By solving a quadratic equation, we will see that the width of the walkway is 3ft.
Remember that for a rectangle of length L and width W, the area is:
A = W*L
Now, we know that for our garden we have:
W = 8ft
L = 12ft
If we add a walkway of width x around the garden, then the new measures are:
W' = 8ft + 2x
L' = 12ft + 2x
So the area is:
A' = ( 8ft + 2x)*( 12ft + 2x)
And we know that it is equal to 192 ft^2, then:
192 ft^2 = ( 8ft + 2x)*( 12ft + 2x)
Now we can solve this for x.
192ft^2 = 96ft^2 + 4x^2 + 20ft*x
0 = 96ft^2 - 192ft^2 + 20ft*x + 4x^2
0 = -96ft^2 + 20ft*x + 4x^2
This is a quadratic equation, and the solutions are given by the Bhaskara's formula:
[tex]x = \frac{-20ft \pm \sqrt{(20ft)^2 - 4*4*(-96ft^2)} }{2*4} \\\\x = \frac{-20ft \pm 44ft}{8}[/tex]
We only take the positive solution, so we get:
x = (-20ft + 44ft)/8 = 3ft
So the width of the walkway is 3ft.
If you want to learn more about quadratic equations, you can read:
https://brainly.com/question/1214333
