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.
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.)