document.write( "Question 320605: The length of a rectangular carpet is 5.5' greater than twice the width. The area is 90. Find the dimensions \n" ); document.write( "

Algebra.Com's Answer #229626 by mananth(16946) $\"\"$ $\"About$

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let the width be x feet
\n" ); document.write( "length = 2x+5.5
\n" ); document.write( "Area = 90
\n" ); document.write( "..
\n" ); document.write( "x(2x+11/2)=90
\n" ); document.write( "x(4x+11 /2 ) =90
\n" ); document.write( "4x^2+11x/2 =90
\n" ); document.write( "4x^2+11x=180
\n" ); document.write( "4x^2+11x-180=0
\n" ); document.write( "x1, x2 the roots of the equation are
\n" ); document.write( "x1=(-11+sqrt((11^2)+4*4*180)) /( 2*8)
\n" ); document.write( "x1=5.47 feet the width
\n" ); document.write( "..
\n" ); document.write( "x2=(-11-sqrt((11^2)+4*4*180)) /( 2*8)
\n" ); document.write( "x2=-8.22 (ignore)
\n" ); document.write( "Length = 2x+5.5
\n" ); document.write( "=2*5.47+5.5
\n" ); document.write( "=16.44 feet
\n" ); document.write( "..
\n" ); document.write( "CHECK
\n" ); document.write( "5.47*16.44=89.92 \n" ); document.write( "

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