document.write( "Question 759799: Three consecutive positive integers are such that the square of their sum exceeds the sum of their squares by 94. Find these three integers. \n" ); document.write( "

Algebra.Com's Answer #462230 by htmentor(1343) $\"\"$ $\"About$

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Let the 3 integers be n, n+1 and n+2
\n" ); document.write( "The square of their sum is (n+n+1+n+2)^2 and this is larger than the sum of their squares n^2 + (n+1)^2 + (n+2)^2 by 94
\n" ); document.write( "In equation form this is
\n" ); document.write( "(n+n+1+n+2)^2 = n^2 + (n+1)^2 + (n+2)^2 + 94
\n" ); document.write( "If you perform the multiplication, collect terms and simplify you should get:
\n" ); document.write( "n^2 + 2n - 15 = 0
\n" ); document.write( "Factor:
\n" ); document.write( "(n-3)(n+5) We are told the integers are positive, so the solution is n=3
\n" ); document.write( "Ans: 3,4,5
\n" ); document.write( "Check:
\n" ); document.write( "12^2 = 9 + 16 + 25 + 94 -> 144 = 50 + 94 = 144 \n" ); document.write( "

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