SOLUTION: find all x in R for which : 1) absolute value of (x-1)<1/2 2) absolute value of (x^2-1)<1/2

Algebra ->  Absolute-value -> SOLUTION: find all x in R for which : 1) absolute value of (x-1)<1/2 2) absolute value of (x^2-1)<1/2      Log On


   



Question 230700: find all x in R for which :
1) absolute value of (x-1)<1/2
2) absolute value of (x^2-1)<1/2

Found 2 solutions by stanbon, jsmallt9:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
find all x in R for which :
1) absolute value of (x-1)<1/2
-1/2 < x-1 < 1/2
-------------------------
Add 1 along the line to get:
1/2 < x < 3/2
--------------------------
2) absolute value of (x^2-1)<1/2
-1/2 < x^2-1 < 1/2
---
Add 1 along the line to get:
1/2 < x^2 < 3/2
Take the square root along the line:
[sqrt(2)/2 < x < sqrt(6)/2
===============================
Cheers,
Stan H.

Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
First let's review the idea of absolute value. The absolute value of a number is its distance from zero on the number line, without regard to direction.

Your first inequality says that
abs%28x-1%29+%3C+1%2F2
This says "the distance of x-1 from zero is less than 1/2". Now picture a number line and picture the numbers that would be within 1/2 of 0. Now, how do we describe these numbers in the form of inequalities. I hope the following makes sense. It is saying "x-1" is between -1/2 and 1/2":
x-1+%3E+-1%2F2 and x-1+%3C+1%2F2
Notice that the absolute values are gone. This is the key step in solving absolute value problems: learning how to remove the absolute values by writing equivalent inequalities (or equations). Now we just solve these inequalities by adding 1 to each side of each inequality:
x+%3E+1%2F2 and x+%3C+3%2F2
This describes the solution set: "All numbers between 1/2 and 3/2 (not including 1/2 and 3/2)"

abs%28x%5E2-1%29%3C1%2F2
This one says that the distance of x%5E2-1 is less than 1/2. Using the same logic as above to rewrite it without absolute values we get:
x%5E2-1+%3E+-1%2F2 and x%5E2-1+%3C+1%2F2
Since there is no x term (just x%5E2 terms, we'll isolate the squared terms"
x%5E2+%3E+1%2F2 and x%5E2+%3C+3%2F2
Now we'll find the square root of each side:
x+%3E+sqrt%281%2F2%29 and x+%3C+sqrt%283%2F2%29
Rationalizing the denominators we get:
x+%3E+sqrt%282%29%2F2 and x+%3C+sqrt%286%29%2F2
So our solution to this problem is "all numbers between sqrt%282%29%2F2 and sqrt%286%29%2F2 (exclusive of sqrt%282%29%2F2 and sqrt%286%29%2F2).