Question 184471
A rectangular garden is to be surrounded by a walkway of constant width.
 The garden's dimensions are 15ft by 25ft. 
The total area, garden plus walkway, is to be 650ft^2. 
What must be the width of the walkway? 
:
Let x = width of the walkway
:
Draw a diagram of this and it will be apparent that the total area dimensions are:
(2x+15) by (2x+25) and that equals 650 sq/ft
:
A simple area equation
(2x+15)*(2x+25) = 650
FOIL and subtract 650 from both sides:
4x^2 + 50x + 30x + 375 - 650 = 0
:
4x^2 + 80x - 275 = 0; our old friend, the quadratic equation!
:
Use the quadratic formula to find x
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
In this equation, a=4, b=80, c=-275
{{{x = (-80 +- sqrt(80^2 - 4 * 4 * -275 ))/(2*4) }}} 
:
{{{x = (-80 +- sqrt(6400 + 4400 ))/(8) }}}
:
{{{x = (-80 +- sqrt(10800))/(8) }}}
Two solutions, but we are ignoring the negative solution
{{{x = (-80 +103.923)/(8) }}}
x = {{{23.923/8}}}
x = 2.99 ft wide is the walkway
:
:
Check solution in the area equation, add 2 * 2.99 = 5.98 to each dimension
30.98 + 20.98 = 649.96 ~ 650