x3 – 3x+ 0  in the interval [0,1] using the Regula Falsi Method(Method of False Position) up to two iterations.

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1 Answer

x^3-3x+1=0; x(x^2-3)=-1; x=-1/(x^2-3) or 1/(3-x^2).

Use iterative process, starting with x=0 on RHS:

1st iteration gives x=1/3;

2nd iteration gives x=1/(26/9)=9/26;

3rd iteration gives x=676/1947=0.347.
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