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Find a root of the equation $x^{3}+x-1=0$ correct to three decimal places, by the method of False Position.
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Solution:

$ \begin{aligned} &\longrightarrow \text { Here } f(x)=x^{3}+x-1=0 \\\\ &\text { At } x=0, f(0)=-1(\text {-ve) } \\\\ &x=1, f(1)=1(\text { tve }) \\ \end{aligned} $

$\therefore$, The root lies between 0 and 1.

$ \text { Take } \begin{array}{rl} x_{0}=0 & x_{1}=1 \\\\ f\left(x_{1}\right)=-1 & f\left(x_{1}\right)=1 \\ \end{array} $

sr.no $x_(i)$ $f(x_i)$ $x_(i-1)$ $ f(x_(i-1))$ $x_(i+1)$ $f(x_(i+1))$
1 1 1 0 -1 0.5 -0.375 < 0
2 1 1 0.5 -0.375 0.636 -0.1067 < 0
3 1 1 0.636 -0.1067 0.6711 -0.02665 < 0
4 1 1 0.6711 -0.02665 0.6796 -0.00652
5 1 1 0.6796 -0.00652 0.68168 -0.0015517
6 1 1 0.68168 -0.0013517 0.68217 -0.000378
7 1 1 0.68217 -0.000378 0.6822

The desired root correct to three decimal places is 0.6822.

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