Let's solve these questions one by one.

16. What quadratic equation can be formed if the two roots are -3 and -4?

If the roots of a quadratic equation are $r_1$ and $r_2$, the equation can be written as: $(x - r_1)(x - r_2) = 0$ Given the roots $r_1 = -3$ and $r_2 = -4$, the quadratic equation becomes: $(x + 3)(x + 4) = 0$ Expanding this, we get: $x^2 + 7x + 12 = 0$ Answer: $x^2 + 7x + 12 = 0$

17. The length of the garden is 7m longer than its width, and the area is $18 \, m^2$. If the width is represented as $x$, what equation can be formed from the situation?

Let the width be $x$ meters. Then the length is $x + 7$ meters. The area $A$ of the garden is given by: $A = \text{length} \times \text{width} = x(x + 7)$ Given that the area is $18 \, m^2$: $x(x + 7) = 18$ Answer: $x(x + 7) = 18$

18. The length of the garden is 7m longer than its width and the area is $18 \, m^2$. How long is the garden?

From the equation $x(x + 7) = 18$, we solve for $x$: $x^2 + 7x - 18 = 0$ We solve this quadratic equation using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Here, $a = 1$, $b = 7$, and $c = -18$: $x = \frac{-7 \pm \sqrt{7^2 + 72}}{2}$ $x = \frac{-7 \pm \sqrt{121}}{2}$ $x = \frac{-7 \pm 11}{2}$ This gives us two possible values for $x$: $x = 2 \quad \text{or} \quad x = -9 \quad (\text{which is not physically meaningful for a length})$ So, the width $x = 2$, and the length $x + 7 = 9$ meters.

Answer: 9m

19. The length of a rectangular playground is 25 meters more than its width, and the area is $3000 \, m^2$. What is the correct equation to represent the problem?

Let the width be $w$ meters. Then the length is $w + 25$ meters. The area $A$ is given by: $A = \text{length} \times \text{width} = w(w + 25)$ Given that the area is $3000 \, m^2$: $w(w + 25) = 3000$ Answer: $w(w + 25) = 3000$

20. The length of a rectangular playground is 25 meters more than its width, and the area is $3000 \, m^2$. What are the dimensions of the playground?

We already have the equation: $w(w + 25) = 3000$ Let's solve for $w$: $w^2 + 25w - 3000 = 0$ Using the quadratic formula: $w = \frac{-25 \pm \sqrt{25^2 + 12000}}{2}$ $w = \frac{-25 \pm \sqrt{625 + 12000}}{2}$ $w = \frac{-25 \pm \sqrt{12625}}{2}$ $w \approx \frac{-25 \pm 112.36}{2}$ This gives us two possible values for $w$: $w \approx 43.68 \quad \text{or} \quad w \approx -68.68 \quad (\text{not physically meaningful})$ Taking $w \approx 43.68$ meters, the length is $w + 25 \approx 68.68$ meters. The closest answer to this is $40m$ by $75m$.

Answer: $40m$ by $75m$

Do you want any more details or explanations?

Here are five related questions:

1. How do you derive the quadratic formula from a general quadratic equation?
2. What happens if a quadratic equation has a discriminant less than zero?
3. Can you explain why negative roots might be discarded in physical problems?
4. How do you solve a quadratic equation by completing the square?
5. What are the general steps to factor a quadratic equation?

Tip: When solving word problems involving quadratic equations, carefully define your variables and translate the problem's conditions into an equation step by step.