document.write( "Question 1036139: The area of a rectangle is 91 square centimeters. Its length is x+2 and its width is x-4. Find the perimeter of the rectangle. \n" ); document.write( "
Algebra.Com's Answer #650748 by addingup(3677)\"\" \"About 
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L*W = 91
\n" ); document.write( "L = x+2
\n" ); document.write( "W = x-4
\n" ); document.write( "(x+2)*(x-4) = 91
\n" ); document.write( "x^2-2x-8 = 91
\n" ); document.write( "x^2-2x = 99
\n" ); document.write( "x^2-2x+1 = 100
\n" ); document.write( "(x-1)^2 = 100
\n" ); document.write( "x-1 = 10 or x-1 = -10
\n" ); document.write( "x = 11 or x = -9
\n" ); document.write( "Toss out the -9, we can't use it, and you have:
\n" ); document.write( "x = 11
\n" ); document.write( "The length is 11+2 = 13
\n" ); document.write( "The width is 11-4 = 7
\n" ); document.write( "The perimeter is:
\n" ); document.write( "2(13)+2(7)= 40
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\n" ); document.write( "Check:
\n" ); document.write( "13+7 = 91 Correct
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