document.write( "Question 34172This question is from textbook Algebra Beginning and Intermediate
\n" ); document.write( ": I need help with the following 2 questions...\r
\n" ); document.write( "\n" ); document.write( "1. Solve by extracing square roots: 3x^2-80=0
\n" ); document.write( "I figured this was it, can you check the work?
\n" ); document.write( "3x^2 - 80 = 0
\n" ); document.write( "3x^2 = 80
\n" ); document.write( "3x^2/3 = 80/3
\n" ); document.write( "x^2 = 80/3
\n" ); document.write( "sq root of x^2 = sq root of 80/3
\n" ); document.write( "Answer - x=sq root of (= or -)80/3?? Does the (+ or -) have to be there?\r
\n" ); document.write( "\n" ); document.write( "2. Solve by extracting square roots: (2x-2)^2 = 9
\n" ); document.write( "sq root of (2x-2)^2 = sq root of 9
\n" ); document.write( "2x-2 = (+ or -)3
\n" ); document.write( "This is where I am confused... is the next step dividing it all by 2, or taking the -2 to the other side of the problem? Could you please help me get though this one?
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Algebra.Com's Answer #20466 by Cintchr(481)\"\" \"About 
You can put this solution on YOUR website!
This is a duplicate of Problem number 34173\r
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\n" ); document.write( "\n" ); document.write( "Not too bad ... you just need to keep going ...\r
\n" ); document.write( "\n" ); document.write( "Problem 1:
\n" ); document.write( "\"+3x%5E2+-+80+=+0+\"
\n" ); document.write( "\"+3x%5E2+=+80+\"
\n" ); document.write( "\"+3x%5E2%2F3+=+80%2F3+\"
\n" ); document.write( "\"+x%5E2+=+80%2F3+\"
\n" ); document.write( "now when you square root the fraction .. you will get a denominator with a square root in it ... That is a BIG NO-NO!!!
\n" ); document.write( "\"+sqrt%28x%5E2%29+=+sqrt%2880%2F3%29+\"
\n" ); document.write( "\"+x+=+sqrt%2880%29%2Fsqrt%283%29+\"
\n" ); document.write( "To rationalize the denominator we need to multiply the numerator and denominator by root3
\n" ); document.write( "\"+x+=+%28sqrt%2880%29%29%28sqrt%283%29%29%2F%28sqrt%283%29%29%28sqrt%283%29%29+\"
\n" ); document.write( "\"+x+=+sqrt%28240%29%2F3+\"
\n" ); document.write( "\"+x+=+4%28sqrt%2815%29%29%2F3+\"
\n" ); document.write( "And yes ... it is +-
\n" ); document.write( "\"+x+=+4%28sqrt%2815%29%29%2F3+\"
\n" ); document.write( "Done.\r
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\n" ); document.write( "\n" ); document.write( "Problem 2:
\n" ); document.write( "\"+%282x-2%29%5E2+=+9+\"
\n" ); document.write( "yes ... root both sides
\n" ); document.write( "here you will have two roots... one positive ... one negative
\n" ); document.write( "\"+2x-2+=+3+\" or \"+2x-2+=+-3+\"
\n" ); document.write( "ADD 2
\n" ); document.write( "\"+2x+=+5+\" or \"+2x+=+-1+\"
\n" ); document.write( "Divide by 2
\n" ); document.write( "\"+x+=+5%2F2+\" or \"+x+=+-1%2F2+\"
\n" ); document.write( "When solving for a given variable .... we work PEMDAS backwards ... doing the adding and subtracting first ... and then the multiplying and dividing
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