It's possible.

Let x = width of the grove

Then

2x = length

:

Area = L * W:

x(2x) = 800

2x^2 = 800

x^2 = 800/2

x^2 = 400

x = Sqrt(400)

x = 20 m by 40 m

:

:

b>

The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48. Determine their present ages.

:

It would simply things if we let x and y be the ages 4 years ago

Then

(x+4) and (y+4) = ages now

:

Present age sum = 20:

(x+4) + (y+4) = 20

x + y + 8 = 20

x + y = 20 - 8

x + y = 12

y = 12-x

:

Product of age 4 yrs ago

x*y = 48

y = 48/x

Replace y with 12-x

12-x = 48/x

12x - x^2 = 48

-x^2 + 12x - 48 = 0

This equation has no real roots, (discriminant less than 0)

There is no solution

If we plot these equations, you can see they don't intersect

y = 12-x and y = 48/x

:

+graph%28+300%2C+200%2C+-2%2C+10%2C+-2%2C+20%2C+12-x%2C+48%2Fx%29+

:

:

c>

Is it possible to design a rectangular park of perimeter 80 m and 400 m2? If so, find its length and breadth.

:

It is: let the sides = x and y

Perimeter: 2x + 2y = 80

Simplify, divide by 2

x + y = 40

y = (40-x)

:

Area:

x * y = 400

Replace y with (40-x)

x(40-x) = 400

40x - x^2 = 400

-x^2 + 40x - 400 = 0

Easier to factor if we multiply by -1

x^2 - 40x + 400 = 0

Factor

(x-20)(x-20) = 0

x = 20,

:

The park will be a square; 20 by 20