document.write( "Question 34172This question is from textbook Algebra Beginning and Intermediate
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document.write( ": I need help with the following 2 questions...\r
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document.write( "1. Solve by extracing square roots: 3x^2-80=0
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document.write( "I figured this was it, can you check the work?
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document.write( "3x^2 - 80 = 0
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document.write( "3x^2 = 80
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document.write( "3x^2/3 = 80/3
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document.write( "x^2 = 80/3
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document.write( "sq root of x^2 = sq root of 80/3
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document.write( "Answer - x=sq root of (= or -)80/3?? Does the (+ or -) have to be there?\r
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document.write( "2. Solve by extracting square roots: (2x-2)^2 = 9
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document.write( "sq root of (2x-2)^2 = sq root of 9
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document.write( "2x-2 = (+ or -)3
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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)![]() ![]() ![]() You can put this solution on YOUR website! This is a duplicate of Problem number 34173\r \n" ); document.write( " \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( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \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( " \n" ); document.write( " \n" ); document.write( "To rationalize the denominator we need to multiply the numerator and denominator by root3 \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "And yes ... it is +- \n" ); document.write( " \n" ); document.write( "Done.\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Problem 2: \n" ); document.write( " \n" ); document.write( "yes ... root both sides \n" ); document.write( "here you will have two roots... one positive ... one negative \n" ); document.write( " \n" ); document.write( "ADD 2 \n" ); document.write( " \n" ); document.write( "Divide by 2 \n" ); document.write( " \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 \n" ); document.write( " \n" ); document.write( " |