document.write( "Question 1102703: The product of two consecutive odd integers is 31 more than twice there sum find the integers \n" ); document.write( "

Algebra.Com's Answer #717414 by ikleyn(52778) $\"\"$ $\"About$

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document.write( "As always in such problems, let me start with the \r\n" );
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document.write( "Answer.  The numbers are (-5,-3)  or  (7,9). \r\n" );
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document.write( "Check.   (-5)*(-3) = 15;  2*((-5)+ (-3) = 2*(-8) = - 16,  and  15 = -16 + 31.  ! Correct !\r\n" );
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document.write( "         7*9 = 63;  2*(7+9) = 2*16 = 32,  and  63 = 32 + 31.                   ! Correct !\r\n" );
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document.write( "Let n be the integer number exactly half-way between two consecutive odd integers.\r\n" );
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document.write( "Then the conditions says:\r\n" );
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document.write( "n^2 -1 = 4n + 31  ====>  n^2 - 4n - 32 = 0  ====>  (n+4)*(n-8) = 0  ====>  there are TWO solutions for n:  n= -4  and  n= 8,\r\n" );
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document.write( "what give you the answer.\r\n" );
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