Questions: Application of Quadratic Equations Question 5 of 11 (2 points) Question Attempt: 1 of Unlimited The height of a rectangular box is 8 ft. The length is 2 ft longer than thrice the width x. The volume is 680 ft^3. Part 1 of 4 (a) Write an equation in terms of x that represents the given relationship. The equation is 24x^2+16x-680=0. Part 2 of 4 (b) Solve the equation to find the dimensions of the given shape. The length is ft. Part 3 of 4 The width is ft.

Find the dimensions of the rectangular box given the volume and relationships between its dimensions.

Write the equation representing the relationship.

The height of the box is $8 \, \text{ft}$, and the length $L$ is $2 \, \text{ft}$ longer than thrice the width $x$. The volume $V$ is given by the equation:
$$V = L \cdot W \cdot H = (3x + 2) \cdot x \cdot 8 = 680$$
This simplifies to:
$$24x^2 + 16x - 680 = 0$$

Factor the polynomial.

The polynomial $24x^2 + 16x - 680$ can be factored as:
$$8 \left(x - 5\right) \left(3x + 17\right) = 0$$

Solve for $x$.

Setting each factor to zero gives:
\[ x - 5 = 0 \quad \Rightarrow \quad x = 5
\]
$$3x + 17 = 0 \quad \Rightarrow \quad x = -\frac{17}{3} \quad (\text{not valid since width cannot be negative})$$
Thus, the width is:
$$\boxed{x = 5} \, \text{ft}$$

Calculate the length.

Using the width to find the length:
$$L = 3x + 2 = 3(5) + 2 = 15 + 2 = 17 \, \text{ft}$$
Thus, the length is:
$$\boxed{L = 17} \, \text{ft}$$

The width of the box is $\boxed{5} \, \text{ft}$ and the length is $\boxed{17} \, \text{ft}$.

The width of the box is $\boxed{5} \, \text{ft}$.
The length of the box is $\boxed{17} \, \text{ft}$.