document.write( "Question 74859: The length of a rectangle is 5 cm more than 2 times its width. If the area of the rectangle is 75 cm2, find the dimensions of the rectangle to the nearest thousandth. \n" ); document.write( "
Algebra.Com's Answer #53783 by psbhowmick(878)\"\" \"About 
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Let the width = x cm.
\n" ); document.write( "Two times width = 2x cm.
\n" ); document.write( "5cm more than this value = (2x + 5) cm
\n" ); document.write( "So, the length = (2x + 5) cm.\r
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\n" ); document.write( "\n" ); document.write( "Hence area = Length X Width = \"%282x+%2B+5%29x+=+2x%5E2+%2B+5x\" \"cm%5E2\".\r
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\n" ); document.write( "\n" ); document.write( "But, according to the problem this area is 75 \"cm%5E2\".\r
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\n" ); document.write( "\n" ); document.write( "So, \"2x%5E2+%2B+5x+=+75\"
\n" ); document.write( "\"2x%5E2+%2B+5x+-+75+=+0\"
\n" ); document.write( "\"2x%5E2+%2B+15x+-+10x+-+75+=+0\"
\n" ); document.write( "\"x%282x+%2B+15%29+-+5%282x+%2B+15%29+=+0\"
\n" ); document.write( "\"%28x-5%29%282x%2B15%29=0\"\r
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\n" ); document.write( "\n" ); document.write( "So either (x-5) = 0 or (2x+15) = 0
\n" ); document.write( "i.e. either x = 5 or x = -7.5\r
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\n" ); document.write( "\n" ); document.write( "But width of a rectangle (x) cannot be negative.
\n" ); document.write( "So negative answer is discarded.
\n" ); document.write( "Hence x = 5.\r
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\n" ); document.write( "\n" ); document.write( "So the width = 5 cm and length = 2x5 + 5 = 15 cm.
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