document.write( "Question 1204189: The product of the first and third of three negative consecutive integers is 59 more than 4 times the second integer. Find the integers.
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Algebra.Com's Answer #840275 by mananth(16946) $\"\"$ $\"About$

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The product of the first and third of three negative consecutive integers is 59 more than 4 times the second integer. Find the integers.\r
\n" ); document.write( "\n" ); document.write( "Let the integers be x-1 , x , x+1\r
\n" ); document.write( "\n" ); document.write( "(x-1)(x+1)=59+4x\r
\n" ); document.write( "\n" ); document.write( "x^2-1-4x =59\r
\n" ); document.write( "\n" ); document.write( "x^2-4x-60=0\r
\n" ); document.write( "\n" ); document.write( "x^2-10x +6x-60=0\r
\n" ); document.write( "\n" ); document.write( "x(x-10)+6(x-10)=0\r
\n" ); document.write( "\n" ); document.write( "(x-10)(x+6)=0\r
\n" ); document.write( "\n" ); document.write( "x=-6 or 10\r
\n" ); document.write( "\n" ); document.write( "the numbers are negative\r
\n" ); document.write( "\n" ); document.write( "so x=-6\r
\n" ); document.write( "\n" ); document.write( "the integers are -7,-6,-5 as per given condition\r
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\n" ); document.write( "(x-1)(x+1)=59+4x\r
\n" ); document.write( "\n" ); document.write( "(-7)(-5) = 59 -4*(-6)\r
\n" ); document.write( "\n" ); document.write( "35 = 35\r
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