Solving a <em>system of equations</em>, it is found that since the <u>quadratic equation has two positive roots</u>, they can be the values of the length and the width, and the design is possible.
The perimeter of a <u>rectangle of length l and width w</u> is given by:
![P = 2(l + w)](https://tex.z-dn.net/?f=P%20%3D%202%28l%20%2B%20w%29)
The area is:
![A = lw](https://tex.z-dn.net/?f=A%20%3D%20lw)
In this problem, perimeter of 63.5 m, hence:
![2l + 2w = 63.5](https://tex.z-dn.net/?f=2l%20%2B%202w%20%3D%2063.5)
![2l = 63.5 - 2w](https://tex.z-dn.net/?f=2l%20%3D%2063.5%20-%202w)
![l = 31.75 - w](https://tex.z-dn.net/?f=l%20%3D%2031.75%20-%20w)
Area of 225 m², hence:
![lw = 225](https://tex.z-dn.net/?f=lw%20%3D%20225)
![(31.75 - w)(w) = 225](https://tex.z-dn.net/?f=%2831.75%20-%20w%29%28w%29%20%3D%20225)
![w^2 - 31.75w + 225 = 0](https://tex.z-dn.net/?f=w%5E2%20-%2031.75w%20%2B%20225%20%3D%200)
Which is a quadratic equation with coefficients
.
Then:
![\Delta = b^2 - 4ac = (-31.75)^2 - 4(1)(225) = 108.0625](https://tex.z-dn.net/?f=%5CDelta%20%3D%20b%5E2%20-%204ac%20%3D%20%28-31.75%29%5E2%20-%204%281%29%28225%29%20%3D%20108.0625)
![w_1 = \frac{-b + \sqrt{\Delta}}{2a} = \frac{31.75 + \sqrt{108.0625}}{2} = 21.1](https://tex.z-dn.net/?f=w_1%20%3D%20%5Cfrac%7B-b%20%2B%20%5Csqrt%7B%5CDelta%7D%7D%7B2a%7D%20%3D%20%5Cfrac%7B31.75%20%2B%20%5Csqrt%7B108.0625%7D%7D%7B2%7D%20%3D%2021.1)
![w_2 = \frac{-b - \sqrt{\Delta}}{2a} = \frac{31.75 - \sqrt{108.0625}}{2} = 10.68](https://tex.z-dn.net/?f=w_2%20%3D%20%5Cfrac%7B-b%20-%20%5Csqrt%7B%5CDelta%7D%7D%7B2a%7D%20%3D%20%5Cfrac%7B31.75%20-%20%5Csqrt%7B108.0625%7D%7D%7B2%7D%20%3D%2010.68)
Since the <u>quadratic equation has two positive roots</u>, they can be the values of the length and the width, and the design is possible.
A similar problem is given at brainly.com/question/10489198