Question

The third approximation of roots of $x^3-9x+1=0$ in the interval $(2,4)$ by the method of false position is?

• A. $8.23$
• B. $1.25$
• C. $2.85$
• D. $2.12$

Answer 1

The correct option is C $2.85$

Here, $x^3-9x+1=0$

Let $f\left(x\right)=x^3-9x+1$

First Iteration:

Here, $f\left(2\right)=-9<0$ and $f\left(4\right)=29>0$

Now, Root lies between $x_0=2$ and $x_1=4$

$x_2=x_0-f\left(x_0\right)\times \frac{x_1-x_0}{f\left(x_1\right)-f\left(x_0\right)}=2-(-9)\times \frac{4-2}{29-(-9)}=2.47368$

Second Iteration:

Here, $f\left(2.47368\right)=-6.1264$ and $f\left(2\right)=29>0$

Now, Root lies between $x_0=2.47368$ and $x_1=4$

$x_3=x_0-f\left(x_0\right)\times \frac{x_1-x_0}{f\left(x_1\right)-f\left(x_0\right)}=2.47-(-6.13)\times \frac{4-2.47}{29-(-6.13)}=2.73989$

Third Iteration:

Here, $f\left(2.73989\right)=-3.09067$ and $f\left(4\right)=29>0$

Now, Root lies between $x_0=2.73989$ and $x_1=4$

$x_4=x_0-f\left(x_0\right)\times \frac{x_1-x_0}{f\left(x_1\right)-f\left(x_0\right)}=2.74-(-3.09)\times \frac{4-2.74}{29-(-3.09)}=2.86125$