Question 45004
Let "w" represent the width of the path. After the path has been layed, the new dimensions of the garden area are:
{{{length = 30 -2w}}} (we must subtract 2w because there is a path at both ends of the lawn),
{{{width = 20 -2w}}}
Now the area is {{{Area=(30-2w)(20-2w)}}}
This area is equal to 400 sq ft:
{{{400=(30-2w)(20-2w)}}}
{{{400=600-60w-40w+4w^2}}}
{{{0=200-60w-40w+4w^2}}}
{{{4w^2-100w+200=0}}}
{{{w^2-25w+50=0}}}
Now use the quadratic solver {{{w = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
where
a = 1
b = -25
c = 50
To give:
{{{w =(25 +- sqrt( 25^2-200 ))/2   }}}
{{{w =(25 +- 20.61)/2   }}}
{{{w =(25 +- 20.61)/2   }}}
so w=2.191 or w=22.808.
22.808 ft seems ridiculously wide for a garden path, so the answer must be w=2.191ft.

I hope this helps.
P.S. I am trying to start up my own homework help website. I would be extremely grateful if you would e-mail me some feedback on the help you received to adam.chapman@student.manchester.ac.uk
Adam