document.write( "Question 281243: The length of a photograph is 1 cm less than twice the width. The area is 45cm squared. Find the dimensions of the photograph. \r
\n" ); document.write( "\n" ); document.write( "the problem must be solved algebracly using the quadratic formula, factoring or anything like that. thanks
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Algebra.Com's Answer #204290 by vleith(2983)\"\" \"About 
You can put this solution on YOUR website!
Let width be w and length be l.
\n" ); document.write( "\"l+=+2w-1\"
\n" ); document.write( "\"Area+=+Length+%2A+Width\"
\n" ); document.write( "\"45+=+%282w-1%29w\"
\n" ); document.write( "\"45+=+2w%5E2+-w\"
\n" ); document.write( "\"0+=+2w%5E2+-+w+-+45\"
\n" ); document.write( "\"0+=+%282w+%2B9%29%28w+-+5%29\"
\n" ); document.write( "so either w-5 =0 or 2w+9=o
\n" ); document.write( "w = 5 or w = -9/2
\n" ); document.write( "since a distance can't be nagative, width = 5
\n" ); document.write( "so length = 2*5-1 = 9\r
\n" ); document.write( "\n" ); document.write( "Does 9*5 = 45??
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