A​ person's rectangular dog pen for his dog must have an area of 200 square feet.​ also, the length must be 20 feet longer than the width. find the dimensions of the pen.

Let the width be x.
Width = x
Length = x + 20

The area is 200 ft²
x(x + 20) = 200
x² + 20x - 200 = 0

a = 1, b = 20, c = -200

[tex]x = \dfrac{-b\pm \sqrt{b^2-4ac} }{2a} [/tex]

[tex]x = \dfrac{-20 \pm \sqrt{20^2-4(1)(-200)} }{2(1)} [/tex]

[tex]x = \dfrac{-20 \pm \sqrt{400+800} }{2} [/tex]

[tex]x = \dfrac{-20 \pm \sqrt{1200} }{2} [/tex]

[tex]x = -27.32 \text { (rejected, length cannot be negative) } \or \ x = 7.32[/tex]

Width = 7.32

Length = 7.32 + 20 = 27.32

Answer: The pen is 7.32 ft by 27.32 ft