use the bisection method to approximate, accurate to two decimal places, the value of the root of the given…

use the bisection method to approximate, accurate to two decimal places, the value of the root of the given function in the given interval.\n$f(x)=x^{2}+2x - 4$ on the interval $1,1.5$
Answer
Explanation:
Step1: Recall bisection - method principle
The bisection method finds the root of a continuous function $y = f(x)$ in an interval $[a,b]$ by repeatedly bisecting the interval and narrowing down the sub - interval that contains the root.
Step2: Initial function and interval
We have $f(x)=x^{2}+2x - 4$ and the interval $[a_1,b_1]=[1,1.5]$. First, calculate $f(1)=1^{2}+2\times1 - 4=1 + 2-4=-1$ and $f(1.5)=1.5^{2}+2\times1.5 - 4=2.25 + 3-4 = 1.25$.
Step3: First iteration
The mid - point $c_1=\frac{1 + 1.5}{2}=1.25$. Calculate $f(1.25)=1.25^{2}+2\times1.25 - 4=1.5625+2.5 - 4=-0.9375$. Since $f(1.25)$ and $f(1.5)$ have opposite signs, the new interval is $[1.25,1.5]$.
Step4: Second iteration
The mid - point $c_2=\frac{1.25 + 1.5}{2}=1.375$. Calculate $f(1.375)=1.375^{2}+2\times1.375 - 4=1.890625+2.75 - 4=0.640625$. Since $f(1.25)$ and $f(1.375)$ have opposite signs, the new interval is $[1.25,1.375]$.
Step5: Third iteration
The mid - point $c_3=\frac{1.25 + 1.375}{2}=1.3125$. Calculate $f(1.3125)=1.3125^{2}+2\times1.3125 - 4=1.72265625+2.625 - 4=0.34765625$. Since $f(1.25)$ and $f(1.3125)$ have opposite signs, the new interval is $[1.25,1.3125]$.
Step6: Fourth iteration
The mid - point $c_4=\frac{1.25 + 1.3125}{2}=1.28125$. Calculate $f(1.28125)=1.28125^{2}+2\times1.28125 - 4=1.6412109375+2.5625 - 4=0.2037109375$. Since $f(1.25)$ and $f(1.28125)$ have opposite signs, the new interval is $[1.25,1.28125]$.
Step7: Fifth iteration
The mid - point $c_5=\frac{1.25 + 1.28125}{2}=1.265625$. Calculate $f(1.265625)=1.265625^{2}+2\times1.265625 - 4=1.6017089844+2.53125 - 4=0.1329589844$. Since $f(1.25)$ and $f(1.265625)$ have opposite signs, the new interval is $[1.25,1.265625]$.
Step8: Sixth iteration
The mid - point $c_6=\frac{1.25 + 1.265625}{2}=1.2578125$. Calculate $f(1.2578125)=1.2578125^{2}+2\times1.2578125 - 4=1.58203125+2.515625 - 4=0.09765625$. Since $f(1.25)$ and $f(1.2578125)$ have opposite signs, the new interval is $[1.25,1.2578125]$.
Step9: Seventh iteration
The mid - point $c_7=\frac{1.25 + 1.2578125}{2}=1.25390625$. Calculate $f(1.25390625)=1.25390625^{2}+2\times1.25390625 - 4=1.57234375+2.5078125 - 4=0.07015625$. Since $f(1.25)$ and $f(1.25390625)$ have opposite signs, the new interval is $[1.25,1.25390625]$.
Step10: Eighth iteration
The mid - point $c_8=\frac{1.25 + 1.25390625}{2}=1.251953125$. Calculate $f(1.251953125)=1.251953125^{2}+2\times1.251953125 - 4=1.5673828125+2.50390625 - 4=0.0712890625$. Since $f(1.25)$ and $f(1.251953125)$ have opposite signs, the new interval is $[1.25,1.251953125]$.
Step11: Approximation to two decimal places
Rounding the mid - point of the final interval to two decimal places, we get $1.25$.
Answer:
$1.25$