]

in Both questions X should be undefined

 

1 : |x|>|x+1| => x > -1/2 and x

 

2. (x*|x|)^1/2 given x

 

if x

so equation will be, (x * -x)^1/2 => sqrt of a negetive num is not a real number so undefined

[dilipcrangan]

for question 1, 5 is the most reasonable choice since x

For question 2, the value is undefined as x*|x|

 

Let me know if OA thinks different

[GMAT-HELP]

in Both questions X should be undefined

 

1 : |x|>|x+1| => x > -1/2 and x

 

 

Ans : x

x > -1/2 is incorrect (take x = -1/4 and substitute in the eqn)

[a_sin]

1) Undefined..

|x|>|x+1| => square them => x^2 > (x+1)^2 => x^2 > x^2+2x+1

=>-2x > 1 or 2x x

 

2) I got undefined but, Somehow GMATPrep says answer is -x

 

for x

lets say x = -2

 

y= sqrt(-2*|-2|) => sqrt(-4) => undefined..

 

I did mistake..

[hitzs]

GMAT-HELP, I did mistake while solving inequality...now i cross checked.

 

wat i did was I stated one of the possibility as |x| as positive but at the same time |x+1| as negative , which can not be possible. But for this need to actually look into the variables present into the equation.

I many times commit this mistake...but don't know how to arrive the actual possible solutions of a variable. So can you plz explain me how should I go to solve a inequality equation....i have phobia for inequality equation :)

 

At present wat I do is just put +- on both sides of equation and solve for all 4 possibilites like

 

|x| >|x+1|

+-(x) > +-(x+1)

take all 4 options one by one

+x > +(x+1) => x undefined

+x > -(x+1) => x>-1/2 (THIS IS WHERE I COMMITED MISTAKE, i didn't noticed that x+1 can't be negetive but there is very bis possibility that i'll miss these in real exams too so i just want to go through a hard core mathematical approach)

-x > (x+1) => x

-x > -(x+1) => undefined

 

plz suggest a way to avoid these silly mistakes..thanx a lot

[GMAT-HELP]

GMAT-HELP, I did mistake while solving inequality...now i cross checked.

 

wat i did was I stated one of the possibility as |x| as positive but at the same time |x+1| as negative , which can not be possible. But for this need to actually look into the variables present into the equation.

I many times commit this mistake...but don't know how to arrive the actual possible solutions of a variable. So can you plz explain me how should I go to solve a inequality equation....i have phobia for inequality equation :)

 

At present wat I do is just put +- on both sides of equation and solve for all 4 possibilites like

 

|x| >|x+1|

+-(x) > +-(x+1)

take all 4 options one by one

+x > +(x+1) => x undefined

+x > -(x+1) => x>-1/2 (THIS IS WHERE I COMMITED MISTAKE, i didn't noticed that x+1 can't be negetive but there is very bis possibility that i'll miss these in real exams too so i just want to go through a hard core mathematical approach)

-x > (x+1) => x

-x > -(x+1) => undefined

 

plz suggest a way to avoid these silly mistakes..thanx a lot

 

I think your approach is valid but may be time consuming.. When there is a modulus on both sides as in this case (where u have to compare +/- of both sides) just square both sides..

In this case,

x^2 > (x+1)^2

=> x^2 > x^2 + 1 + 2x

=> 0 > 1+2x

=> -1/2 > x or x

This is valid only when there is |mod| on both sides.

[Macintosh]

how is x

[hitzs]

GMAT-HELP,

 

thanx for inouts!!

 

I think its a nice approach...but how did u got when 0>1+2x => -1/2 > X or x

 

Also wat do u say when one side have |x| and other have a variable or constant...i think in that case we need to take that lengthy approach

 

in DS these equations trouble me lot....i think it'll certainly gonna help me

[GMAT-HELP]

GMAT-HELP,

 

thanx for inouts!

 

I think its a nice approach...but how did u got when 0>1+2x => -1/2 > X or x

 

Also wat do u say when one side have |x| and other have a variable or constant...i think in that case we need to take that lengthy approach

 

in DS these equations trouble me lot....i think it'll certainly gonna help me

They both mean the same thing: (0 > 1+2x) = (1+2x

=> -1 > 2x => -1/2 > x = (x

[sdasar]

how is x

 

X can be negative inifinity. Still the equation is satisfied. Hence it is undefined.

[shud]

I think your approach is valid but may be time consuming.. When there is a modulus on both sides as in this case (where u have to compare +/- of both sides) just square both sides..

In this case,

x^2 > (x+1)^2

=> x^2 > x^2 + 1 + 2x

=> 0 > 1+2x

=> -1/2 > x or x

This is valid only when there is |mod| on both sides.

 

Amazing Rule. I think, this is very true for all the cases.