document.write( "Question 872843: The length of a rectangle is 8 feet longer than its width. The area is 3500 sq ft.
\n" ); document.write( "I have to find the width of the rectangle and just one length because this is for a fence that is connected to the back of a house. The back of the house serves as one fenced area so I just need the length to the part I have to fence.
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Algebra.Com's Answer #526461 by lwsshak3(11628)\"\" \"About 
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The length of a rectangle is 8 feet longer than its width. The area is 3500 sq ft.
\n" ); document.write( "I have to find the width of the rectangle and just one length because this is for a fence that is connected to the back of a house. The back of the house serves as one fenced area so I just need the length to the part I have to fence.
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\n" ); document.write( "let x=width
\n" ); document.write( "x+8=length
\n" ); document.write( "length*width=area
\n" ); document.write( "x(x+8)=3500
\n" ); document.write( "x^2+8x=3500
\n" ); document.write( "x^2+8x-3500=0
\n" ); document.write( "solve for x using quadratic formula:
\n" ); document.write( "\"x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+\"
\n" ); document.write( "a=1, b=8, c=-3500
\n" ); document.write( "ans:
\n" ); document.write( "x≈55.30
\n" ); document.write( "x+8≈63.30
\n" ); document.write( "Fencing required=twice the width+length=2x+(x+8)=3x+8≈174 ft \n" ); document.write( "
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