Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 12 - Sequences; Induction; the Binomial Theorem - 12.1 Sequences - 12.1 Assess Your Understanding - Page 809: 96

Answer

$\approx 9.434$

Work Step by Step

We start with an approximation: $a_0=9$ Compute $a_1,a_2,a_3,a_4,a_5$ using the formula: $a_n=\dfrac{1}{2}\left(a_{n-1}+\dfrac{p}{a_{n-1}}\right)$, where $p=89$. $a_1=\dfrac{1}{2}\left(a_0+\dfrac{89}{a_0}\right)=\dfrac{1}{2}\left(4+\dfrac{89}{4}\right)\approx 9.4444444$ $a_2=\dfrac{1}{2}\left(a_1+\dfrac{89}{a_1}\right)=\dfrac{1}{2}\left(9.4444444+\dfrac{89}{9.4444444}\right)\approx 9.4339869$ $a_3=\dfrac{1}{2}\left(a_2+\dfrac{89}{a_2}\right)=\dfrac{1}{2}\left(9.4339869+\dfrac{89}{9.4339869}\right)\approx 9.4339811$ $a_4=\dfrac{1}{2}\left(a_3+\dfrac{89}{a_3}\right)=\dfrac{1}{2}\left(9.4339811+\dfrac{89}{9.4339811}\right)\approx 9.4339811$ $a_5=\dfrac{1}{2}\left(a_4+\dfrac{89}{a_4}\right)=\dfrac{1}{2}\left(9.4339811+\dfrac{89}{9.4339811}\right)\approx 9.4339811$ So $\sqrt {89}\approx 9.434$
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