Solve the inequality :

This is what i did :

squarring both sides .

then

???

What does that mean ? so wat 's the range of x ?

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- August 27th 2009, 04:36 AMthereddevilsinequality
Solve the inequality :

This is what i did :

squarring both sides .

then

???

What does that mean ? so wat 's the range of x ? - August 27th 2009, 06:03 AMKrahl
your solution means that the equation is valid for all x.

notice the x's cancel out no matter what x is. plugging x=200 in to it you will see that the 200's will cancel out giving you -2. so whatever x is you will get or because the first absolute value will always be 2 units away from the second absolute value depending on the sign of x.

The way i would solve these is to solve the equations;

1.

2.

3.

4.

and take the union of the solutions. because if x solves any one of these then it solves the original inequality. - August 27th 2009, 08:05 AMred_dog
Use the inequality:

Then

Therefore the inequality holds for every real x. - August 27th 2009, 08:07 AMthereddevils
- August 27th 2009, 10:53 AMnikhilcorrection
you have not included the interval of x for which you opened the mod and you said to take the union of solution obtained from the inequations. this can be misleading sometimes for example

|x-1|-|x-3|>=4

let 1<=x<3 so we have

x-1+(x-3)>=4 or

2x-4>=4 or

x>=4 which contradicts with our interval

you can check this by putting x=4.you will get LHS=2 which is not >=4 so such solution should be discarded.

but if you will not write the interval of x for which you opened mod and took union of the intervals obtained then you might get wrong result. - August 27th 2009, 11:02 AMnikhil
hi thereddevils!!!

|x|=x if x>=0

=-x if x<0

|x-3|-|x-1|>=2

consider two points 1 and 3

take case i.e

1) x>=3

2) 1<=x<3

3)x<1

you will get 3 inequations. solve them and take the union of solution.

discard the solution if its interval does not match your interval i.e you took x>1 but your corresponding inequation gave x<3.(though you will not have this situation in this problem.) - August 27th 2009, 11:07 AMnikhilTIP
whenever equation is like |x-a|+|x-b|+|x-c|....and so on >=n

then draw points a,b,c.... on number line.

now start taking interval i.e x<a then a<=x<b then b<=x<c and so on

this will help to solve inequality in an organised way