Question 204074
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The problem states the gravel is enough for 192 sq. ft. ---> uniform width built around the garden.


Let us see the garden then,
{{{drawing(400,400,-7,8,4,-4,line(-5,2,5,2),line(5,2,5,-2),line(5,-2,-5,-2),line(-5,-2,-5,2),green(line(-5.7,2.5,5.7,2.5),green(line(5.7,2.5,5.7,-2.5)),green(line(5.7,-2.5,-5.7,-2.5)),green(line(-5.7,-2.5,-5.7,2.5)),green(arrow(-5,0,-5.7,0)),green(arrow(5,0,5.7,0)),green(arrow(0,2,0,2.5)),green(arrow(0,-2,0,-2.5))),red(locate(-5.4,0,x)),red(locate(5.3,0,x)),red(locate(0,2.15,x)),red(locate(0,-2.35,x)),green(locate(-.7,0,GARDEN)),green(locate(0,1.55,15ft)),green(locate(0,-1.75,15ft)),green(locate(-4.75,0,11ft)),green(locate(3.85,0,11ft)),red(locate(0,2.7,15+x+x)),red(locate(5.75,0,11+x+x)),green(locate(0,.5,A[1])),red(locate(0,-3,A[2])))}}}


As you can see, we need to SUBTACT {{{red(A[2])}}} to {{{green(A[1])}}} for us to get 192 sq. ft of gravel used for the uniform width.


It follows,
{{{red(A[2])-green(A[1])=192ft^2}}}
{{{(15+x+x)(11+x+x)-(15)(11)=192ft^2}}}
{{{(15+2x)(11+2x)-(15)(11)=192}}}
{{{cross(165)+22x+30x+4x^2-cross(165)=192}}}
{{{4x^2+52x-192=0}}}, where{{{system(a=4,b=52,c=-192)}}}


By Quadratic Eqn,
Solving discriminant, {{{b^2-4ac=52^2-4(4)(-192)=2704+3072=red(5776)}}}
Then,
{{{x= (-52+-sqrt(5776))/(2*4)=(-52+-76)/8}}}
{{{x=(-52-76)/8=-128/8=-16)}}},disregard (-)
{{{x=(-52+76)/8=24/8}}}
{{{highlight(red(x=3ft))}}}, (Answer) ---> the uniform width


Let us check,
{{{red(A[2])-green(A[1])=192ft^2}}}
{{{(15+red(3)+red(3))(11+red(3)+red(3))-(15)(11))=192ft^2}}}
{{{(21)(17)-(15)(11)=192}}}
{{{357-165=192}}}
{{{192ft^2=192ft^2}}}


Thank you,
Jojo</font>