Question

The third approximation of root of $x^3-x^2-1=0$ in the interval $(1,2)$ using successive bisection method is?

• A. $1.475$
• B. $1.375$
• C. $2.213$
• D. $1.564$

Answer 1

Answer: (B)

$1.375$

We have to find the third approximation of root of the equation $x^3-x^2-1=0$ in the interval $(1,2)$ using successive Bisection method.
$\text{Iteration 1: k=0}$
$c_0=\frac{a_0+b_0}{2}=\frac{1+2}{2}=1.5$
Since $f\left(c_0\right)f\left(a_0\right)=f\left(1.5\right)f\left(1\right)<0$
Therefore set $a_1=a_0,b_1=c_0$
$\text{Iteration 2: k=1}$
$c_1=\frac{a_1+b_1}{2}=\frac{1+1.5}{2}=1.25$
Since $f\left(c_1\right)f\left(a_1\right)=f\left(1.25\right)f\left(1\right)>0$
Therefore set $a_2=c_1,b_2=b_1$
$\text{Iteration 3: k=2}$
$c_2=\frac{a_2+b_2}{2}=\frac{1.25+1.5}{2}=1.375$
Thus the third approximation of the root is $1.375$ respectively.