Let's solve the problem step by step.

1. Define the variables:

   • Let $w$ be the width of the frame.
   • The outside dimensions of the frame are 8 inches by 10 inches, so the picture inside the frame will be smaller due to the width of the frame.

2. Calculate the dimensions of the picture showing:

   • The actual width of the picture (inside the frame) will be $8 - 2w$ inches (since the frame reduces the width on both sides by $w$).
   • The actual height of the picture (inside the frame) will be $10 - 2w$ inches (since the frame reduces the height on both sides by $w$).

3. Set up the area equation:

   • The area of the picture showing is given as 48 square inches. The area of the picture is the product of its width and height, so: $(8 - 2w)(10 - 2w) = 48$

4. Expand and simplify the equation: $(8 - 2w)(10 - 2w) = 80 - 16w - 20w + 4w^2 = 80 - 36w + 4w^2$ Therefore, the equation becomes: $4w^2 - 36w + 80 = 48$ Simplifying further: $4w^2 - 36w + 32 = 0$ Divide the entire equation by 4: $w^2 - 9w + 8 = 0$

5. Solve the quadratic equation: We can factor this equation: $(w - 8)(w - 1) = 0$ So, the possible solutions for $w$ are: $w = 8 \quad \text{or} \quad w = 1$

6. Interpret the solution: Since the frame width $w$ must be smaller than the overall dimensions of the frame, $w = 8$ is not possible (the frame would completely cover the picture), so the only feasible solution is $w = 1$.

Thus, the width of the frame is 1 inch.

Would you like more details on any part of this solution?

Relative Questions:

1. How do you solve quadratic equations by factoring?
2. What are some alternative methods to solve quadratic equations?
3. How does changing the dimensions of the frame affect the area of the picture?
4. Can the quadratic formula always be used for problems like this one?
5. What happens if the picture frame's area is less than 48 square inches?

Tip: Always check the physical feasibility of solutions after solving mathematical equations!