Question 165093
Let x = width of the path



With these type of problems, it helps to draw a picture. So draw two rectangles, one inside of the other, and label the inside rectangle's dimensions:





<img src="http://i150.photobucket.com/albums/s91/jim_thompson5910/Algebra%20dot%20com/rectangle_garden_path1.png" alt="Photobucket - Video and Image Hosting">



Now the label the width of the path "x" (denoted in red)


<img src="http://i150.photobucket.com/albums/s91/jim_thompson5910/Algebra%20dot%20com/rectangle_garden_path2.png" alt="Photobucket - Video and Image Hosting">


Since there are 2 "x" lengths per side, this means that you need to add "2x" to both the length and width of the inner rectangle to get the length and width of the outer rectangle. If this makes no sense at all, here's a visual:



<img src="http://i150.photobucket.com/albums/s91/jim_thompson5910/Algebra%20dot%20com/rectangle_garden_path4.png" alt="Photobucket - Video and Image Hosting">


So the length and width of the outer rectangle is {{{34+2x}}} and {{{10+2x}}} respectively. This means that for the outer rectangle {{{A=640}}} (the given area of both the walkway and the garden), {{{L=34+2x}}} and {{{W=10+2x}}}



Remember, the area of any rectangle is {{{A=LW}}}



{{{A=LW}}} Start with the area of a rectangle formula



{{{640=(34+2x)(10+2x)}}} Plug in {{{A=640}}}, {{{L=34+2x}}} and {{{W=10+2x}}}



{{{640=340+68x+20x+4x^2}}} FOIL



{{{0=340+68x+20x+4x^2-640}}} Subtract 640 from both sides.



{{{0=4x^2+88x-300}}} Combine like terms.



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=4}}}, {{{b=88}}}, and {{{c=-300}}}



Let's use the quadratic formula to solve for x.



{{{x = (-b +- sqrt( b^2-4ac ))/(2a)}}} Start with the quadratic formula



{{{x = (-(88) +- sqrt( (88)^2-4(4)(-300) ))/(2(4))}}} Plug in  {{{a=4}}}, {{{b=88}}}, and {{{c=-300}}}



{{{x = (-88 +- sqrt( 7744-4(4)(-300) ))/(2(4))}}} Square {{{88}}} to get {{{7744}}}. 



{{{x = (-88 +- sqrt( 7744--4800 ))/(2(4))}}} Multiply {{{4(4)(-300)}}} to get {{{-4800}}}



{{{x = (-88 +- sqrt( 7744+4800 ))/(2(4))}}} Rewrite {{{sqrt(7744--4800)}}} as {{{sqrt(7744+4800)}}}



{{{x = (-88 +- sqrt( 12544 ))/(2(4))}}} Add {{{7744}}} to {{{4800}}} to get {{{12544}}}



{{{x = (-88 +- sqrt( 12544 ))/(8)}}} Multiply {{{2}}} and {{{4}}} to get {{{8}}}. 



{{{x = (-88 +- 112)/(8)}}} Take the square root of {{{12544}}} to get {{{112}}}. 



{{{x = (-88 + 112)/(8)}}} or {{{x = (-88 - 112)/(8)}}} Break up the expression. 



{{{x = (24)/(8)}}} or {{{x =  (-200)/(8)}}} Combine like terms. 



{{{x = 3}}} or {{{x = -25}}} Simplify. 



So the possible answers are {{{x = 3}}} or {{{x = -25}}} 



However, since a negative width is not possible, this means that {{{x = -25}}} is NOT a solution.



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Answer:


So the solution is  {{{x = 3}}} which means that the width of the path is 3 meters.