The length of a 200 square foot rectangular vegetable garden is 4feet less than twice the width. Find the length and width of the garden

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

Ver imagen Cricetus
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