The 2${}^{nd}$ approximation to a root of the equation x² - x - 1 = 0 in the interval (1, 2) by Bisection method will be:

Bisection Method: Finding the 2nd Approximation of a Root

The Bisection method is a popular numerical technique used to find the roots of a continuous function $f(x)$ within a specified interval. The fundamental principle of this method is to repeatedly divide the interval in half and then determine which sub-interval contains the root. This iterative process narrows down the range until the desired accuracy for the root is achieved. A key condition for applying the Bisection method is that the function must have opposite signs at the endpoints of the initial interval, ensuring that at least one root lies within that interval.

We are tasked with finding the 2nd approximation to a root of the equation $x^2-x-1=0$ in the interval $(1,2)$ using the Bisection method. Let's define our function as $f(x)=x^2-x-1$.

Initial Interval Check for the Bisection Method

Before proceeding with the Bisection method, it is crucial to confirm that a root indeed exists within the given interval $(1,2)$. This is done by evaluating the function $f(x)$ at both endpoints of the interval to check for a sign change.

- Let the initial interval be $[a_0,b_0]=[1,2]$.
- Calculate $f(a_0)$: $$f(1)=(1{)}^2-(1)-1=1-1-1=-1$$
- Calculate $f(b_0)$: $$f(2)=(2{)}^2-(2)-1=4-2-1=1$$

Since $f(1)=-1$ (a negative value) and $f(2)=1$ (a positive value), there is a sign change. As $f(x)$ is a continuous function, we can confidently say that a root of the equation $x^2-x-1=0$ lies within the interval $(1,2)$.

**First Approximation ($x_1$) using Bisection**

The first approximation, denoted as $x_1$, is calculated as the midpoint of the initial interval $[a_0,b_0]$.

- The general formula for the midpoint in the Bisection method is $x_k=\frac{a_{k-1}+b_{k-1}}{2}$.
- For the 1st approximation: $$x_1=\frac{a_0+b_0}{2}=\frac{1+2}{2}=\frac{3}{2}=1.5$$
- Next, we evaluate the function $f(x)$ at this first approximation $x_1$: $$f(x_1)=f(1.5)=(1.5{)}^2-(1.5)-1=2.25-1.5-1=0.75-1=-0.25$$

Interval Update for Second Approximation

To determine the new sub-interval that contains the root for the next iteration, we compare the sign of $f(x_1)$ with the signs of $f(a_0)$ and $f(b_0)$.

- We found $f(x_1)=f(1.5)=-0.25$ (negative).
- Recall $f(a_0)=f(1)=-1$ (negative).
- Recall $f(b_0)=f(2)=1$ (positive).

Since $f(1.5)$ is negative and $f(2)$ is positive, the root must lie in the interval between 1.5 and 2. This means we replace the old lower bound $a_0$ with $x_1$.

- Therefore, for the next iteration, our new interval becomes $[a_1,b_1]=[1.5,2]$.

**Second Approximation ($x_2$) using Bisection**

The second approximation, $x_2$, is calculated as the midpoint of the newly refined interval $[a_1,b_1]$.

- Using the midpoint formula for the second iteration: $$x_2=\frac{a_1+b_1}{2}=\frac{1.5+2}{2}=\frac{3.5}{2}=1.75$$

Thus, the 2nd approximation to the root of the equation $x^2-x-1=0$ in the interval $(1,2)$ by the Bisection method is 1.75.

Summary of Bisection Method Iterations

The steps performed for the Bisection method can be summarized in the table below:Based on the calculations, the 2nd approximation obtained using the Bisection method is 1.75.