document.write( "Question 663691: Find three consecutive integers such that three times the sum of all three equals the product of the larger two. \n" ); document.write( "

Algebra.Com's Answer #412885 by kevwill(135) $\"\"$ $\"About$

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Let x be the first integer. Then the next two consecutive integers are x+1 and x+2.
\n" ); document.write( "Three times the sum of all three is 3*(x + (x+1) + (x+2))
\n" ); document.write( "The product of the larger two is (x+1) * (x+2)
\n" ); document.write( "So we have
\n" ); document.write( "(x+1) * (x+2) = 3*(x + (x+1) + (x+2))
\n" ); document.write( "x^2 + 3x + 2 = 3*(3x +3)
\n" ); document.write( "x^2 + 3x + 2 = 9x + 9
\n" ); document.write( "x^2 + 3x + 2 - 9x - 9 = 9x + 9 - 9x - 9
\n" ); document.write( "x^2 - 6x - 7 = 0
\n" ); document.write( "We can solve for x using the quadratic equation $\"x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+\"$ with a=1, b=-6 and c=-7
\n" ); document.write( " $\"x+=+%28-%28-6%29+%2B-+sqrt%28+%28-6%29%5E2-4%2A1%2A%28-7%29+%29%29%2F%282%2A1%29+\"$
\n" ); document.write( " $\"x+=+%286+%2B-+sqrt%28+36%2B28+%29%29%2F2+\"$
\n" ); document.write( " $\"x+=+%286+%2B-+sqrt%28+64+%29%29%2F2+\"$
\n" ); document.write( " $\"x+=+%286+%2B-+8%29%2F2+\"$
\n" ); document.write( " $\"x+=+14%2F2+=+7\"$ or $\"x+=+-2%2F2+=+-1\"$
\n" ); document.write( "So there are two solutions to this problem:
\n" ); document.write( "7, 8, 9 and -1, 0, 1\r
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