document.write( "Question 108566: If the sides of a square are increased by 3 in., the area is increased by
\n" ); document.write( "39 in.2. What were the dimensions of the original square? \n" ); document.write( "
Algebra.Com's Answer #79204 by ankor@dixie-net.com(22740)\"\" \"About 
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If the sides of a square are increased by 3 in., the area is increased by
\n" ); document.write( "39 in.2. What were the dimensions of the original square?
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\n" ); document.write( "Let x = the side of the original square
\n" ); document.write( "Then
\n" ); document.write( "x^2 = the area of the original square
\n" ); document.write( "and
\n" ); document.write( "(x+3) = the side of the new square
\n" ); document.write( "then
\n" ); document.write( "(x+3)^2 = the area of the new square
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\n" ); document.write( "New square area - old square area = 39
\n" ); document.write( "(x+3)^2 - x^2 = 39
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\n" ); document.write( "FOIL (x+2)(x+3)
\n" ); document.write( "x^2 + 6x + 9 - x^2 = 39
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\n" ); document.write( "Conveniently the x^2's eliminate each other so we have:
\n" ); document.write( "6x + 9 = 39
\n" ); document.write( "6x = 39-9
\n" ); document.write( "x = 30/6
\n" ); document.write( "x = 5 in, side of the original square
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\n" ); document.write( "Check our solution, (new square side would be 8\")
\n" ); document.write( "8^2 - 5^2
\n" ); document.write( "64 - 25 = 39 sq/in as given
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\n" ); document.write( "Did you understand what went on here? Any questions? \n" ); document.write( "
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