document.write( "Question 1033247: please help me solve: If the length of a rectangle is 2 more than its width, and its area is 35 square feet, what are the measures of the length and width? \n" ); document.write( "
Algebra.Com's Answer #647800 by addingup(3677)\"\" \"About 
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begin with the formula for the area
\n" ); document.write( "L*W = area
\n" ); document.write( "And the problem says:
\n" ); document.write( "L = W+2
\n" ); document.write( "So now we have:
\n" ); document.write( "(W+2)*W = 35
\n" ); document.write( "W^2+2W-35 = 0
\n" ); document.write( "(W-5)(W+7)= 0
\n" ); document.write( "Split into 2 equations:
\n" ); document.write( "W-5 = 0
\n" ); document.write( "or
\n" ); document.write( "W+7 = 0
\n" ); document.write( "W = 5
\n" ); document.write( "or
\n" ); document.write( "W= -7 We know we're looking for a positive number, so toss this one out. Your answer is:
\n" ); document.write( "W = 5
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\n" ); document.write( "Check:
\n" ); document.write( "(5+2)*5 = 35
\n" ); document.write( "7*5 = 35 Correct
\n" ); document.write( "John
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