document.write( "Question 268570: Tiles are 1 -square inch.
\n" ); document.write( "Made a retangular picture with 36 tiles.
\n" ); document.write( "The length is 2 inches longer than the width.
\n" ); document.write( "What are the dimensions of the picture?\r
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Algebra.Com's Answer #196797 by stanbon(75887)\"\" \"About 
You can put this solution on YOUR website!
Tiles are 1 -square inch.
\n" ); document.write( "Made a retangular picture with 36 tiles.
\n" ); document.write( "The length is 2 inches longer than the width.
\n" ); document.write( "What are the dimensions of the picture?
\n" ); document.write( "-----------
\n" ); document.write( "Let width be \"x\" inches
\n" ); document.write( "Then length = \"x+2\" inches
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\n" ); document.write( "Area = x(x+2) = 36 sq in.
\n" ); document.write( "x^2+2x = 36
\n" ); document.write( "x^2 + 2x - 36 = 0
\n" ); document.write( "x = [-2 +- sqrt(4 - 4*1*-36)]/2
\n" ); document.write( "---
\n" ); document.write( "x = [-2 +- sqrt(148)]/2
\n" ); document.write( "Positive solution:
\n" ); document.write( "x = [-2 + 2sqrt(37)]/2
\n" ); document.write( "x = -1 + sqrt(37) is approximately = 5.08 inches (width)
\n" ); document.write( "x + 2 is appoximately = 7.08 inches (length)
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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