## Intermediate Algebra for College Students (7th Edition)

$x=2.42$ $meters$ This also indicates that at least 2-meter-wide border around the pool can be done, in accordance with the zoning law, using the available tile.
The shaded region is the tile border with uniform width of $x$. Set up the equation. $A_{pool} =$ $LW$ $A_{pool} =$ $(12)(8)$ $A_{pool} =96$ $A_{tile border} =L_{tb}W_{tb}-A_{pool}$ $120 =(12+x)(8+x)-96$ $120 =96+24x+16x+4x^2-96$ $120 =40x+4x^2$ Divide both sides by $4$. The equation becomes: $30 =10x+x^2$ $x^2+10x-30=0$ Use the quadratic formula: $x = \frac{-b±\sqrt{b^2-4ac}}{2a}$ $a=1$, $b=10$, $c=-30$ $x = \frac{-10±\sqrt{10^2-(4⋅1⋅-30)}}{2⋅1}$ $x = \frac{-10±\sqrt{100-(-120)}}{2}$ $x = \frac{-10±\sqrt{100+120}}{2}$ $x = \frac{-10±\sqrt{220}}{2}$ $x = \frac{-10±\sqrt{4⋅55}}{2}$ $x = \frac{-10±2\sqrt{55}}{2}$ $x=-5±\sqrt {55}$ Since measurements cannot be negative, therefore: $x=-5+\sqrt {55}$ $x=2.42$ $meters$ This also indicates that at least 2-meter-wide border around the pool can be done, in accordance with the zoning law, using the available tile. 