## Calculus with Applications (10th Edition)

Published by Pearson

# Chapter 12 - Sequences and Series - 12.6 Newton's Method - 12.6 Exercises - Page 652: 22

#### Answer

$$17.320$$

#### Work Step by Step

\eqalign{ & {\text{We can note that }}\sqrt {300} {\text{ is a solution of the equation }}{x^2} - 300 = 0.{\text{ }} \cr & {\text{Let }}f\left( x \right) = {x^2} - 300,\,\,\,\,\,\,f'\left( x \right) = 2x \cr & {\text{We have that }}{\left( {17} \right)^2} = 289{\text{ and }}{\left( {18} \right)^2} = 1324 \cr & {\text{Since 17 < }}\sqrt {300} < 18,{\text{ we can use }}{c_1} = 17{\text{ as the first approximation }} \cr & {\text{Then using Newton's method we have}}: \cr & {c_2} = {c_1} - \frac{{f\left( {{c_1}} \right)}}{{f'\left( {{c_1}} \right)}} = 17 - \frac{{{{\left( {15} \right)}^2} - 300}}{{2\left( {17} \right)}} = 17.3235 \cr & {c_3} = {c_2} - \frac{{f\left( {{c_2}} \right)}}{{f'\left( {{c_2}} \right)}} = 17.3235 - \frac{{{{\left( {17.3235} \right)}^2} - 300}}{{2\left( {17.3235} \right)}} = 17.3205 \cr & {\text{In the same way}} \cr & {c_4} = 17.3205 - \frac{{{{\left( {17.3205} \right)}^2} - 300}}{{2\left( {17.3205} \right)}} = 17.3205 \cr & {\text{Since }}{c_3} = {c_4} = 17.320,{\text{ to the nearest thousand}}{\text{, }}\sqrt {300} = 17.320 \cr}

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