Answer
Both length and width measure $5in$
Work Step by Step
The exercise describes a flat square carton that has 3x3 squares cut off from all corners to form flaps on all sides. If we arbitrarily assign the variable $x$ to describe the measurement of the flaps formed, we can express the length and width of the square carton as $(x + 6)$. The exercise further says that, when the flaps are folded, the carton becomes a box with volume $75$ $in^3$. Since the flaps were formed by cutting off 3x3 corners, we can deduce that the width of the flaps, and therefore, the height of the box, measures $3in$. We can now express the volume ($V_{box} = l \times w \times h$) of the box as follows: $$V_{box} = 75 = (x+6)(x+6)(3)$$ $$75 = 3(x+6)^2$$ $$75 = 3(x^2 + 12x + 36)$$ $$0 = 3x^2 + 36x + 108 - 75$$ $$0 = 3x^2 + 36x + 33$$ $$0 = 3(x^2 + 12x + 11)$$ $$0 = x^2 + 12x + 11$$ which is a quadratic equation that can be factorized by identifying two factors of 11 that, when added, give 12: $$0 = (x + 11)(x + 1)$$ $$x = -1$$ $$or$$ $$x = -11$$ By substituting in the values in the initial expression established ($x + 6$), we get the following possible solutions for length and width: $$(-1 + 6) = 5$$ $$or$$ $$(-11 + 6) = -5$$. Since we're dealing with measurements, we cannot use negative values. Therefore, the only possible solution is that both length and width measure $5in$
