Question

Kelly wants to create a rectangular playpen for his cats in his backyard, and estimates that the cats will need 500 square feet. He also wants the length of the playpen to be 20 feet longer than the width, so they have room to run. What dimensions should the playpen be?

Answer 1

Option C is correct.

Explanation:

We have been given that Kelly wants to create a rectangular playpen for his cats backyard

And cats need 500 square feet

And length of playpen to be 20 feet longer than the width.

Let the width of playpen be x.

Then the length of playpen be (x+20).

Then the equation becomes:

x(x+20) =500

We will solve above for x.

`x^2+20x-500=0`

Using discriminant
`D=b^2-4ac`

`D=20^2-4(1)(-500)`

`\Rightarrow D=2400`

Now, we will use formula for x which is:

`x=(-b\pm√(D))/(2a)`

`x=(-20\pm√(2400))/(2\cdot 1)`

`x=(-20\pm20√(6))/(2)`

Taking 2 as common factor from numerator and denominator.

`x=-10\pm\10sqrt{6}`

Or
`10√(6)+10\text{and}10√(6)-10`

Option c is correct.