document.write( "Question 248998: The product of two consecutive positive odd integers is one less than three times their sum. Find the integers.\r
\n" ); document.write( "\n" ); document.write( "I tried numerous ways to write this and I am just not sure how to. I have x as on integer and x+2 as the other. I tried writing it like 3x(x + 2)- 1 = 0 or x(x + 2) = 3x(x + 2)- 1 but these don't seem to work out. Help me please. \n" ); document.write( "
Algebra.Com's Answer #181416 by dabanfield(803)\"\" \"About 
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The product of two consecutive positive odd integers is one less than three times their sum. Find the integers.
\n" ); document.write( "I tried numerous ways to write this and I am just not sure how to. I have x as on integer and x+2 as the other. I tried writing it like 3x(x + 2)- 1 = 0 or x(x + 2) = 3x(x + 2)- 1 but these don't seem to work out. Help me please.\r
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\n" ); document.write( "\n" ); document.write( "You're on the right track :)\r
\n" ); document.write( "\n" ); document.write( "If x and x+2 are the two integers then\r
\n" ); document.write( "\n" ); document.write( "x*(x+2) = 3(x+x+2) - 1 or
\n" ); document.write( "x^2 +2*x = 3*(2x+2) - 1
\n" ); document.write( "x^2 + 2*x = 6*x + 6 -1
\n" ); document.write( "x^2 + 2*x = 6*x + 5\r
\n" ); document.write( "\n" ); document.write( "x^2 - 4*x - 5 = 0 =
\n" ); document.write( "(x-5)*(x+1) = 0\r
\n" ); document.write( "\n" ); document.write( "Now you can solve for positive x.
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