SOLUTION: A rectangular garden has dimensions of 15 feet by 11 feet. A gravel path of uniform width is to be built around the garden. How wide can the path be if there is enough gravel for 1

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: A rectangular garden has dimensions of 15 feet by 11 feet. A gravel path of uniform width is to be built around the garden. How wide can the path be if there is enough gravel for 1      Log On


   

Question 180189: A rectangular garden has dimensions of 15 feet by 11 feet. A gravel path of uniform width is to be built around the garden. How wide can the path be if there is enough gravel for 192 square feet?
Iam having a mental block


Answer by ptaylor(2198) About Me  (Show Source):
You can put this solution on YOUR website!
FIRST, DRAW THE PICTURE
Let x=width of the path
Then the outside dimensions of the rectangular garden plus path are:
Length =15+2x
Width=11+2x
Now we are told that Length*Width=192 sq ft or:
(2x+15)(2x+11)=192 expand the left side using FOIL
4x^2+22x+30x+165=192 subtract 192 from each side
4x^2+22x+30x+165-192=0 collect like terms
4x^2+52x-27=0 quadratic in standard form. Solve using the quadratic formula:
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+=
x+=+%28-52+%2B-+sqrt%28+52%5E2-4%2A4%2A%28-27%29+%29%29%2F%288%29+=
x+=+%28-52+%2B-+sqrt%28+2704%2B432+%29%29%2F%288%29+=
x+=+%28-52+%2B-+sqrt%28+3136+%29%29%2F%288%29+=
x+=+%28-52%2B+or+-56%29%2F%288%29+=
Disregard the negative value for x. Path widths are positive:
x+=+%28-52+%2B56%29%2F%288%29+=
x+=+%284%29%2F%288%29+= 0.5 ft---width of path
CK
(12)(16)= 192
192=192

I BET YOU COULD HAVE SOLVED THE QUADRATIC!!!!!!!!!!
HOPE THIS HELPS----------PTAYLOR