OEQuote:
Call the uniform width x, and the outer dimensions of the picture together with the frame become 8 + 2x and 12 + 2x, because the frame’s width adds to both the top and bottom and the left and right of the picture. The full area of the picture together with frame is then (8 + 2x)(12+ 2x) = 96 + 16x + 24x + 4x^2 = 4x^2 + 40x + 96. This expression must equal 2 × 8 × 12 = 192, because the frame’s area is equal to the picture’s, implying that the area enclosed by the frame and picture together must equal twice the area of the picture alone. So 4x^2 + 40x + 96 = 192. Subtract 192 to obtain 4x^2 + 40x – 96 = 0, and divide by 4 on both sides: x^2 + 10x – 24 = 0.
The left side factors as (x + 12)(x – 2), since 12x – 2x = 10x, and so x = 2 or –12. Only x = 2, (D), works in context of the problem, because a length cannot be negative. You can also get this result using a trial and error approach with the answer choices. When x = 2, the outer dimensions become 12 × 16, for a total area of 192. This makes the picture’s area, 8 × 12 = 96, the same as the frame’s area, 192 – 96 = 96