Answer: 1unit.
Step-by-step explanation:
Area of the rectangle
A = length × width
Since the area = 20 and the width is x and the length is x + 1.
We now substitute for the values in the above formula and solve for x
20 = x × x + 1
20 = x( x + 1 ), we now open
20 = x² + x , then re arrange.
x² + x - 20 = 0, this is a quadratic equation.we solve for x using quadratic means
Here, I am going to solve using factorization by grouping
x² + x - 20 = 0
x² + 5x - 4x - 20 = 0
x( x + 5 ) - 4( x + 5 ) = 0
( x + 5 )( x - 4 ). = 0
Therefore, the solution for x will be
x = -5 or 4, but x can not be -5 ( negative), so x = 4.
Now the difference between width and the length can easily be calculated from the above,
Width = 4 (x) , and length = 5 (4 + 1),
Now difference will be
5 - 4 = 1unit.
![54, a_2, a_3, 128 \\ \\ a_1=54 \\ a_4=128 \\ \\ a_n=a_1 \times r^{n-1} \\ a_4=a_1 \times r^3 \\ 128=54 \times r^3 \\ \frac{128}{54}=r^3 \\ \frac{128 \div 2}{54 \div 2}=r^3 \\ \frac{64}{27}=r^3 \\ \sqrt[3]{\frac{64}{27}}=\sqrt[3]{r^3} \\ \frac{\sqrt[3]{64}}{\sqrt[3]{27}}=r \\ r=\frac{4}{3}](https://tex.z-dn.net/?f=54%2C%20a_2%2C%20a_3%2C%20128%20%5C%5C%20%5C%5C%0Aa_1%3D54%20%5C%5C%0Aa_4%3D128%20%5C%5C%20%5C%5C%0Aa_n%3Da_1%20%5Ctimes%20r%5E%7Bn-1%7D%20%5C%5C%0Aa_4%3Da_1%20%5Ctimes%20r%5E3%20%5C%5C%0A128%3D54%20%5Ctimes%20r%5E3%20%5C%5C%0A%5Cfrac%7B128%7D%7B54%7D%3Dr%5E3%20%5C%5C%20%5Cfrac%7B128%20%5Cdiv%202%7D%7B54%20%5Cdiv%202%7D%3Dr%5E3%20%5C%5C%0A%5Cfrac%7B64%7D%7B27%7D%3Dr%5E3%20%5C%5C%0A%5Csqrt%5B3%5D%7B%5Cfrac%7B64%7D%7B27%7D%7D%3D%5Csqrt%5B3%5D%7Br%5E3%7D%20%5C%5C%0A%5Cfrac%7B%5Csqrt%5B3%5D%7B64%7D%7D%7B%5Csqrt%5B3%5D%7B27%7D%7D%3Dr%20%5C%5C%0Ar%3D%5Cfrac%7B4%7D%7B3%7D)
