document.write( "Question 142751: The area of a room is 20 square metre. If the length is increased by 3 metre and the width is increased by 1 metre, the room will double in area. Determine the original dimensions of the room. \n" ); document.write( "
Algebra.Com's Answer #103871 by stanbon(75887)\"\" \"About 
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The area of a room is 20 square metre. If the length is increased by 3 metre and the width is increased by 1 metre, the room will double in area. Determine the original dimensions of the room.
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\n" ); document.write( "lw = 20
\n" ); document.write( "(l+3)(w+1) = 40
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\n" ); document.write( "lw + l + 3w + 3 = 40
\n" ); document.write( "20 + l + 3w + 3 = 40
\n" ); document.write( "l + 3w = 17
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\n" ); document.write( "But l = 20/w
\n" ); document.write( "(20/w) + 3w = 17
\n" ); document.write( "20 + 3w^2 = 17w
\n" ); document.write( "3w^2 - 17w + 20 = 0
\n" ); document.write( "w = [17 +- sqrt(17^2 - 4*3*20)]/6
\n" ); document.write( "w = [17 +- sqrt(49)]/6
\n" ); document.write( "w = [17+-7]/6
\n" ); document.write( "w = 4 or w = 5/3
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\n" ); document.write( "If width = 4, then length = 5
\n" ); document.write( "This answer checks out.
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\n" ); document.write( "If width = 5/3,
\n" ); document.write( "lw = 20
\n" ); document.write( "l(5/3) = 20
\n" ); document.write( "l = 12
\n" ); document.write( "This is wrong.
\n" ); document.write( "Therefore w = 5/3 is an extraneous answer.
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H \n" ); document.write( "
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