Answer :
To solve for [tex]\( x_3 \)[/tex] using the iterative formula [tex]\( x_{n+1} = \frac{18}{x_n^2 - 4} \)[/tex] with the initial value [tex]\( x_1 = 5 \)[/tex], follow these steps:
1. Calculate [tex]\( x_2 \)[/tex]:
- [tex]\( x_2 = \frac{18}{x_1^2 - 4} \)[/tex]
- Substitute [tex]\( x_1 = 5 \)[/tex] into the formula:
[tex]\[ x_2 = \frac{18}{5^2 - 4} = \frac{18}{25 - 4} = \frac{18}{21} \][/tex]
- Simplify:
[tex]\[ x_2 = \frac{18}{21} \approx 0.8571 \][/tex]
2. Calculate [tex]\( x_3 \)[/tex]:
- Use the iterative formula with [tex]\( x_2 \)[/tex]:
[tex]\[ x_3 = \frac{18}{x_2^2 - 4} \][/tex]
- First, find [tex]\( x_2^2 \)[/tex]:
[tex]\[ x_2^2 = (0.8571)^2 \approx 0.7347 \][/tex]
- Continue with the formula:
[tex]\[ x_3 = \frac{18}{0.7347 - 4} = \frac{18}{-3.2653} \][/tex]
- Simplify:
[tex]\[ x_3 \approx -5.51 \][/tex]
3. Round [tex]\( x_3 \)[/tex] to 2 decimal places:
- The value of [tex]\( x_3 \)[/tex] is already approximately -5.51, which is rounded to two decimal places.
So, the calculated value of [tex]\( x_3 \)[/tex] is
[tex]\[ x_3 \approx -5.51 \][/tex]
1. Calculate [tex]\( x_2 \)[/tex]:
- [tex]\( x_2 = \frac{18}{x_1^2 - 4} \)[/tex]
- Substitute [tex]\( x_1 = 5 \)[/tex] into the formula:
[tex]\[ x_2 = \frac{18}{5^2 - 4} = \frac{18}{25 - 4} = \frac{18}{21} \][/tex]
- Simplify:
[tex]\[ x_2 = \frac{18}{21} \approx 0.8571 \][/tex]
2. Calculate [tex]\( x_3 \)[/tex]:
- Use the iterative formula with [tex]\( x_2 \)[/tex]:
[tex]\[ x_3 = \frac{18}{x_2^2 - 4} \][/tex]
- First, find [tex]\( x_2^2 \)[/tex]:
[tex]\[ x_2^2 = (0.8571)^2 \approx 0.7347 \][/tex]
- Continue with the formula:
[tex]\[ x_3 = \frac{18}{0.7347 - 4} = \frac{18}{-3.2653} \][/tex]
- Simplify:
[tex]\[ x_3 \approx -5.51 \][/tex]
3. Round [tex]\( x_3 \)[/tex] to 2 decimal places:
- The value of [tex]\( x_3 \)[/tex] is already approximately -5.51, which is rounded to two decimal places.
So, the calculated value of [tex]\( x_3 \)[/tex] is
[tex]\[ x_3 \approx -5.51 \][/tex]
