Question

Using successive Bisection method find the second, third and fourth approximation of root of the equation $x^3-3x-5=0$ in the interval $(2,2.5)$

• A. $2.375,2.135\&2.2815$
• B. $1.25,1.375\&1.4375$
• C. $4.23,3.214\&2.135$
• D. $2.4475,2.175\&3.2815$

Answer 1

The correct option is A $2.375,2.135\&2.2815$

We have to find the second,third and fourth approximation of root of the equation $x^3-3x-5=0$ in the interval $(2,2.5)$ using successive Bisection method.
$\text{Iteration 1: k=0}$
$c_0=\frac{a_0+b_0}{2}=\frac{2+2.5}{2}=2.25$
Since $f\left(c_0\right)f\left(a_0\right)=f\left(2.25\right)f\left(2\right)>0$
Therefore set $a_1=2.25,b_1=b_0$
$\text{Iteration 2: k=1}$
$c_1=\frac{a_1+b_1}{2}=\frac{2.25+2.5}{2}=2.375$
Since $f\left(c_1\right)f\left(a_1\right)=f\left(2.375\right)f\left(2.25\right)<0$
Therefore set $a_2=a_1,b_2=c_1$
$\text{Iteration 3: k=2}$
$c_2=\frac{a_2+b_2}{2}=\frac{2.25+2.375}{2}=2.3125$
Since $f\left(c_2\right)f\left(a_2\right)=f\left(2.3125\right)f\left(2.25\right)<0$
Therefore set $a_3=a_2,b_3=c_2$
$\text{Iteration 4: k=3}$
$c_3=\frac{a_3+b_3}{2}=\frac{2.25+2.3125}{2}=2.28125$
Thus the second,third and fourth approximations are $2.375,2.3125,2.28125$ respectively.