1. Given that the width of the rectangle is x, and the area of the rectangle may be represented by the equation x^2 + 5x = 300, we can solve this equation for the width (x) as such:
x^2 + 5x = 300
x^2 + 5x - 300 = 0 (Subtract 300 from both sides)
(x - 15)(x + 20) = 0 (Factorise x^2 + 5x - 300)
From this, we get: x = 15 or x = -20
Since the width must be a positive length (ie. more than 0), -20 would be an invalid answer in the given context and thus the width is given by x = 15.
2. If we know that the length is 5 inches more than the width, we simply need to add 5 to the width we found above to obtain the length:
Length = x + 5
Length = 15 + 5 = 20
Thus, the width of the rectangle is 15 inches and the length of the rectangle is 20 inches.
Answer:
![\large\boxed{x=1-\sqrt5\ \vee\ x=1+\sqrt5}](https://tex.z-dn.net/?f=%5Clarge%5Cboxed%7Bx%3D1-%5Csqrt5%5C%20%5Cvee%5C%20x%3D1%2B%5Csqrt5%7D)
Step-by-step explanation:
![x^2-2x-4=0\qquad\text{add 4 to both sides}\\\\x^2-2x=4\\\\x^2-2(x)(1)=4\qquad\text{add}\ 1^2\ \text{to both sides}\\\\x^2-2(x)(1)+1^2=4+1^2\qquad\text{use}\ (a-b)^2=a^2-2ab+b^2\\\\(x-1)^2=4+1\\\\(x-1)^2=5\to x-1=\pm\sqrt5\qquad\text{add 1 to both sides}\\\\\boxed{x=1-\sqrt5}\ \vee\ \boxed{x=1+\sqrt5}](https://tex.z-dn.net/?f=x%5E2-2x-4%3D0%5Cqquad%5Ctext%7Badd%204%20to%20both%20sides%7D%5C%5C%5C%5Cx%5E2-2x%3D4%5C%5C%5C%5Cx%5E2-2%28x%29%281%29%3D4%5Cqquad%5Ctext%7Badd%7D%5C%201%5E2%5C%20%5Ctext%7Bto%20both%20sides%7D%5C%5C%5C%5Cx%5E2-2%28x%29%281%29%2B1%5E2%3D4%2B1%5E2%5Cqquad%5Ctext%7Buse%7D%5C%20%28a-b%29%5E2%3Da%5E2-2ab%2Bb%5E2%5C%5C%5C%5C%28x-1%29%5E2%3D4%2B1%5C%5C%5C%5C%28x-1%29%5E2%3D5%5Cto%20x-1%3D%5Cpm%5Csqrt5%5Cqquad%5Ctext%7Badd%201%20to%20both%20sides%7D%5C%5C%5C%5C%5Cboxed%7Bx%3D1-%5Csqrt5%7D%5C%20%5Cvee%5C%20%5Cboxed%7Bx%3D1%2B%5Csqrt5%7D)
6m+6n=-30
6m-5n=14
11n=-44
n=-4
m=-1
Ordered pair is (-1,-4)
Answer:
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it’s 100 all sides are even on a square