- #1

sylent33

- 39

- 5

- Homework Statement
- Find all the real solutions

- Relevant Equations
- Equation solving

Im a having trouble understanding how this exactly works.

$$ |x^2 - 4| < |x^2+2| $$ So I know the usual thing to do when you have absolute values,here it is even simpler since the right part of the inequality is always positive so I just have these 2 cases.

1. ## x^2-4 >= 0 ## and 2. ## x^2-4 < 0 ##

Out of that 1. ##x >=\pm 2 ## and 2. ##x<\pm 2 ##

For the first case the term ##x^2 -4 ## stays positive and for the solution I get 0 < 6 which is a true statement for all real number so the solution should be R. Now the part that I dont get. We have this boundary set by the fact that ##x^2 -4## is positve and it says that x has to be great or equal than -2 and +2. So it should be all the real numbers from 2 so ##L_1 = [2,+\inf] ## For the second part case

$$ -x^2+4 < x^2+2 $$ I get ## x>\pm 1 ## and using the bondary I have for case 2 that it has to be less than -2 and +2 I get ##L_2 = (-2,-1) \cup (1,2) ##

Now for the final solution I put ## L = (-2,-1) \cup [2, \infty) ## That is not correct,the solution sheet says it should be $$ L =(-\infty,1) \cup (1, \infty) $$

I dont understand how they got here so I checked what they got for L1 and L2. L2 matches but for L1 they have ##L_1 = R \backslash {(2,-2)} ##

The way I am interpreting this is,all real numbers except 2 and -2. I do not understand how this can be their solution. It is great EQUAL shouldnt 2 and -2 be included,Also we are looking for solutions GREATER than -2 and 2 so I though since 2 is greater than -2 I only need to go from 2 to infinity. Why do I have to include minus infinity?

$$ |x^2 - 4| < |x^2+2| $$ So I know the usual thing to do when you have absolute values,here it is even simpler since the right part of the inequality is always positive so I just have these 2 cases.

1. ## x^2-4 >= 0 ## and 2. ## x^2-4 < 0 ##

Out of that 1. ##x >=\pm 2 ## and 2. ##x<\pm 2 ##

For the first case the term ##x^2 -4 ## stays positive and for the solution I get 0 < 6 which is a true statement for all real number so the solution should be R. Now the part that I dont get. We have this boundary set by the fact that ##x^2 -4## is positve and it says that x has to be great or equal than -2 and +2. So it should be all the real numbers from 2 so ##L_1 = [2,+\inf] ## For the second part case

$$ -x^2+4 < x^2+2 $$ I get ## x>\pm 1 ## and using the bondary I have for case 2 that it has to be less than -2 and +2 I get ##L_2 = (-2,-1) \cup (1,2) ##

Now for the final solution I put ## L = (-2,-1) \cup [2, \infty) ## That is not correct,the solution sheet says it should be $$ L =(-\infty,1) \cup (1, \infty) $$

I dont understand how they got here so I checked what they got for L1 and L2. L2 matches but for L1 they have ##L_1 = R \backslash {(2,-2)} ##

The way I am interpreting this is,all real numbers except 2 and -2. I do not understand how this can be their solution. It is great EQUAL shouldnt 2 and -2 be included,Also we are looking for solutions GREATER than -2 and 2 so I though since 2 is greater than -2 I only need to go from 2 to infinity. Why do I have to include minus infinity?