Question

Using successive Bisection method find the second, third and fourth approximation of root of the equation $x^3+x^2-1$ in the interval $(0,1)$

• A. $0.75,1.875,0.8125$
• B. $1.75,0.875,0.8125$
• C. $0.75,0.875,0.8125$
• D. $0.75,0.875,1.8125$

Answer 1

Answer: (C)

$0.75,0.875,0.8125$

We have to find the second,third and fourth approximation of root of the equation $x^3+x^2-1=0$ in the interval $(0,1)$ using successive Bisection method.
$\text{Iteration 1: k=0}$
$c_0=\frac{a_0+b_0}{2}=\frac{0+1}{2}=0.5$
Since $f\left(c_0\right)f\left(a_0\right)=f\left(0.5\right)f\left(0\right)>0$
Therefore set $a_1=0.5,b_1=b_0$
$\text{Iteration 2: k=1}$
$c_1=\frac{a_1+b_1}{2}=\frac{0.5+1}{2}=0.75$
Since $f\left(c_1\right)f\left(a_1\right)=f\left(0.75\right)f\left(0.5\right)>0$
Therefore set $a_2=c_1,b_2=b_1$
$\text{Iteration 3: k=2}$
$c_2=\frac{a_2+b_2}{2}=\frac{0.75+1}{2}=0.875$
Since $f\left(c_2\right)f\left(a_2\right)=f\left(0.875\right)f\left(0.75\right)<0$
Therefore set $a_3=a_2,b_3=c_2$
$\text{Iteration 4: k=3}$
$c_3=\frac{a_3+b_3}{2}=\frac{0.75+0.875}{2}=0.8125$
Thus the second,third and fourth approximations are $0.75,0.875,0.8125$ respectively.