Let the width of the pen be denoted by $w$ feet. Since the length is 10 feet longer than the width, the length can be written as $w + 10$ feet.

The area of a rectangle is given by:

$\text{Area} = \text{length} \times \text{width}$

Given that the area of the pen is 300 square feet, we can set up the following equation:

$w \times (w + 10) = 300$

Now, expand the equation:

$w^2 + 10w = 300$

Subtract 300 from both sides to set the equation to zero:

$w^2 + 10w - 300 = 0$

This is a quadratic equation. We can solve it using the quadratic formula:

$w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

For the equation $w^2 + 10w - 300 = 0$, the coefficients are $a = 1$, $b = 10$, and $c = -300$. Plug these values into the quadratic formula:

$w = \frac{-10 \pm \sqrt{10^2 - 4(1)(-300)}}{2(1)}$

$w = \frac{-10 \pm \sqrt{100 + 1200}}{2}$

$w = \frac{-10 \pm \sqrt{1300}}{2}$

$w = \frac{-10 \pm 36.06}{2}$

Now, calculate the two possible values for $w$:

$w = \frac{-10 + 36.06}{2} \quad \text{or} \quad w = \frac{-10 - 36.06}{2}$

$w = \frac{26.06}{2} = 13.03 \quad \text{(positive solution)}$

$w = \frac{-46.06}{2} = -23.03 \quad \text{(negative solution, not possible since width cannot be negative)}$

Thus, the width of the pen is approximately 13 feet.

Now, the length is 10 feet longer than the width:

$\text{Length} = 13 + 10 = 23 \text{ feet}$

So, the dimensions of the pen are approximately:

• Width: 13 feet
• Length: 23 feet

Would you like more details or have any questions?

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Here are 5 questions related to this topic:

1. How do we derive the quadratic formula?
2. What are other methods to solve quadratic equations?
3. How would the problem change if the area was different?
4. What are the properties of quadratic functions?
5. Can we solve the same problem graphically?

Tip: Always check if a quadratic equation has real solutions by first evaluating the discriminant $b^2 - 4ac$. If the discriminant is negative, there are no real solutions.