Respuesta :
Answer:
Length = 18.099 ft
Width = 11.049 ft
Step-by-step explanation:
let the length of the field be x ft
and the width be y ft
as per the condition given in problem
x=2y-4 -----------(A)
Also the area is given as 200 sqft
Hence
xy=200
Hence from A we get
y(2y-4)=200
taking 2 as GCF out
2y(y-2)=200
Dividing both sides by 2 we get
[tex]y(y-2)=100[/tex]
[tex]y^2-2y=100[/tex]
subtracting 100 from both sides
[tex]y^2-2y-100=0[/tex]
Now we solve the above equation with the help of Quadratic formula which is given in the image attached with this for any equation in form
[tex]ax^2+bx+c=0[/tex]
Here in our case
a=1
b=-1
c=-100
Putting those values in the formula and solving them for y
[tex]y=\frac{-(-2)+\sqrt{(-2)^2-4 \times (-1) \times 100}}{2 \time 1}[/tex]
[tex]y=\frac{-(-2)-\sqrt{(-2)^2-4 \times (-1) \times 100}}{2 \time 1}[/tex]
Solving first
[tex]y=\frac{2+\sqrt{4+400}{2}[/tex]
[tex]y=\frac{2+\sqrt{404}{2}[/tex]
[tex]y=\frac{2+20.099}{2}[/tex]
[tex]y=\frac{22.099}{2}[/tex]
[tex]y=11.049[/tex]
Solving second one
[tex]y=\frac{-(-2)-\sqrt{(-2)^2-4 \times (-1) \times 100}}{2 \time 1}[/tex]
[tex]y=\frac{2-\sqrt{4+400}{2}[/tex]
[tex]y=\frac{2-\sqrt{404}{2}[/tex]
[tex]y=\frac{2-20.099}{2}[/tex]
[tex]y=\frac{-18.99}{2}[/tex]
[tex]y=-9.045[/tex]
Which is wrong as the width can not be in negative
Our width of the field is
y=11.099
Hence the length will be
x=2y-4
x=2(11.049)-4
x=22.099-4
x=18.099
Hence our length x and width y :
Length = 18.099 ft
Width = 11.049 ft
