Thread: inequality

Solve the inequality :



This is what i did :

squarring both sides .





then

???

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

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.

Use the inequality:



Then

Therefore the inequality holds for every real x.

Originally Posted by Krahl

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.

Thanks but how did you get the 4 equations ? Can u explain that to me ?

Originally Posted by Krahl

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.

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.

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.)

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