Is it possible to design a rectangular mango grove whose length is twice its breadth,and the area is $800 \, m^2$? If so,find its length and breadth.

(A) Let the breadth of the mango grove be $x \, m$.
Then,the length of the mango grove will be $2x \, m$.
Area of the rectangular mango grove $= \text{Length} \times \text{Breadth} = (2x)(x) = 2x^2 \, m^2$.
Given that the area is $800 \, m^2$,we have $2x^2 = 800$.
$x^2 = 400$.
$x^2 - 400 = 0$.
Comparing this with the standard quadratic equation $ax^2 + bx + c = 0$,we get $a = 1, b = 0, c = -400$.
The discriminant $D = b^2 - 4ac = (0)^2 - 4(1)(-400) = 1600$.
Since $D > 0$,the equation has two distinct real roots,meaning such a design is possible.
$x^2 = 400 \implies x = \pm 20$.
Since the breadth cannot be negative,we take $x = 20$.
Thus,the breadth is $20 \, m$ and the length is $2(20) = 40 \, m$.