3. REASON QUANTITATIVELY A rectangular box has dimensions of \( x \) inches, \( (x+5) \) inches, and \( (x-2) \) inches. If the volume of the box is \( 30 x \) cubic inches, explain how to find the dimensions of the box. Show your work. \[ (x+5)^{2}+(x-2)^{2}=30 x \]

To determine the dimensions of the rectangular box, we'll use the information provided: - **Dimensions of the box:** \( x \) inches, \( (x + 5) \) inches, and \( (x - 2) \) inches. - **Volume of the box:** \( 30x \) cubic inches. ### Step 1: Set Up the Volume Equation The volume \( V \) of a rectangular box is calculated by multiplying its length, width, and height. Given the dimensions, the volume can be expressed as: \[ V = x \times (x + 5) \times (x - 2) \] According to the problem, the volume is also equal to \( 30x \) cubic inches. Therefore, we can write the equation: \[ x \times (x + 5) \times (x - 2) = 30x \] ### Step 2: Simplify the Equation First, observe that both sides of the equation have a factor of \( x \). Assuming \( x \neq 0 \) (since a dimension of length 0 doesn't make physical sense), we can divide both sides by \( x \): \[ (x + 5)(x - 2) = 30 \] ### Step 3: Expand the Left Side Now, expand the left side of the equation using the distributive property (also known as the FOIL method for binomials): \[ (x + 5)(x - 2) = x \cdot x + x \cdot (-2) + 5 \cdot x + 5 \cdot (-2) \] \[ = x^2 - 2x + 5x - 10 \] \[ = x^2 + 3x - 10 \] So, the equation becomes: \[ x^2 + 3x - 10 = 30 \] ### Step 4: Form a Quadratic Equation Subtract \( 30 \) from both sides to set the equation to zero: \[ x^2 + 3x - 10 - 30 = 0 \] \[ x^2 + 3x - 40 = 0 \] ### Step 5: Solve the Quadratic Equation We can solve the quadratic equation \( x^2 + 3x - 40 = 0 \) using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] For our equation, \( a = 1 \), \( b = 3 \), and \( c = -40 \). Plugging these values in: \[ x = \frac{-3 \pm \sqrt{3^2 - 4 \times 1 \times (-40)}}{2 \times 1} \] \[ x = \frac{-3 \pm \sqrt{9 + 160}}{2} \] \[ x = \frac{-3 \pm \sqrt{169}}{2} \] \[ x = \frac{-3 \pm 13}{2} \] This gives us two potential solutions: 1. **Positive Solution:** \[ x = \frac{-3 + 13}{2} = \frac{10}{2} = 5 \] 2. **Negative Solution:** \[ x = \frac{-3 - 13}{2} = \frac{-16}{2} = -8 \] Since a physical dimension cannot be negative, we discard \( x = -8 \) and accept \( x = 5 \) inches. ### Step 6: Find the Other Dimensions Now that we've found \( x = 5 \) inches, we can find the other dimensions: 1. **Second Dimension:** \[ x + 5 = 5 + 5 = 10 \text{ inches} \] 2. **Third Dimension:** \[ x - 2 = 5 - 2 = 3 \text{ inches} \] ### Step 7: Verify the Volume To ensure our solution is correct, let's verify the volume with the found dimensions: \[ \text{Volume} = x \times (x + 5) \times (x - 2) = 5 \times 10 \times 3 = 150 \text{ cubic inches} \] According to the problem, the volume should be \( 30x = 30 \times 5 = 150 \) cubic inches. Both calculations agree, confirming that our solution is correct. ### **Final Dimensions of the Box:** - **Length:** \( 5 \) inches - **Width:** \( 10 \) inches - **Height:** \( 3 \) inches