Answer:
![x^2 + 15x - 60 = 0](https://tex.z-dn.net/?f=x%5E2%20%2B%2015x%20-%2060%20%3D%200)
The actual dimension is 18.28 by 3.28
Step-by-step explanation:
Given
Dimension:
![(x + 15)\ by\ x](https://tex.z-dn.net/?f=%28x%20%2B%2015%29%5C%20by%5C%20x)
![Area = 60in^2](https://tex.z-dn.net/?f=Area%20%3D%2060in%5E2)
Required
Determine the quadratic equation and get the possible values of x
Solving (a): Quadratic Equation.
The cardboard is rectangular in shape.
Hence, Area is calculated as thus:
![Area = Length * Width](https://tex.z-dn.net/?f=Area%20%3D%20Length%20%2A%20Width)
![60= (x + 15) * x](https://tex.z-dn.net/?f=60%3D%20%28x%20%2B%2015%29%20%2A%20x)
Open Bracket
![60= x^2 + 15x](https://tex.z-dn.net/?f=60%3D%20x%5E2%20%2B%2015x)
Subtract 60 from both sides
![x^2 + 15x - 60 = 0](https://tex.z-dn.net/?f=x%5E2%20%2B%2015x%20-%2060%20%3D%200)
<em>Hence, the above represents the quadratic equation</em>
Solving (b): The actual dimension
First, we need to solve for x
This can be solved using quadratic formula:
![x = \frac{-b \± \sqrt{b^2 - 4ac}}{2a}](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7B-b%20%5C%C2%B1%20%5Csqrt%7Bb%5E2%20-%204ac%7D%7D%7B2a%7D)
Where
![a = 1](https://tex.z-dn.net/?f=a%20%3D%201)
![b = 15](https://tex.z-dn.net/?f=b%20%3D%2015)
![c = -60](https://tex.z-dn.net/?f=c%20%3D%20-60)
So:
![x = \frac{-b \± \sqrt{b^2 - 4ac}}{2a}](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7B-b%20%5C%C2%B1%20%5Csqrt%7Bb%5E2%20-%204ac%7D%7D%7B2a%7D)
![x = \frac{-15 \± \sqrt{15^2 - 4*1*-60}}{2*1}](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7B-15%20%5C%C2%B1%20%5Csqrt%7B15%5E2%20-%204%2A1%2A-60%7D%7D%7B2%2A1%7D)
![x = \frac{-15 \± \sqrt{225 + 240}}{2}](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7B-15%20%5C%C2%B1%20%5Csqrt%7B225%20%2B%20240%7D%7D%7B2%7D)
![x = \frac{-15 \± \sqrt{465}}{2}](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7B-15%20%5C%C2%B1%20%5Csqrt%7B465%7D%7D%7B2%7D)
![x = \frac{-15 \± \21.56}{2}](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7B-15%20%5C%C2%B1%20%5C21.56%7D%7B2%7D)
Split:
or ![x = \frac{-15 - 21.56}{2}](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7B-15%20-%2021.56%7D%7B2%7D)
or ![x = \frac{-36.36}{2}](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7B-36.36%7D%7B2%7D)
or ![x = -18.18](https://tex.z-dn.net/?f=x%20%3D%20-18.18)
But length can't be negative;
So:
![x = 3.28](https://tex.z-dn.net/?f=x%20%3D%203.28)
The actual dimensions:
is
![Length =3.28 +15](https://tex.z-dn.net/?f=Length%20%3D3.28%20%2B15)
![Length =18.28](https://tex.z-dn.net/?f=Length%20%3D18.28)
![Width = x](https://tex.z-dn.net/?f=Width%20%3D%20x)
![Width =3.28](https://tex.z-dn.net/?f=Width%20%3D3.28)
<em>The actual dimension is 18.28 by 3.28</em>