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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written 2.6 years ago by

prajapatijaimin • 3.2k

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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