Full solution

12\,\text{cm} \times 8\,\text{cm}

picture with a equal frame around it. The frame has an area of

64\,\text{cm}^2

. How wide is the frame?

Calculate Picture Area: Calculate the area of the picture. $\newline$ Area of picture = length $\times$ width = $12\,\text{cm}$ $\times$ $8\,\text{cm}$ = $96\,\text{cm}^2$ .
Calculate Total Area: Calculate the total area of the picture with the frame. $\newline$ Area of frame = $64\,\text{cm}^2$ (given). $\newline$ Total area = Area of picture + Area of frame = $96\,\text{cm}^2 + 64\,\text{cm}^2 = 160\,\text{cm}^2$ .
Find Frame Width: Let $x$ be the width of the frame. The dimensions of the new rectangle (picture + frame) will be $(12 + 2x)\,\text{cm}$ by $(8 + 2x)\,\text{cm}$ .
Set Up Equation: Set up the equation for the total area using the new dimensions. $\newline$ Total area = $(12 + 2x)(8 + 2x) = 160\,\text{cm}^2$ .
Expand Equation: Expand the equation. $96 + 24x + 16x + 4x^2 = 160$ .
Simplify Equation: Combine like terms and move all terms to one side to form a quadratic equation. $\newline$ $4x^2 + 40x + 96 - 160 = 0$ . $\newline$ $4x^2 + 40x - 64 = 0$ .
Factor Quadratic: Divide the entire equation by $4$ to simplify. $\newline$ $x^2 + 10x - 16 = 0$ .
Solve for $x$ : Factor the quadratic equation. $(x + 16)(x - 1) = 0.$
Final Frame Width: Solve for $x$ . $\newline$ $x + 16 = 0$ or $x - 1 = 0$ . $\newline$ $x = -16$ or $x = 1$ .
Final Frame Width: Solve for $x$ . $\newline$ $x + 16 = 0$ or $x - 1 = 0$ . $\newline$ $x = -16$ or $x = 1$ .Since a frame width can't be negative, the width of the frame is $1\,\text{cm}$ .