Question

Pat wants to create s rectangular playpen for her dogs in her backyard. And estimates the dogs will need 600 square feet. She also wants the length of the playpen to be 30 feet longer than the width, so they have room to run around. What dimensions should the playpen be?
CORRECT ANSWER IS A

Answer 1

Lets calculate the parameters of the problem. Suppose that the width is x feet. THen, the length is x+30 feet. The area of this parallelogram is given by the product of length and width, hence x(x+30). This has to be equal to 600.
Hence, x is the solution to the equation x(x+30)=600.
x*x+30x=600 hence x*x+30x-600=0. This is a second degree equation and we can calculate Δ=b*b-4ac=3300. The general solution is given by:
$x= \frac{-b+ \sqrt{\Delta} }{2a}$.
Substituting, we get that x=-30/4+$\sqrt{3300} /4$ or x=-7.5+25*$\sqrt{33}$. This is the width; the length is given by x+30.

Answer 2

Answer:

Length $= 13.72$ feet

Width $= 43.72$ feet

Explanation:

Given

The area of rectangular playpen for her dogs is $600$ square feet

Let the length of the rectangle be represented by "$X$" feet

then width of the rectangle would be "$X+30$" feet

The area of the rectangle in terms of  "$X$" is equal to

$X* (X+30)$

$X^2+30X = 600$

This becomes a quadratic equation which can be written as

$X^2 + 30X -600 = 0$

On solving this equation, we get -

$X = 13.72$feet

Thus,

Length $= 13.72$ feet

Width $= 13.72+30\\= 43.72$feet