document.write( "Question 1144267: The length of a rectangle is 2 feet more than the width. The area of the rectangle is 2 square feet. What are the dimensions of the rectangle? \n" ); document.write( "
Algebra.Com's Answer #765338 by ikleyn(52803)\"\" \"About 
You can put this solution on YOUR website!
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\n" ); document.write( "\n" ); document.write( "            For this problem,  I will show you two ways  (two methods)  of solution.\r
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\n" ); document.write( "\n" ); document.write( "            First way is  TRADITIONAL.  You may find it everywhere,  and it is boring.\r
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document.write( "If x is the width, then the length is (x+2) and the area equation is\r\n" );
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document.write( "    x*(x+2) = 2  square feet,\r\n" );
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document.write( "    x^2 + 2x - 2 = 0.\r\n" );
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document.write( "Use the quadratic formula\r\n" );
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document.write( "    \"x%5B1%2C2%5D\" = \"%28-2+%2B-+sqrt%28%28-2%29%5E2+-+4%2A1%2A%28-2%29%29%29%2F2\" = \"%28-2+%2B-+sqrt%2812%29%29%2F2\" = \"%28-2+%2B-+2%2Asqrt%283%29%29%2F2\" = \"-1+%2B-+sqrt%283%29\".\r\n" );
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document.write( "The width should be positive, so only positive root  x = \"-1+%2B+sqrt%283%29\"  is the solution for the width.\r\n" );
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document.write( "ANSWER.  The width is  W = \"sqrt%283%29-1\".  The length is  L = W+2 = \"sqrt%283%29%2B1\".\r\n" );
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\n" ); document.write( "\n" ); document.write( "            The other method is fresh as a matutinal dawn in May,  unexpected and elegant. \r
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\n" ); document.write( "\n" ); document.write( "            You may learn it only from me at this forum and in this site -- and practically nowhere else.\r
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document.write( "Let \"x\" be an unknown value on number line exactly half-way between the length L and the width W values of the rectangle.\r\n" );
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document.write( "Then, OBVIOUSLY,  x = W + 1 = L - 1, and the area is\r\n" );
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document.write( "    L*W = (x+1)*(x-1) = 2,    or\r\n" );
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document.write( "          \"x%5E2+-+1\" = 2,  i.e.\r\n" );
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document.write( "          \"x%5E2\" = 2 + 1 = 3;\r\n" );
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document.write( "     hence,  x = \"sqrt%283%29\".  \r\n" );
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document.write( "Thus the dimensions of the rectangle are  W = x-1 = \"sqrt%283%29-1\"  (the width)  and  L = x+1 = \"sqrt%283%29%2B1\"  (the length).\r\n" );
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document.write( "You got the same answer, in a quick and simple manner.\r\n" );
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\n" ); document.write( "\n" ); document.write( "See the lessons\r
\n" ); document.write( "\n" ); document.write( "    - HOW TO solve the problem on quadratic equation mentally and avoid boring calculations \r
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\n" ); document.write( "\n" ); document.write( "    - Problems on the area and the dimensions of a rectangle\r
\n" ); document.write( "\n" ); document.write( "    - Three methods to find the dimensions of a rectangle when its perimeter and the area are given\r
\n" ); document.write( "\n" ); document.write( "    - Three methods to find the dimensions of a rectangle when its area and the difference of two dimensions are given \r
\n" ); document.write( "\n" ); document.write( "in this site,\r
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\n" ); document.write( "\n" ); document.write( "where you will find many other similar solved problems  (your  TEMPLATES)  with detailed explanations.\r
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\n" ); document.write( "\n" ); document.write( "            Come again soon to this forum to learn more (!)\r
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