document.write( "Question 266103: find three consecutive positive integers such that the product of the first and the third is 29 more than the second. \n" ); document.write( "

Algebra.Com's Answer #195552 by roseo(33) $\"\"$ $\"About$

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Let x= 1st consecutive integer
\n" ); document.write( " x+1 = 2nd consecutive integer
\n" ); document.write( " x+2 = 3rd consecutive integer
\n" ); document.write( "The product of the first and the third would be x(x+2) and 29 more that the second would be 29+x+1
\n" ); document.write( "Then your equation would be x(x+2) = 29+x+1
\n" ); document.write( " x^2 +2x=30+x
\n" ); document.write( "Put the terms on one side and factor the quadratic equation x^2+x-30=0
\n" ); document.write( "(x+6)(x-5)=0
\n" ); document.write( "x+6=0 x-5=0
\n" ); document.write( "x=-6 and x=5
\n" ); document.write( "Because your answers must be positive they would be 5,6, and 7. \n" ); document.write( "

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