document.write( "Question 536392: The length of a photograph is 4 in. greater than its width. The area of the photograph is 52 1/4 in.^2. What are the dimensions of the photograph? \n" ); document.write( "
Algebra.Com's Answer #352344 by fcabanski(1391)\"\" \"About 
You can put this solution on YOUR website!
A=52 and 1/4=52.25=L*W


\n" ); document.write( "W = width.


\n" ); document.write( "L is 4 inches greater than W = W + 4.


\n" ); document.write( "W(W+4)=52.25


\n" ); document.write( "W^2 + 4W = 52.25


\n" ); document.write( "Subtract 52.25 from both sides.


\n" ); document.write( "W^2 + 4W - 52.25=0


\n" ); document.write( "This doesn't appear easy to factor, so use the quadratic equation.


\n" ); document.write( "=


\n" ); document.write( "\"%28-4+%2B-+sqrt%28+225+%29%29%2F%282%29=+%28-4+%2B-+15%29%2F2\"=


\n" ); document.write( "-19/2 or 11/2.


\n" ); document.write( "Throw away the negative answer, because a photo can't have negative dimensions.


\n" ); document.write( "W=11/2=5 and 1/2 inches.


\n" ); document.write( "L=W+4=9 and 1/2 inches
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