Question

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

• A. $1.78049$
• B. $1.276$
• C. $2.123$
• D. $0.726$

Answer 1

The correct option is A $1.78049$

Here, $f\left(x\right)=x^3-x-4=0$

first iteration:

]Here $f\left(1\right)=-4<0$ and $f\left(2\right)=2>0$

Now, root lies between $x_0=1$ and $x_1=2$

$x_2=x_0-f\left(x_0\right)\times \frac{x_1-x_0}{f\left(x_1\right)-f\left(x_0\right)}=1-(-4)\times \frac{2-1}{2-(-4)}=1.66667$

$f\left(x_2\right)=f\left(1.66667\right)=-1.03704<0$

2nd iteration:

Here $f\left(1.66667\right)=-1.03704<0$ and $f\left(2\right)=2>0$

Now, root lies between $x_0=1.66667$ and $x_1=2$

$x_3=x_0-f\left(x_0\right)\times \frac{x_1-x_0}{f\left(x_1\right)-f\left(x_0\right)}=1.67-(-1.04)\times \frac{2-1.67}{2-(-1.04)}=1.78049$