document.write( "Question 30418: The Keypad and the viewing windowof the TI83 graphing calculator is rectangular. The length of the rectangle is 2cm more than twice the width, and the area of the rectangle is 144cm2. Find the length and the width. \n" ); document.write( "
Algebra.Com's Answer #17089 by sdmmadam@yahoo.com(530)\"\" \"About 
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Let the length be L cms and
\n" ); document.write( "let the width be B cms
\n" ); document.write( "The length of the rectangle is 2cm more than twice the widt
\n" ); document.write( "That is L =2 B+2 ----(1)
\n" ); document.write( " the area of the rectangle is 144cm2
\n" ); document.write( "That is LB = 144 ----(2)
\n" ); document.write( "(2B+2)XB = 144
\n" ); document.write( "Dividing by 2
\n" ); document.write( "(B+1)B = 72
\n" ); document.write( "B^2+B-72 =0
\n" ); document.write( "(B+9)(B-8) = 0
\n" ); document.write( "(product = -72 and sum is 1 gives the quantiites to be 9 and (-8) )
\n" ); document.write( "(B+9) =0 gives B = -9 which is negative and hence not applicable as we are talking about the solids
\n" ); document.write( "(B-8) = 0 gives B = 8 ----(*)
\n" ); document.write( "Putting B = 8 in (2)
\n" ); document.write( "LB = 144
\n" ); document.write( "LX8 = 144
\n" ); document.write( "L=144/8 = 18
\n" ); document.write( "Answer: Length = 18 cms and Width = 8 cms
\n" ); document.write( "Verification: Length should be 2 greater than twice width
\n" ); document.write( "And ofcourse 18 is 2 greater than twice 8.
\n" ); document.write( "Hene our values are correct. \n" ); document.write( "
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