Solution:

Nature of the roots of the following quadratic equations:

(i) $2x^2 – 3x + 5 = 0$
The discriminant is given by $D = b^2 - 4ac$
Here, $a = 2$, $b = -3$, and $c = 5$
$D = (-3)^2 - 4(2)(5) = 9 - 40 = -31$
Since $D < 0$, the roots are imaginary.

(ii) $3x^2 – 4\sqrt{3}x + 4 = 0$
The discriminant is given by $D = b^2 - 4ac$
Here, $a = 3$, $b = -4\sqrt{3}$, and $c = 4$
$D = (-4\sqrt{3})^2 - 4(3)(4) = 48 - 48 = 0$
Since $D = 0$, the roots are real and equal.
The roots are given by $x = \frac{-b}{2a} = \frac{-(-4\sqrt{3})}{2(3)} = \frac{4\sqrt{3}}{6} = \frac{2\sqrt{3}}{3}$

(iii) $2x^2 – 6x + 3 = 0$
The discriminant is given by $D = b^2 - 4ac$
Here, $a = 2$, $b = -6$, and $c = 3$
$D = (-6)^2 - 4(2)(3) = 36 - 24 = 12$
Since $D > 0$, the roots are real and distinct.
The roots are given by $x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-(-6) \pm \sqrt{12}}{2(2)} = \frac{6 \pm 2\sqrt{3}}{4} = \frac{3 \pm \sqrt{3}}{2}$

Solution:

(i) $2x^2 + kx + 3 = 0$

The quadratic equation is $2x^2 + kx + 3 = 0$. For the equation to have two equal roots, the discriminant must be zero. The discriminant $\Delta$ of a quadratic equation $ax^2 + bx + c = 0$ is given by: $$\Delta = b^2 - 4ac$$ Here, $a = 2$, $b = k$, and $c = 3$. Substituting these values into the discriminant formula: $$\Delta = k^2 - 4(2)(3)$$ $$\Delta = k^2 - 24$$ For equal roots, $\Delta = 0$: $$k^2 - 24 = 0$$ $$k^2 = 24$$ $$k = \pm \sqrt{24}$$ $$k = \pm 2\sqrt{6}$$ Thus, the values of $k$ are $k = 2\sqrt{6}$ and $k = -2\sqrt{6}$.

(ii) kx(x – 2) + 6 = 0

The quadratic equation is $kx(x - 2) + 6 = 0$. Expanding the equation: $$kx^2 - 2kx + 6 = 0$$ For the equation to have two equal roots, the discriminant must be zero. The discriminant $\Delta$ of the quadratic equation $kx^2 - 2kx + 6 = 0$ is given by: $$\Delta = (-2k)^2 - 4(k)(6)$$ $$\Delta = 4k^2 - 24k$$ For equal roots, $\Delta = 0$: $$4k^2 - 24k = 0$$ $$4k(k - 6) = 0$$ So, $k = 0$ or $k = 6$. Thus, the values of $k$ are $k = 0$ and $k = 6$.

Solution:

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.

Let the breadth of the rectangular mango grove be $b$ meters. Then, the length of the mango grove is $2b$ meters, as given in the problem.

The area of a rectangle is given by the formula:

Area = Length $\times$ Breadth

Substituting the values, we get:

Area = $2b \times b = 800$

This simplifies to:

$2b^2 = 800$

Now, divide both sides by 2:

$b^2 = 400$

Taking the square root of both sides:

$b = \sqrt{400}$

$b = 20$ meters

Since the length is twice the breadth, the length is:

Length = $2b = 2 \times 20 = 40$ meters

Therefore, the breadth of the mango grove is 20 meters, and the length is 40 meters.

Solution:

Problem:

The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48. Is this situation possible? If so, determine their present ages.

Solution:

Let the present ages of the two friends be $x$ and $y$ years.

According to the first condition, the sum of their ages is 20 years:

$x + y = 20$

According to the second condition, four years ago, the product of their ages was 48 years:

$(x - 4)(y - 4) = 48$

Expanding the second equation:

$xy - 4x - 4y + 16 = 48$

$xy - 4x - 4y = 32$

Now, substitute $x + y = 20$ into the equation:

Since $x + y = 20$, we can write $4x + 4y = 80$. Substituting this into the equation:

$xy - 80 = 32$

$xy = 112$

We now have the system of equations:

1) $x + y = 20$

2) $xy = 112$

This is a quadratic equation in terms of $x$ and $y$. The quadratic equation can be written as:

$t^2 - (x + y)t + xy = 0$

Substitute $x + y = 20$ and $xy = 112$ into the quadratic equation:

$t^2 - 20t + 112 = 0$

Now, solve this quadratic equation using the quadratic formula:

$t = \frac{-(-20) \pm \sqrt{(-20)^2 - 4(1)(112)}}{2(1)}$

$t = \frac{20 \pm \sqrt{400 - 448}}{2}$

$t = \frac{20 \pm \sqrt{-48}}{2}$

Since the discriminant is negative ($\sqrt{-48}$), the quadratic equation has no real solutions.

Therefore, this situation is not possible.

Solution:

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

Let the length of the rectangular park be $l$ meters and the breadth be $b$ meters.

We are given the perimeter of the rectangle as 80 m and the area as 400 m². The formulas for perimeter and area of a rectangle are:

Perimeter, $P = 2(l + b)$

Area, $A = l \times b$

From the given data:

1. $P = 80 \Rightarrow 2(l + b) = 80 \Rightarrow l + b = 40$

2. $A = 400 \Rightarrow l \times b = 400$

Now, solve these two equations:

From equation (1), $l + b = 40$, we can express $l$ as:

$l = 40 - b$

Substitute this value of $l$ into equation (2):

$(40 - b) \times b = 400$

$40b - b^2 = 400$

Rearrange the terms:

$b^2 - 40b + 400 = 0$

This is a quadratic equation. Solve it using the quadratic formula:

$b = \frac{-(-40) \pm \sqrt{(-40)^2 - 4(1)(400)}}{2(1)}$

$b = \frac{40 \pm \sqrt{1600 - 1600}}{2}$

$b = \frac{40 \pm \sqrt{0}}{2}$

$b = \frac{40}{2} = 20$

Therefore, the breadth of the park is $b = 20$ meters.

Now, substitute $b = 20$ into $l + b = 40$:

$l + 20 = 40 \Rightarrow l = 40 - 20 = 20$

Thus, the length of the park is $l = 20$ meters.

The length and breadth of the park are both 20 meters.