The walkway is 1.5 m wide.
The area of the pool is 12(6) = 72 m².
Adding a walkway of unknown width, x, around all 4 sides of the pool increases the width by 2x and the length by 2x; thus the area of the entire pool and walkway together would be given by
(12+2x)(6+2x)
We know that the area of just the walkway is 9 m² less than the area of the pool. This means that:
(12+2x)(6+2x)-72 = 72-9
Multiplying through we have:
12*6+12*2x+2x*6+2x*2x - 72 = 63
72 + 24x + 12x + 4x² - 72 = 63
24x + 12x + 4x² = 63
36x + 4x² = 63
Writing in standard form we have:
4x² + 36x = 63
We want to set it equal to 0 to solve, so subtract 63 from both sides:
4x² + 36x - 63 = 63 - 63
4x² + 36x - 63 = 0
Using the quadratic formula,
[tex]x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}
\\
\\=\frac{-36\pm \sqrt{36^2-4(4)(-63)}}{2(4)}
\\
\\=\frac{-36\pm \sqrt{1296--1008}}{8}=\frac{-36\pm \sqrt{1296+1008}}{8}
\\
\\=\frac{-36\pm \sqrt{2304}}{8}=\frac{-36\pm 48}{8}=\frac{-36+48}{8}\text{ or }\frac{-36-48}{8}
\\
\\=\frac{12}{8} \text{ or }\frac{-84}{8}=1.5 \text{ or }-10.5[/tex]
Since a negative width makes no sense, the walkway is 1.5 m wide.