Question 11 A rectangular garden bed has an area of 40 square feet. The length of the garden bed is 6 feet more than the width. What are the dimensions of the garden bed? Important: Be sure to submit in order from least to greatest! Length \( =\underline{\text { Blank } 1} \) Width \( =\underline{\text { Blank } \mathbf{2}} \)

To solve for the dimensions of the garden bed, let's denote the width as \( w \) feet. Since the length is 6 feet more than the width, we can express the length as \( w + 6 \) feet. The area of the rectangle can be represented by the equation: \[ w \times (w + 6) = 40 \] Expanding this gives: \[ w^2 + 6w - 40 = 0 \] Using the quadratic formula, \( w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 6 \), and \( c = -40 \): 1. Calculate the discriminant: \[ b^2 - 4ac = 6^2 - 4 \cdot 1 \cdot (-40) = 36 + 160 = 196 \] 2. Find the width: \[ w = \frac{-6 \pm \sqrt{196}}{2 \cdot 1} = \frac{-6 \pm 14}{2} \] This results in two possible widths: \[ w = \frac{8}{2} = 4 \quad \text{(width)} \quad \text{and} \quad w = \frac{-20}{2} = -10 \quad \text{(not valid)} \] 3. The width is 4 feet, and substituting back to find the length gives: \[ l = w + 6 = 4 + 6 = 10 \text{ feet} \] Finally, the dimensions of the garden bed are: Length \( = \underline{10} \) Width \( = \underline{4} \)