## College Algebra (6th Edition)

$x = 3$
In accordance with the figure, we let $x$ be the measurement of the width of the path surrounding the garden. According to the exercise, the area of the combined measurements (both garden and path) measures 378 square meters, which we can write as the following equation: $$Area = 378 = (15 + 2x)(12 + 2x)$$ since we have the garden's width plus the path's width flanking both sides. To solve for $x$, we first use the FOIL method to expand the equation: $$378 = 15(12) + 15(2x) + 2x(12) + 2x(2x)$$ $$378 = 180 + 30x + 24x + 4x^2$$ $$378 = 180 +54x + 4x^2$$ $$0 = 4x^2 + 54x + 180 - 378$$ $$0 = 4x^2 + 54x - 198$$ which is a quadratic equation. Now, we can use the Quadratic Formula $x = \frac{-b\frac{+}{} \sqrt{b^2 - 4ac}}{2a}$ to solve the problem: $$x = \frac{-(54)\frac{+}{} \sqrt{(54)^2 - 4(4)(-198)}}{2(4)}$$ $$x = \frac{-54\frac{+}{} \sqrt{2,916 + 3,168}}{8}$$ $$x = \frac{-54\frac{+}{} \sqrt{6,084}}{8}$$ $$x = \frac{-54\frac{+}{} 78}{8}$$ $$x = 3$$ $$OR$$ $$x = -\frac{33}{2}$$ But since we're dealing with measurements, and negative values are not adequate, the only viable solution for the exercise is $x = 3$.