document.write( "Question 548513: A picture frame measures 26 in. by 24 in.. There is 575 in. squared of picture showing. The frame is of uniform thickness. Find the thickness of the frame. \n" ); document.write( "
Algebra.Com's Answer #357042 by ankor@dixie-net.com(22740)\"\" \"About 
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A picture frame measures 26 in. by 24 in..
\n" ); document.write( " There is 575 in. squared of picture showing.
\n" ); document.write( " The frame is of uniform thickness. Find the thickness of the frame.
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\n" ); document.write( "Let x = the width of the frame
\n" ); document.write( "then
\n" ); document.write( "(26-2x) by (24-2x) is the dimensions of the pictures, which is 575 sq/in
\n" ); document.write( "therefore, the area equation:
\n" ); document.write( "(26-2x)*(24-2x) = 575
\n" ); document.write( "FOIL
\n" ); document.write( "624 - 52x - 48x + 4x^2 = 575
\n" ); document.write( "Combine like terms to form a quadratic equation:
\n" ); document.write( "4x^2 - 100x + 624 - 575 = 0
\n" ); document.write( "4x^2 - 100x + 49 = 0
\n" ); document.write( "We can us the quadratic formula, but this will factor to
\n" ); document.write( "(2x-49)(2x-1) = 0
\n" ); document.write( "two solutions, but only one will make sense
\n" ); document.write( "2x = 49
\n" ); document.write( "x = 49/2
\n" ); document.write( "x = 24.5\", obviously this could not be the width of frame
\n" ); document.write( "and
\n" ); document.write( "2x = 1
\n" ); document.write( "x = .5 inch is the reasonable width of the frame
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\n" ); document.write( ":
\n" ); document.write( "We can confirm this by finding the area using x=.1, then 2x = 1, so we have
\n" ); document.write( "(26+1)(24+1) = 675 for the overall area, like was given
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