A school plans to fence a rectangular playground with an area of 600 m². The length of the playground is 5 metres more than its width. (a) Form a quadratic equation representing the situation. (b) Solve the equation by completing the square. (c) Find the dimensions of the playground.

Let the width of the rectangular playground be $w$ meters.
The length of the playground is 5 meters more than its width, so the length $l = w + 5$ meters.
The area of the playground is given as $600\,m^2$.

The area of a rectangle is given by the formula $Area = length \times width$.
Substitute the expressions for length, width, and the given area:

Rearrange the equation into the standard quadratic form $ax^2 + bx + c = 0$:

The quadratic equation representing the situation is $\boxed{w^2 + 5w - 600 = 0}$.

Start with the equation from part (a):

Move the constant term to the right side of the equation:

To complete the square for $w^2 + 5w$, take half of the coefficient of $w$ (which is 5), square it, and add it to both sides of the equation.
Half of 5 is $\frac{5}{2}$.
Squaring it gives $(\frac{5}{2})^2 = \frac{25}{4}$.

The left side is now a perfect square trinomial:

Note that $\sqrt{2425} = \sqrt{25 \times 97} = 5\sqrt{97}$.

Subtract $\frac{5}{2}$ from both sides:

The solutions to the equation are $\boxed{w = \frac{-5 + 5\sqrt{97}}{2} \text{ or } w = \frac{-5 - 5\sqrt{97}}{2}}$.

From part (b), we have two possible values for $w$:
$w_1 = \frac{-5 + 5\sqrt{97}}{2}$
$w_2 = \frac{-5 - 5\sqrt{97}}{2}$
Since width cannot be a negative value, we must choose the positive solution.
$\sqrt{97} \approx 9.8488$.
$w_1 = \frac{-5 + 5(9.8488)}{2} = \frac{-5 + 49.244}{2} = \frac{44.244}{2} \approx 22.122\,m$.
$w_2 = \frac{-5 - 5(9.8488)}{2} = \frac{-5 - 49.244}{2} = \frac{-54.244}{2} \approx -27.122\,m$.
Therefore, the width $w = \frac{-5 + 5\sqrt{97}}{2}\,m$.

The length $l = w + 5$.

To add 5, express it with a denominator of 2: $5 = \frac{10}{2}$.

The dimensions of the playground are:
Width: $\boxed{\frac{-5 + 5\sqrt{97}}{2}\,m}$
Length: $\boxed{\frac{5 + 5\sqrt{97}}{2}\,m}$

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