document.write( "Question 326928: The length of a rectangle is 4 feet more than twice the length. The area of the rectangle is 48 ft square. Find the length and the width of the rectangle. \n" ); document.write( "
Algebra.Com's Answer #234177 by AAfter Search(61)![]() ![]() ![]() You can put this solution on YOUR website! Let the length and width of the rectangle be 2w + 4 and w. \n" ); document.write( "Area of the rectangle is Length X Breadth i.e., \n" ); document.write( "(2w + 4) x w = 48 \n" ); document.write( "=> 2w^2 + 4w = 48 \n" ); document.write( "=> 2w^2 + 4w - 48 = 0 \n" ); document.write( "=> w^2 + 2w - 24 = 0 \n" ); document.write( "=> w^2 + 6w - 4w - 24 = 0 \n" ); document.write( "=> w(w+ 6) - 4(w + 6) = 0 \n" ); document.write( "=> (w - 4)(w + 6) = 0 \n" ); document.write( "=> w = 4, -6. \n" ); document.write( "Negative value of w is inappropriate. \n" ); document.write( "Hence, width w is 4 feet. \n" ); document.write( "Length is 2 x 4 + 4 = 8 + 4 = 12 feet. \n" ); document.write( " |