The length of a rectangle is 2 feet more than it's width. How do you find the dimensions of the rectangle if its area is 63 square feet?

Question

Eva Abramson · Accepted Answer

#7# by #9# feet.

We let the length be #x + 2# and the width be #x#. The area of a rectangle is given by #A = l * w#. #A = l * w# #63 = x(x + 2)# #63 = x^2 + 2x# #0 = x^2 + 2x - 63# #0 = (x + 9)(x - 7)# #x = -9 and 7# A negative answer is impossible here, so the width is #7# feet and the length is #9# feet. Hopefully this helps!

Genesis Allen · Answer

undefined Let $x$ represent the width of the rectangle. Since the length is 2 feet more than the width, the length can be represented as $x + 2$. The area of the rectangle is given by the formula $A = \text{length} \times \text{width}$. Substituting the given values, we have $63 = (x + 2) \times x$. Solving this quadratic equation, we find $x = 7$ or $x = -9$. Since width cannot be negative, the width is 7 feet, and the length is $7 + 2 = 9$ feet. Thus, the dimensions of the rectangle are 7 feet by 9 feet.