A field has a length 4 m longer than the width. Set up a quadratic equation to find the dimension of the field if it's area is \( 165 \mathrm{~m}^{2} \) and solve the quadratic equation to find the length and the width.

To find the dimensions of the field, we can define the width as \( w \) meters. Therefore, the length would be \( w + 4 \) meters. Given that the area is \( 165 \, \mathrm{m}^2 \), we can set up the equation: \[ w(w + 4) = 165 \] Expanding that gives: \[ w^2 + 4w - 165 = 0 \] Now, we will apply the quadratic formula: \[ w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-4 \pm \sqrt{16 + 660}}{2} = \frac{-4 \pm \sqrt{676}}{2} = \frac{-4 \pm 26}{2} \] Calculating the possible values for \( w \): 1. \( w = \frac{22}{2} = 11 \) 2. \( w = \frac{-30}{2} \) (not valid since width can't be negative) So, the width is \( 11 \, \mathrm{m} \), making the length \( 15 \, \mathrm{m} \). Therefore, the dimensions are 11 meters wide and 15 meters long. To ensure you don’t make common mistakes when solving such problems, always check to make sure your values make sense within the context of the problem. In this case, a negative width would be physically impossible. Write down your equations clearly and double-check your arithmetic to avoid errors when calculating the discriminant or when applying the quadratic formula – it’s a classic pitfall! For further exploration, you might enjoy delving into the world of quadratic equations through various sources. Check out Khan Academy, which provides excellent video tutorials and practice problems on quadratic equations and their applications. Another fun read could be "Algebra and Trigonometry" by Michael Sullivan, which provides practical applications and examples that can enhance your understanding of the subject while making learning engaging!