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Use the iterative formula below to work out the value of [tex]x_3[/tex], starting with:

[tex]\[
\begin{aligned}
x_1 & = 9 \\
x_{n+1} & = \frac{14}{x_n^2 - 2}
\end{aligned}
\][/tex]

Give your answer to 2 decimal places.


Sagot :

To find the value of [tex]\( x_3 \)[/tex] using the given iterative formula, let's follow each step carefully:

1. Starting Value: [tex]\( x_1 = 9 \)[/tex].

2. Calculate [tex]\( x_2 \)[/tex] using the formula [tex]\( x_{n+1} = \frac{14}{x_n^2 - 2} \)[/tex]:
[tex]\[ x_2 = \frac{14}{x_1^2 - 2} \][/tex]
Substituting [tex]\( x_1 = 9 \)[/tex] into the formula:
[tex]\[ x_2 = \frac{14}{9^2 - 2} = \frac{14}{81 - 2} = \frac{14}{79} \approx 0.17721518987341772 \][/tex]

3. Calculate [tex]\( x_3 \)[/tex] using the formula [tex]\( x_{n+1} = \frac{14}{x_n^2 - 2} \)[/tex]:
[tex]\[ x_3 = \frac{14}{x_2^2 - 2} \][/tex]
Substituting [tex]\( x_2 \approx 0.17721518987341772 \)[/tex] into the formula:
[tex]\[ x_3 = \frac{14}{(0.17721518987341772)^2 - 2} \][/tex]
Find [tex]\( 0.17721518987341772^2 \)[/tex]:
[tex]\[ (0.17721518987341772)^2 \approx 0.031413611967418683 \][/tex]
Now, substitute this value back into the formula for [tex]\( x_3 \)[/tex]:
[tex]\[ x_3 = \frac{14}{0.031413611967418683 - 2} = \frac{14}{-1.968586388032581317} \approx -7.111671821585545 \][/tex]

4. Round [tex]\( x_3 \)[/tex] to 2 decimal places:
[tex]\[ x_3 \approx -7.11 \][/tex]

Therefore, the value of [tex]\( x_3 \)[/tex] to 2 decimal places is [tex]\(-7.11\)[/tex].