document.write( "Question 282703: A certain rectangle's length is 5 feet longer than its width. If the area of the rectangle is 50 square feet, find its larger dimension \n" ); document.write( "
Algebra.Com's Answer #205234 by oberobic(2304)\"\" \"About 
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L = length
\n" ); document.write( "W = width
\n" ); document.write( "Area= L*W, for a rectangle
\n" ); document.write( "A = 50 ft^2 :: given
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\n" ); document.write( "L = W + 5 :: Length is 5 ft longer than the width, so you have to add 5 to W to make them equal
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\n" ); document.write( "Substitute L=W+5
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\n" ); document.write( "(W+5) * W = 50
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\n" ); document.write( "Collect and simplify
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\n" ); document.write( "W^2 + 5W = 50
\n" ); document.write( "W^2 + 5W - 50 = 0
\n" ); document.write( "(W + 10)(W - 5) = 0
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\n" ); document.write( "So our candidate solutions are W= -10 and W=5.
\n" ); document.write( "But a negative width is nonsensical, so our only practical option is W=5.
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\n" ); document.write( "Substituting W=5, we can find L.
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\n" ); document.write( "L = W+5
\n" ); document.write( "L = 5+5 = 10
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\n" ); document.write( "Checking the answer, we see
\n" ); document.write( "5*10 = 50, which matches the setup
\n" ); document.write( "10 = 5+5, which also matches the setup
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\n" ); document.write( "Answer:
\n" ); document.write( "The larger dimension is the length, which is 10.
\n" ); document.write( "The width is 5.
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\n" ); document.write( "Done.
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