document.write( "Question 1102792: Find three consecutive even integers such that the square of the first is 80 less than the square of the third. (Only an algebraic solution will be accepted. ) \n" ); document.write( "
Algebra.Com's Answer #717531 by ikleyn(52782)\"\" \"About 
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document.write( "Let n be the central of these tree consecutive integers, so that the numbers are\r\n" );
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document.write( "n-2, n and n+2.\r\n" );
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document.write( "Then the condition says\r\n" );
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document.write( "(n+2)^2 - (n-2)^2 = 80,   or, expanding parentheses,\r\n" );
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document.write( "n^2 + 4n + 4 - (n^2 -4n +4) = 80,\r\n" );
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document.write( "which is simplified to\r\n" );
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document.write( "8n = 80.\r\n" );
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document.write( "Hence, n = \"80%2F8\" = 10.\r\n" );
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document.write( "Answer.  The numbers are 8, 10 and 12.\r\n" );
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document.write( "Check.   12^2 - 8^2 = 144 - 64 = 80.   ! Correct !\r\n" );
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