Question 1030133
Let {{{ w }}} = the width of the pool
{{{ 2w }}} = the length of the pool
There is 2 ft of walkway on BOTH sides of
the pool, so add {{{ 4 }}} ft to both width and length
{{{ 1056 = ( w + 4 )*( 2w + 4 ) }}}
{{{ 2w^2 + 8w + 4w + 16 = 1056 }}}
{{{ 2w^2 + 12w + 16 = 1056 }}}
{{{ 2w^2 + 12w = 1040 }}}
{{{ w^2 + 6w = 520 }}}
Complete the square:
{{{ w^2 + 6w + ( 6/2 )^2 = 520 + ( 6/2 )^2 }}}
{{{ w^2 + 6w + 9 = 520 + 9 }}}
{{{ ( w + 3 )^2 = 529 }}}
{{{ ( w + 3 )^2 = 23^2 }}}
{{{ w + 3 = 23 }}}
{{{ w = 20 }}}
and
{{{ 2w = 40 }}}
(a)
The length is 40 ft 
The width is 20 ft
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(b)
The area of the walkway is:
{{{ 1056 - w*(2w) }}}
{{{ 1056 - 20*40 }}}
{{{ 1056 - 800 = 256 }}} ft2
and
{{{ 256*8 = 2048 }}}
The material costs $2,048 
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check:
{{{  ( w + 4 )*( 2w + 4 ) = 1056 }}}
{{{ ( 20 + 4 )*( 2*20 + 4 ) = 1056 }}}
{{{ 24*44 = 1056 }}}
{{{ 1056 = 1056 }}}
OK