document.write( "Question 68462: I'm having trouble solving the following equation:
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document.write( "square root of (3x-2) + square root of (2x+2) + 1 = 0. My major problem is coming from the fact that I'm getting stuck on isoloating the radical, and I'm a little confused on where to place 1. Please help. \n" );
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Algebra.Com's Answer #48700 by ptaylor(2198)![]() ![]() You can put this solution on YOUR website! \n" ); document.write( "\n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( "\n" ); document.write( "This quadratic can be factored:\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( "\n" ); document.write( "and\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( "\n" ); document.write( "ck\r \n" ); document.write( "\n" ); document.write( "substitute 1 in the original equation:\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( "(+or-1)+(+or-2)+1=0 \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "verify that it also checks for x=17\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Hope this helps---ptaylor\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |