Question

|x+12|<12 find the solution set of the inequality. use interval notation to write your answer.

Answer 1

if x+12 = -12 or 12 ; then we got something to compare to

Answer 2

to me that suggest that when x = 0 and when x = -24; we get an absolute value of 12

Answer 3

so id say it between -24 and 0

Answer 4

-12<x+12<12
-24<x<0

Answer 5

Using the definition of absolute value, the inequality becomes:\[\sqrt{(x+12)^2}<12\]
Square both sides:\[(x+12)^2 < 144\]
Expand left-hand side:\[x^2 + 24x + 144 < 144\]
Subtract 144 from both sides:\[x^2+24x < 0\]
Make this an equality first: \[x^2 + 24x = 0\]
Your critical points are: \[x=-24, 0\]

The solution set is (-24, 0). (Try plugging all the values in, they should fit the inequality)

Answer 6

They changed it so it now ignores whitespace :|

Answer 7

Now it's impossible to set things out on separate lines,

Answer 8

what do you mean by seperate lines?

Answer 9

huh, it must just look like that to me.

Answer 10

refresh, or use a different browser ...

Answer 11

i know internet explorer has lots of issues with this site

Answer 12

its designed to run on google chrome; but even that has its issues

Answer 13

Using interval notation, wouldn't the answer look something like this?
\[(-\infty,-12)U(12,\infty) and since its an inequality, we could evaluate, say, point -13 and 13 \to define where the function is defined.\]

Answer 14

no, that would be everything greater than 12

Answer 15

Surely it can't be till infinity, if you try plugging in a negative or positive big number the inequality would be violated.

Answer 16

Yes amistre64, I just did not evaluate the equation.