A school is building a rectangular soccer field that has an area of 6,000 square yards. The soccer field must be 40 yards longer than its width. Determine algebraically the dimensions of the soccer field, in yards. [Only an algebraic solution!!!]

Here it is given that a school is building a field whose length is 40 yards more than its breadth. So if we assume the breadth to be x then its length will be x +40 .

[tex]\begin{picture}(10,6)\setlength{\unitlength}{1cm}\thicklines\put(0,0){\line(1,0){5}}\put(5,0){\line(0,1){3}}\put(5,3){\line(-1,0){5}}\put(0,3){\line(0,-1){3}}\put(2.5,-0.5){$\sf x +40$}\put(5.3,1.5){$\sf x $}\put(1,1){$\bf Area = 6000\ yard^2$}\end{picture}[/tex]

Now we know that the area of rectangle is ,

[tex]\longrightarrow Area = length\times breadth [/tex]

Substitute,

[tex]\longrightarrow x( x +40)= 6000 \\[/tex]

[tex]\longrightarrow x^2+40x = 6000\\[/tex]

[tex]\longrightarrow x^2+40x -6000=0\\ [/tex]

[tex]\longrightarrow x^2+100x -60x -6000=0\\ [/tex]

[tex]\longrightarrow x(x+100)-60(x+100)=0\\ [/tex]

[tex]\longrightarrow (x+100)(x-60)=0\\ [/tex]

[tex]\longrightarrow \underline{\underline{ x = 60 , -100}}[/tex]

As sides can't be negative, therefore,

[tex]\longrightarrow x = 60 [/tex]

Therefore,

[tex]\longrightarrow Length = x+40 = 100\ yards [/tex]
[tex]\longrightarrow Breadth = x = 60\ yards[/tex]

facundo3141592 facundo3141592 · Answer 2 · 2022-03-30T13:33:52+02:00

By solving a quadratic equation we will see that the width is 60yd and the length is 100yd.

How do find the dimensions of the soccer field?

Remember that for a rectangle of length L and width W, the area is:

A = L*W

In this case, we know that the length is 40 yards larger than the width, then we have:

L = W + 40yd

And the area is 6,000 yd^2

Then we can write:

6,000 yd^2 = (W + 40yd)*W

Now we can solve this for W:

6,000 yd^2 = W^2 + 40yd*W

This is a quadratic equation:

W^2 + 40yd*W - 6,000 yd^2 = 0

The solutions are given by Bhaskara's formula:

[tex]W = \frac{-40 \pm \sqrt{40^2 - 4*1*(-6000)} }{2*1} \\\\W = \frac{-40 \pm 160 }{2}[/tex]

We only care for the positive solution, as the other has no physical sense.

W = (-40yd + 160yd)/2 = 60 yd

And the length is 40yd more than that, so:

L = 60 yd + 40yd = 100yd

If you want to learn more about quadratic equations, you can read:

https://brainly.com/question/1214333

A school is building a rectangular soccer field that has an area of 6,000 square yards. The soccer field must be 40 yards longer than its width. Determine algebraically the dimensions of the soccer field, in yards. [Only an algebraic solution!!!]​

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How do find the dimensions of the soccer field?

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A school is building a rectangular soccer field that has an area of 6,000 square yards. The soccer field must be 40 yards longer than its width. Determine algebraically the dimensions of the soccer field, in yards. [Only an algebraic solution!!!]