Question 922034
The measurements of the rectangular garden are
{{{x}}}= width, in feet, and {{{2x+15}}}= length in feet.
{{{drawing(350,100,-1.75,36.75,-0.5,10.5,
rectangle(0,0,35,10),locate(0.5,5.5,x),locate(16,10,2x+15)
)}}} .
The area of that garden in square feet is
{{{x(2x+15)=350}}}
{{{2x^2+15x=350}}}
{{{2x^2+15x-350=0}}}
We can easily solve this equation by factoring or by using the quadratic formula.
I prefer factoring.
Looking for factors of {{{2*(-350)=-700}}} that add up to {{{15}}} ,
I find that {{{35*(-20)=-700}}} and {{{35+(-20)=15}}} .
{{{2x^2+15x-350=0}}}--->{{{2x^2-20x+35x-350=0}}}--->{{{2x(x-10)+35(x-10)=0}}}--->{{{(2x+35)(x-10)=0}}}--->{{{system(x=10,"or",x=-35/2)}}}
As {{{x}}} is a measure of the width of the garden, in feet, {{{x=10}}} is the only acceptable solution.
So the width is 10 feet,
and since {{{2*10+15=35}}} ,
the length is 35 feet.
The best way to split the garden in 3 sections is with fencing sections parallel to the short sides:
{{{drawing(350,150,-1,37.5,-3,12,
rectangle(0,0,35,10),locate(0.5,5.5,10),
locate(16,-0.2,35),arrow(15.8,-1,0,-1),arrow(18.2,-1,35,-1),
rectangle(35/3,0,70/3,10),locate(11.7,5.5,10),
locate(23.6,5.5,10),locate(35.1,5.5,10)
)}}} .
From the drawing, I see that the gardener needs to set up fencing along the two 35-foot length and along the 4 10-foot sections.
The gardener needs
{{{4*10ft+2*35ft=40ft+70ft=highlight(110ft)}}} of fencing.