document.write( "Question 1170327: A rectangle is 2 inches longer than it is wide. Numerically, its area exceeds its perimeter by 31. Find the perimeter.\r
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Algebra.Com's Answer #795208 by ikleyn(52798) $\"\"$ $\"About$

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document.write( "Let x be the width, in inches;\r\n" );
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document.write( "then the length is (x+2) inches.\r\n" );
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document.write( "The area is  x*(x+2) square inches.\r\n" );
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document.write( "The perimeter is  2x + 2*(x+2) = 4x + 4 inches.\r\n" );
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document.write( "The equation is  \r\n" );
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document.write( "    x*(x+2) - (4x+4) = 31,   or\r\n" );
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document.write( "    x^2 +2x - 4x + 2 - 4 = 31\r\n" );
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document.write( "    x^2 - 2x - 35 = 0\r\n" );
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document.write( "    (x-7)*(x+5) = 0.\r\n" );
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document.write( "Of the two roots  x= 7, x= -5, only value of 7 is positive and is the solution to the problem.\r\n" );
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document.write( "The dimensions of the rectangle are  7 and 9 inches.      ANSWER\r\n" );
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document.write( "CHECK.  The area is 7*9 = 63 sq.inches.  The perimeter is  2*(7+9) = 32 inches.  The difference is  63 - 32 = 31.   ! Correct !\r\n" );
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