SOLUTION: Please help me. I do part of the equation and then I am just lost. A rectangular garden is to be surrounded by a walkway of constant width. The garden's dimensions are 30 ft by

Question 57978: Please help me. I do part of the equation and then I am just lost.
A rectangular garden is to be surrounded by a walkway of constant width. The garden's dimensions are 30 ft by 40 ft. The total area, garden plus walkway, is to be 1800 ft2. What must be the width of the walkway to the nearest thousandth?
I have done
(40+2x)(30+2x)=1800ft^2
1200+80x+60x+4x^2=1800ft^2
4x^2+140x-1284 = 0(Ugh)
2x^2+70x-642=0
x=-70+/-sqrt70^2-4(2)(642)/2(2)
Am I right so far? Now I don't understand what comes next.
Please help

Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
A rectangular garden is to be surrounded by a walkway of constant width. The garden's dimensions are 30 ft by 40 ft. The total area, garden plus walkway, is to be 1800 ft2. What must be the width of the walkway to the nearest thousandth?
I have done
(40+2x)(30+2x) = 1800 ft^2: You made the width = x, this is right
:
1200 + 80x + 60x + 4x^2 = 1800 ft^2: You FOILed, and this is right
:
4x^2 + 140x + 1200 - 1800 = 0; here you should have this:
:
4x^2 + 140x - 600 = 0; last term should be 600
:
Simplify, divide by 4:
x^2 + 35x - 150 = 0;
:
Unfortunately, it will not readily factor so the we will have to use the quadratic formula: a = 1; b = 35; c = -150; solve for x
:
This comes out to: x = 3.860 & - 38.86
:
Obviously the solution; x = 3.86 ft, is the width of the path
:
We can check this:
[30 + 2(3.86)] * [40 + 2(386)] =
[30 + 7.72] * [40 + 7.72] =
37.72 * 47.72 = 1799.9984 ~ 1800
:
Anyway you had the right idea, just a bump in th road threw you off a little bit, hope this helps

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