document.write( "Question 1131622: The diagonal of a rectangle is 3 ft longer than the length of the rectangle and 4 ft longer than twice the width. Find the dimensions of the rectangle. \n" ); document.write( "

Algebra.Com's Answer #748283 by ikleyn(52790) $\"\"$ $\"About$

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document.write( "One equation is \r\n" );
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document.write( "   L = 2W + 1,       (1)\r\n" );
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document.write( "as the tutor @LoverMath1 derived in her post.\r\n" );
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document.write( "The other equation is\r\n" );
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document.write( "     = 2W + 4.\r\n" );
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document.write( "which follows from the condition.\r\n" );
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document.write( "This second condition, due to (1), is the same as\r\n" );
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document.write( "     = 2W + 4.     (2)\r\n" );
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document.write( "Equation (2) is your basic equation to solve.   Square both sides\r\n" );
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document.write( "    W^2 + 4W^2 + 4W + 1 = 4W^2 + 16W + 16\r\n" );
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document.write( "    W^2 - 12W - 15 = 0.\r\n" );
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document.write( "Answer.  The dimensions of the rectangle are  W =   and  L = 2W+1 =  feet.\r\n" );
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