document.write( "Question 604411: The outside measurements of a rectangular frame for a picture are 30 inches by 40 inches. The area of the rectangular picture inside the frame is 936 inches. Find the width of the frame that surrounds the picture.
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Algebra.Com's Answer #381215 by alicealc(293)\"\" \"About 
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assume that the frame has width of x inches
\n" ); document.write( "Picture's width = 30 - 2x (x on both sides)
\n" ); document.write( "Picture's length = 40 - 2x (x on both sides)\r
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\n" ); document.write( "\n" ); document.write( "(30 - 2x)*(40 - 2x) = 936
\n" ); document.write( "\"1200+-+60x+-+80x+%2B+4x%5E2+=+936\"
\n" ); document.write( "\"4x%5E2+-+140x+%2B+1200+=+936\"
\n" ); document.write( "\"4x%5E2+-+140x+%2B+1200+-+936+=+0\"
\n" ); document.write( "\"4x%5E2+-+140x+%2B+264+=+0\"
\n" ); document.write( "(divide all by 4)
\n" ); document.write( "\"x%5E2+-+35x+%2B+66+=+0\"
\n" ); document.write( "(x - 33)*(x - 2) = 0
\n" ); document.write( "x - 33 = 0 or x - 2 = 0
\n" ); document.write( "x = 33 or x = 2\r
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\n" ); document.write( "\n" ); document.write( "because the width of the frame can't be 33 (the picture's width and length will be negative if x = 33), then we take x = 2\r
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\n" ); document.write( "\n" ); document.write( "so, the width of the frame that surrounds the picture is 2 inches
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