Posted by Anonymous on Monday, February 13, 2012 at 5:00pm.
A rectangular parcel of land has an area of 4,000 ft2. A diagonal between opposite corners is measured to be 10 ft longer than one side of the parcel. What are the dimensions of the land, correct to the nearest foot?

pre cal  Damon, Monday, February 13, 2012 at 5:32pm
x^2 + y^2 = diagonal^2 = d^2
x y = 4000 so y = 4000/x
d = x+10 so d^2 = x^2 + 20 x + 100
x^2 + 16*10^6/x^2 = x^2 + 20 x + 100
16*10^6 = 20 x^3 + 100 x^2
5 x^3 + 25 x^2  4*10^6 = 0
I do not see an easy way to solve that. Use iteration, x and 5x^3+25 x^2
x = 10^2 > 5,250,000 too big
x = 80 > 2,720,000 too small
x = 90 > 3,847,500 getting close
x = 92 > 4,333,568 very close
x = 91 > 3,974,880 between 91 and 92
x = 91.5> 4,039,610 between 91 and 91.5
so 91 to nearest foot
91^2 + y^2 = (91+10))^2
8281 +y^2 = 10201
y = 43.8 or 44
so
91 by 44 
pre cal  Reiny, Monday, February 13, 2012 at 5:42pm
let one side be x
then the diagonal is x+10
let the third side by y
x^2 + y^2 = (x+10)^2
x^2 + y^2 = x^2 + 20x + 100
y^2 = 20x + 100
y = √(20x+100)
so the area of the triangle is half the rectangle area
(1/2)xy =(1/2)(4000)
xy = 4000
x√(20x+100) = 4000
x^2(20x+100) = 16000000
20x^3 + 100x^2  16000000 = 0
x^3 + 5x^2  800000 = 0
I must admit that I used Wolfram to solve this
http://www.wolframalpha.com/input/?i=x%5E3+%2B+5x%5E2++800000+%3D+0
and I got x = 91.1947
then y = 43.8622
and the diagonal would be 101.1947
check:
area of rectangle = xy = (91.1947)(43.8622) = 4000.00017 (close enough)
I will let you check if Pythagoras also works out, it does.
I don't know what method you would have to solve that cubic, since it does not factor, and therefore does not have rational roots.
Do you know Newton's Method ?