Question

A regulation tennis court for a doubles match is laid out so that its length is 6 feet more than two
times its width. The area of the doubles court is 2808 square feet. Find the length and width of a
doubles court.

Answer 1

let width = w
length = 2w + 6
area = LW = 2808
(2w + 6)w = 2808
Can you continue now?

Answer 2

So...2w^+6w = 2808

Answer 3

But will I end up having to use the quad. formula?

Answer 4

B/c I've no idea how to solve this using factoring.

Answer 5

first term is 2w^2. yes you'll need the quadratic formula or factoring

Answer 6

Sorry, that was a typo.

Answer 7

I thought so. First factor out a 2

Answer 8

w^2+3 = 1404

Answer 9

Argh...3w

Answer 10

w^2 + 3w - 1404 = 0

Answer 11

Right. So x = -3 +/- sqrt(9 - 4(1)(-1404))/2

Answer 12

x = -3 +/- sqrt(9 - 5616)/2

Answer 13

Okay...there's no way that's right. What have I done? lol

Answer 14

In the root, you have -4(1)(-1404) which is positive, not negative

Answer 15

Okay, then...x = -3 +/- sqrt(5626)/2

Answer 16

right

Answer 17

x = (-3 +/- 75)/2 -> -39 or 36

Answer 18

So w = 36 and l = 78

Answer 19

Good. Now -39 is meaningless as a width of a rectangle, so discard it.
You have w = 36
L = 2w + 6 =...

Answer 20

right

Answer 21

Okay...thank you once again!!

Answer 22

wlcm