Newton's method of approximation is used to approximate values.
The value of [tex]\mathbf{4\sqrt{73}}[/tex] is 34.17601498
Let the number be x.
So, we have:
[tex]\mathbf{x = 4\sqrt{73}}[/tex]
Square both sides
[tex]\mathbf{x^2 = 1168}[/tex]
Subtract 1168 from both sides
[tex]\mathbf{x^2 - 1168 = 0}[/tex]
Express as a function
[tex]\mathbf{f(x) = x^2 - 1168 }[/tex]
Set x1 to 73/2.
So, we have:
[tex]\mathbf{x_1 = \frac{73}{2} = 36.5}[/tex]
Where:
[tex]\mathbf{x_{n+1} = x_n- \frac{x_n^2-1168}{2x_n}}[/tex]
So, we have:
[tex]\mathbf{x_{1+1} = x_1- \frac{x_1^2-1168}{2x_1}}[/tex]
[tex]\mathbf{x_{2} = x_1- \frac{x_1^2-1168}{2x_1}}[/tex]
This gives
[tex]\mathbf{x_{2} = 36.5- \frac{36.5^2-1168}{2 \times 36.5}}[/tex]
[tex]\mathbf{x_{2} = 34.25}[/tex]
[tex]\mathbf{x_{3} = x_2- \frac{x_2^2-1168}{2x_2}}[/tex]
[tex]\mathbf{x_{3} = 34.25- \frac{34.25^2-1168}{2 \times 34.25}}[/tex]
[tex]\mathbf{x_{3} = 34.1760948905}[/tex]
[tex]\mathbf{x_{4} = x_3- \frac{x_3^2-1168}{2x_3}}[/tex]
[tex]\mathbf{x_{4} = 34.1760948905- \frac{34.1760948905^2-1168}{2\times 34.1760948905}}[/tex]
[tex]\mathbf{x_{4} = 34.1760149814}[/tex]
[tex]\mathbf{x_{5} = x_4- \frac{x_4^2-1168}{2x_4}}[/tex]
[tex]\mathbf{x_{5} = 34.1760149814- \frac{34.1760149814^2-1168}{2\times 34.1760149814}}[/tex]
[tex]\mathbf{x_{5} = 34.1760149813}[/tex]
The values of x4 and x5 are the same up till the 8th decimal place.
Hence, the value of [tex]\mathbf{4\sqrt{73}}[/tex] is 34.17601498
Read more about Newton's method of approximation at:
https://brainly.com/question/14279052
