Question

The area of a rectangle is 33ft^2. The length is 8 feet more than the width. What are the dimensions of the rectangle?

Answer 1

`length = (8 + x) \: feet\\ width = x \: feet \\ area = length * width \\ 33 = (8 + x)x \\ 33 = 8x + {x}^(2) \\ {x}^(2) + 8x - 33 = 0 \\ (x - 3)(x + 11) = 0 \\ (x + 11) \: is \: neglected \: since \: it \: is \: negative \\ therefore : x = 3 \\ dimensions \\ length = (8 + 3) = 11 \: feet \\ width = 3 \: feet`