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)![]() ![]() ![]() You can put this solution on YOUR website! L = length \n" ); document.write( "W = width \n" ); document.write( "Area= L*W, for a rectangle \n" ); document.write( "A = 50 ft^2 :: given \n" ); document.write( ". \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 \n" ); document.write( ". \n" ); document.write( "Substitute L=W+5 \n" ); document.write( ". \n" ); document.write( "(W+5) * W = 50 \n" ); document.write( ". \n" ); document.write( "Collect and simplify \n" ); document.write( ". \n" ); document.write( "W^2 + 5W = 50 \n" ); document.write( "W^2 + 5W - 50 = 0 \n" ); document.write( "(W + 10)(W - 5) = 0 \n" ); document.write( ". \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. \n" ); document.write( ". \n" ); document.write( "Substituting W=5, we can find L. \n" ); document.write( ". \n" ); document.write( "L = W+5 \n" ); document.write( "L = 5+5 = 10 \n" ); document.write( ". \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 \n" ); document.write( ". \n" ); document.write( "Answer: \n" ); document.write( "The larger dimension is the length, which is 10. \n" ); document.write( "The width is 5. \n" ); document.write( ". \n" ); document.write( "Done. \n" ); document.write( " |