1. ## Tennis Court

A regulation doubles tennis court has an area of 2808 square feet. If it is 6 feet longer than twice the width, determine the dimension of the court.

Let x = width

Let = 2x + 6

Let area = A = 2808 ft^2

A = LW

2808 = (2x + 6)(x)

Is this the correct equation set up? If not, why?

2. ## Re: Tennis Court Originally Posted by harpazo 2808 = (2x + 6)(x)
Is this the correct equation set up? If not, why?
Correct; 2x^2 + 6x - 2808 = 0
Divide by 2: x^2 + 3x - 1404 = 0

3. ## Re: Tennis Court Originally Posted by DenisB Correct; 2x^2 + 6x - 2808 = 0
Divide by 2: x^2 + 3x - 1404 = 0
Do you suggest the quadratic formula here?

4. ## Re: Tennis Court

Yes! Unless you wanna try factoring...

5. ## Re: Tennis Court

I decided to use the quadratic formula.

Let length = 2x + 6

Let width = x

I found x to be 75.

So, the width is 75 feet.

The length is 2(75) + 6 or 156 feet.

Something is wrong. You see, 156 x 75 does not equal the given area.

6. ## Re: Tennis Court

(2x + 6)(x) = 2808
2x^2 + 6x - 2808 = 0
x^2 + 3x - 1404 = 0

x = [-3 +- sqrt(3^2 - 4(-1404)] / 2

That leads to x = 36 or x = -39, so x = 36

How did you end up with x = 75?

7. ## Re: Tennis Court Originally Posted by DenisB (2x + 6)(x) = 2808
2x^2 + 6x - 2808 = 0
x^2 + 3x - 1404 = 0

x = [-3 +- sqrt(3^2 - 4(-1404)] / 2

That leads to x = 36 or x = -39, so x = 36

How did you end up with x = 75?

8. ## Re: Tennis Court

sqrt{3^2 - 4(-1404)}

sqrt{9 + 5616}

sqrt{5625} = 75

This is how I got 75.

(-3 + 75)/2

72/2 = 36

I now see that x = 36, which is the width's length.

Length = 2(36) + 6 = 72 + 6 = 78.

Width = 36 feet

Length = 78 feet

9. ## Re: Tennis Court Originally Posted by harpazo A regulation doubles tennis court has an area of 2808 square feet. If the length is 6 feet longer than twice the width, determine the dimensions of the court.

Let x = width

Let 2x + 6 = the length
harpazo, it's good to proofread for missing words.

10. ## Re: Tennis Court

I am speeding through the textbook and thus messing up with posted information.