Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 8 - Section 8.2 - The Quadratic Formula - Exercise Set - Page 610: 107

Answer

$x=2.42$ $meters$ 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.

Work Step by Step

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.
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