Mathematics, 29.07.2019 06:00 davidoj13

Arectangular swimming pool is bordered by a concrete patio. the width of the patio is the same on every side. the area of the surface of the pool is equal to the area of the patio. what is the width of the patio?

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$x = \frac{1}{4}\left(-(l + w) + \sqrt{l^2 + 6lw + w^2} \right)$

where

l = length of the pool (w/o the patio)
w = width of the pool (w/o the patio)

Explanation:

Let

x = width of the patio
l = length of the pool (w/o the patio)
w = width of the pool (w/o the patio)

Since the pool is bordered by a complete patio,

Length of the pool (with the patio)
= (length of the pool (w/o the patio)) + 2*(width of the patio)
Length of the pool (with the patio) = l + 2x

Width of the pool (with the patio)
= (width of the pool (w/o the patio)) + 2*(width of the patio)
Width of the pool (with the patio) = w + 2x

Note that

Area of the pool (w/o the patio)
= (length of the pool (w/o the patio))(width of the pool (w/o the patio))
Area of the pool (w/o the patio) = lw

Area of the pool (with the patio)
= (length of the pool (w/o the patio))(width of the pool (w/o the patio))
= (l + 2x)(w + 2x)
= w(l + 2x) + 2x(l + 2x)
= lw + 2xw + 2xl + 4x²
Area of the pool (with the patio) = 4x² + 2x(l + w) + lw

Area of the patio
= (Area of the pool (with the patio)) - (Area of the pool (w/o the patio))
= (4x² + 2x(l + w) + lw) - lw
Area of the patio = 4x² + 2x(l + w)

Since the area of the patio is equal to the area of the surface of the pool, the area of the patio is equal to the area of the pool without the patio. In terms of the equation,

Area of the patio = Area of the pool (w/o the patio)
4x² + 2x(l + w) = lw
4x² + 2x(l + w) - lw = 0 (1)

Let

a = numerical coefficient of x² = 4
b = numerical coefficient of x = 2(l + w)
c = constant term = -lw

Then using quadratic formula, the roots of the equation 4x² + 2x(l + w) - lw = 0 is given by

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\\ = \frac{-2(l + w) \pm \sqrt{(2(l + w))^2 - 4(4)(-lw)}}{2(4)} 
\\ = \frac{-2(l + w) \pm \sqrt{(4(l + w)^2) + 16lw}}{8} 
\\ = \frac{-2(l + w) \pm \sqrt{(4(l^2 + 2lw + w^2) + 4(4lw)}}{8}
\\ = \frac{-2(l + w) \pm \sqrt{(4(l^2 + 2lw + w^2 + 4lw)}}{8}
\\ = \frac{-2(l + w) \pm \sqrt{(4(l^2 + 6lw + w^2)}}{8}$
$= \frac{-2(l + w) \pm 2\sqrt{l^2 + 6lw + w^2}}{8} \\= \frac{2}{8}(-(l + w) \pm \sqrt{l^2 + 6lw + w^2}) \\x = \frac{1}{4}(-(l + w) \pm \sqrt{l^2 + 6lw + w^2}) \\\boxed{x = \frac{1}{4}\left(-(l + w) + \sqrt{l^2 + 6lw + w^2} \right) \text{ or }}
\\\boxed{x = -\frac{1}{4}\left((l + w) + \sqrt{l^2 + 6lw + w^2} \right)}$

Since $(l + w) + \sqrt{l^2 + 6lw + w^2} \ \textgreater \ 0$ , $-\frac{1}{4}\left((l + w) + \sqrt{l^2 + 6lw + w^2}\right)$ is negative. Since x represents the patio width, x cannot be negative. Hence, the patio width is given by

$\boxed{x = \frac{1}{4}\left(-(l + w) + \sqrt{l^2 + 6lw + w^2} \right)}$

Answer from: Quest

3 is greater then both of those numbers

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2 apples

step-by-step explanation:

taking away equals subtraction. 6-4=2

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