Question

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

• A. $2.430$
• B. $1.340$
• C. $1.430$
• D. $1.230$

Answer 1

The correct option is C $1.430$

Here, $x^3-x^2-1=0$

Let $f\left(x\right)=x^3-x^2-1$

First Iteration:

Here, $f\left(1\right)=-1<0$ and $f\left(2\right)=3>0$

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

$x_2=x_0-f\left(x_0\right)\times \frac{x_1-x_0}{f\left(x_1\right)-f\left(x_0\right)}=1-(-1)\times \frac{2-1}{3-(-1)}=$

Second Iteration:

Here, $f\left(1.25\right)=-0.60938<0$ and $f\left(2\right)=3>0$

Now, Root lies between $x_0=1.25$ and $x_1=2$

$x_3=x_0-f\left(x_0\right)\times \frac{x_1-x_0}{f\left(x_1\right)-f\left(x_0\right)}=1.25-(-0.61)\times \frac{2-1.25}{3-(-0.61)}=1.37662$

Third Iteration:

Here, $f\left(1.37662\right)=-0.28626<0$ and $f\left(2\right)=3>0$

Now, Root lies between $x_0=1.37662$ and $x_1=2$

$x_4=x_0-f\left(x_0\right)\times \frac{x_1-x_0}{f\left(x_1\right)-f\left(x_0\right)}=1.38-(-0.29)\times \frac{2-1.38}{3-(-0.29)}=1.43093$