Question
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Quadratic Functions in Context Quick Check
1 of 51 of 5 Items
Question
Hassan wants to add a border to a photo he took before he frames it. The final area of the entire framed picture will be 96 square inches. The length of the picture is 10 inches, and the width is 6 inches. Which of the following is the width of the border?(1 point)
Responses
9 inches
9 inches
1 inch
1 inch
3 inches
3 inches
4 inches
Quadratic Functions in Context Quick Check
1 of 51 of 5 Items
Question
Hassan wants to add a border to a photo he took before he frames it. The final area of the entire framed picture will be 96 square inches. The length of the picture is 10 inches, and the width is 6 inches. Which of the following is the width of the border?(1 point)
Responses
9 inches
9 inches
1 inch
1 inch
3 inches
3 inches
4 inches
Answers
GPT-4o mini
To find the width of the border, we start by determining the area of the entire framed picture and then subtract the area of the photo itself.
1. Calculate the area of the photo:
- Length of the photo = 10 inches
- Width of the photo = 6 inches
- Area of the photo = length × width = 10 inches × 6 inches = 60 square inches.
2. The area of the entire framed picture is 96 square inches.
3. The area of the border can be calculated by subtracting the area of the photo from the area of the entire framed picture:
- Area of the border = Area of the framed picture - Area of the photo
- Area of the border = 96 square inches - 60 square inches = 36 square inches.
4. Let \( w \) be the width of the border. When you add a border of width \( w \) around the photo, the new dimensions of the entire framed picture will be:
- Length = 10 + 2w
- Width = 6 + 2w
5. The area of the entire framed picture can also be expressed in terms of \( w \):
- Area = (10 + 2w)(6 + 2w).
6. Set the two expressions for the area equal to each other:
\[
(10 + 2w)(6 + 2w) = 96
\]
7. Expanding the left side:
\[
60 + 20w + 12w + 4w^2 = 96 \\
4w^2 + 32w + 60 = 96
\]
8. Simplifying the equation:
\[
4w^2 + 32w + 60 - 96 = 0 \\
4w^2 + 32w - 36 = 0
\]
9. Dividing the entire equation by 4:
\[
w^2 + 8w - 9 = 0
\]
10. Now we can factor the quadratic:
\[
(w + 9)(w - 1) = 0
\]
11. Setting each factor equal to zero gives:
\[
w + 9 = 0 \quad \Rightarrow \quad w = -9 \quad (\text{not valid}) \\
w - 1 = 0 \quad \Rightarrow \quad w = 1 \quad (\text{valid})
\]
Thus, the width of the border is **1 inch**.
1. Calculate the area of the photo:
- Length of the photo = 10 inches
- Width of the photo = 6 inches
- Area of the photo = length × width = 10 inches × 6 inches = 60 square inches.
2. The area of the entire framed picture is 96 square inches.
3. The area of the border can be calculated by subtracting the area of the photo from the area of the entire framed picture:
- Area of the border = Area of the framed picture - Area of the photo
- Area of the border = 96 square inches - 60 square inches = 36 square inches.
4. Let \( w \) be the width of the border. When you add a border of width \( w \) around the photo, the new dimensions of the entire framed picture will be:
- Length = 10 + 2w
- Width = 6 + 2w
5. The area of the entire framed picture can also be expressed in terms of \( w \):
- Area = (10 + 2w)(6 + 2w).
6. Set the two expressions for the area equal to each other:
\[
(10 + 2w)(6 + 2w) = 96
\]
7. Expanding the left side:
\[
60 + 20w + 12w + 4w^2 = 96 \\
4w^2 + 32w + 60 = 96
\]
8. Simplifying the equation:
\[
4w^2 + 32w + 60 - 96 = 0 \\
4w^2 + 32w - 36 = 0
\]
9. Dividing the entire equation by 4:
\[
w^2 + 8w - 9 = 0
\]
10. Now we can factor the quadratic:
\[
(w + 9)(w - 1) = 0
\]
11. Setting each factor equal to zero gives:
\[
w + 9 = 0 \quad \Rightarrow \quad w = -9 \quad (\text{not valid}) \\
w - 1 = 0 \quad \Rightarrow \quad w = 1 \quad (\text{valid})
\]
Thus, the width of the border is **1 inch**.