document.write( "Question 230700: find all x in R for which :
\n" ); document.write( "1) absolute value of (x-1)<1/2
\n" ); document.write( "2) absolute value of (x^2-1)<1/2 \n" ); document.write( "
Algebra.Com's Answer #170863 by jsmallt9(3758)\"\" \"About 
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.

\n" ); document.write( "Your first inequality says that
\n" ); document.write( "\"abs%28x-1%29+%3C+1%2F2\"
\n" ); document.write( "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\":
\n" ); document.write( "\"x-1+%3E+-1%2F2\" and \"x-1+%3C+1%2F2\"
\n" ); document.write( "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:
\n" ); document.write( "\"x+%3E+1%2F2\" and \"x+%3C+3%2F2\"
\n" ); document.write( "This describes the solution set: \"All numbers between 1/2 and 3/2 (not including 1/2 and 3/2)\"

\n" ); document.write( "\"abs%28x%5E2-1%29%3C1%2F2\"
\n" ); document.write( "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:
\n" ); document.write( "\"x%5E2-1+%3E+-1%2F2\" and \"x%5E2-1+%3C+1%2F2\"
\n" ); document.write( "Since there is no x term (just \"x%5E2\" terms, we'll isolate the squared terms\"
\n" ); document.write( "\"x%5E2+%3E+1%2F2\" and \"x%5E2+%3C+3%2F2\"
\n" ); document.write( "Now we'll find the square root of each side:
\n" ); document.write( "\"x+%3E+sqrt%281%2F2%29\" and \"x+%3C+sqrt%283%2F2%29\"
\n" ); document.write( "Rationalizing the denominators we get:
\n" ); document.write( "\"x+%3E+sqrt%282%29%2F2\" and \"x+%3C+sqrt%286%29%2F2\"
\n" ); document.write( "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\"). \n" ); document.write( "
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