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The positive root of $x^{2}-78.8=0$ after first approximation by Newton Raphson method assuming initial approximation to the root is 14, is
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Verified Answer
The correct answer is:
$9.814$
Here, $\quad x_{0}=14, f(x)=x^{2}-78.8$
and $f^{\prime}(x)=2 x$
$$
\begin{aligned}
\therefore \quad x_{1}=x_{0}-& \frac{f\left(x_{0}\right)}{f^{\prime}\left(x_{0}\right)} \\
&=14-\frac{(14)^{2}-(78.8)}{2 \times 14}=9.814
\end{aligned}
$$
and $f^{\prime}(x)=2 x$
$$
\begin{aligned}
\therefore \quad x_{1}=x_{0}-& \frac{f\left(x_{0}\right)}{f^{\prime}\left(x_{0}\right)} \\
&=14-\frac{(14)^{2}-(78.8)}{2 \times 14}=9.814
\end{aligned}
$$
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