Answer
Both the length and width of the box measure 10 inches.
Work Step by Step
The exercise is asking us to find length and width of the open box, which is a square with 2x2 corners cut off. Therefore, we can say that both, length and width, measure $(x + 4)$. The flaps used to fold the box, consequently, measure 2 inches and as such, so does the height of the box once folded. We can now say that the volume of the box ($V_{box} = l \times w\times h$) is as follows: $$V = 200 in^3 = (x+4)(x+4)(2)$$ $$200 = 2(x+4)^2$$ $$200 = 2(x^2 + 8x + 16)$$ $$200 = 2x^2 + 16x + 32$$ $$0 = 2x^2 + 16x -168$$ $$0 = 2(x^2 + 8x - 84)$$$$0 = x^2 + 8x - 84$$ This is now a quadratic equation which can be factorized by finding two factors of 84 that, when subtracted, yield +8: $$0 = (x + 14)(x-6)$$ Since we're using measurements, negative values are not adequate, therefore the only viable solution for the exercise must be $$0=x-6$$ $$x = 6$$ And since the original exercise asks us to find the length and width of the box, we end by substituting in our original expression: $$(x + 4)$$ $$6 + 4 = 10$$
