Question

find real of the equation x^3-2x-5=0 by regular falsi method correct to three decimal places​

Answer 1

Regula Falsi Method:

Step-by-step explanation:

Suppose $f(x)=x^{3} -2x-5$

Then, $f(2)=2^{3}-2(2)-5=8-4-5=-1$

And, $f(3)=3^{3}-2(3)-5=27-6-5=16$

Since, $f(2)$ and $f(3)$ are of opposite signs, the real roots lies between $x_{1}=2$ and $x_{2}=3$.

Therefore, the first approximation is obtained from$x_{3}=x_{2}-\frac{x_{2}-x_{1}}{f(x_{2})-f(x_{1})} f(x_{2})$

That means,

$x_{3}=3-\frac{3-2}{16-(-1)} (16)\\x_{3}=3-\frac{16}{17} \\x_{3}=3-0.9411\\x_{3}=2.058$

Therefore, $f(x_{3})=(2.058)^{3}-2(2.058)-5=8.716-4.116-5=-0.4$

Since, $f(x_{2})$ and $f(x_{3})$ are of opposite signs, the real roots lies between $x_{2}$ and $x_{3}$.

Now, the second approximation is given by $x_{4}=x_{3}-\frac{x_{3}-x_{2}}{f(x_{3})-f(x_{2})} f(x_{3})$

That means,

$x_{4}=2.058-\frac{2.058-3}{-0.4-16} (-0.4)\\x_{4}=2.058-(\frac{-0.942}{-16.4} )(-0.4)\\x_{4}=2.058-(\frac{0.3768}{-16.4} )\\x_{4}=2.058+0.0229\\x_{4}=2.081$

Therefore, $f(x_{4})=(2.081)^{3}-2(2.081)-5=-0.15$

Thus, the real root of the given equation, $x^{3}-2x-5=0$ correct to three decimal places is $2.081$