Let.....
x=width of deck around pool
Then the length and width of pool plus deck is...
L=2x+18......with width of deck "x" added to either side L
W=2x+9.......with width of deck "x" added to either side W
A=400ft2.......total area of pool plus deck, prob statement, given
A=L(W)
400=(2x+18)(2x+9)...substitute for "L" and "W"
400=4x2+54x+162.....expand using FOIL
0=2x2+27x-119......subtract 400 each side, divide both sides by 2
Now use the quadratic formula to solve for "x"......
-b±√(b2-4ac)
r= ----------------- where........
2a
a=2, b=27, c=-119
I will leave the substitution and simplification of the quadratic
equation to you. You should arrive at two values for "x"....
x=3.5
x=-17.....disregard this value as we are solving for distance and
distance cannot be negative.
So x=3.5ft is our solution. Lets check our answer by subsitiuting
our value into the dimension expressions and area equation....
L=2(3.5)+18=7+18=25ft
W=2(3.5)+9=7+9=16ft
W=2(3.5)+9=7+9=16ft
A=LxW
400=25(16)
400=400.......√check, our solution "X=3.5ft" is correct.
Always check your work!