Question

Solve x^3+x^2−10x+1=0 for the root lying between 0 and 1 by the method of false position. Perform two iterations only.

Answer 1

Answer on Question #85640 – Math – Quantitative Methods

Question

Solve $x^3 + x^2 - 10x + 1 = 0$ for the root lying between 0 and 1 by the method of false position. Perform two iterations only.

Solution

The regula falsi (false position) method:

$$c_n = \frac{f(a_n)b_n - f(b_n)a_n}{f(a_n) - f(b_n)}$$

At iteration number $n$, the number $c_n$ is calculated as above and then, if $f(a_n)$ and $f(c_n)$ have the same sign, set $a_{n+1} = c_n$ and $b_{n+1} = b_n$, otherwise set $b_{n+1} = c_n$ and $a_{n+1} = a_n$. This process is repeated until the root is approximated sufficiently well.

Step 1:

$$a_1 = 0, \quad b_1 = 1$$ $$f(a_1) = f(0) = 0^3 + 0^2 - 10 \cdot 0 + 1 = 1, \quad f(b_1) = f(1) = 1^3 + 1^2 - 10 \cdot 1 + 1 = -7$$ $$c_1 = \frac{f(a_1)b_1 - f(b_1)a_1}{f(a_1) - f(b_1)} = \frac{1 \cdot 1 - (-7) \cdot 0}{1 - (-7)} = \frac{1}{8}, \quad f(c_1) = \left(\frac{1}{8}\right)^3 + \left(\frac{1}{8}\right)^2 - 10 \left(\frac{1}{8}\right) + 1 = -\frac{119}{512}$$ $$b_2 = c_1 = \frac{1}{8}, \quad a_2 = a_1 = 0$$

Step 2:

$$a _ {2} = 0, \qquad b _ {2} = \frac {1}{8}$$ $$f (a _ {2}) = f (0) = 0 ^ {3} + 0 ^ {2} - 1 0 \cdot 0 + 1 = 1, \qquad f (b _ {2}) = f \left(\frac {1}{8}\right) = \left(\frac {1}{8}\right) ^ {3} + \left(\frac {1}{8}\right) ^ {2} - 1 0 \left(\frac {1}{8}\right) + 1 = - \frac {1 1 9}{5 1 2}$$ $$c _ {2} = \frac {f (a _ {2}) b _ {2} - f (b _ {2}) a _ {2}}{f (a _ {2}) - f (b _ {2})} = \frac {6 4}{6 3 1}, \qquad f (c _ {2}) \cong - 0. 0 0 3 \rightarrow b _ {3} = c _ {2} = \frac {6 4}{6 3 1}, \qquad a _ {3} = a _ {2} = 0$$

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