Question

Find an approximate value of √3
Correct to 2 decimal place using
Bisection method​

Answer 1

Using the bisection method we can find that the approximate value of $\sqrt3$ is 1.73 .

To find an approximate value of √3 correct to 2 decimal places using the bisection method, we need to find the root of the function $f(x) = x^2 - 3 = 0.$

First, we need to select an interval [a, b] that contains the root of the function. We can choose [1, 2], since $f(1) = -2$ and $f(2) = 1$, which means that the root is between 1 and 2.

Next, we will apply the bisection method to this interval. The midpoint of the interval is:

$c = (a + b) \div 2 = (1 + 2) \div 2 = 1.5$

We evaluate the function at the midpoint:

$f(c) = c^2 - 3 = 1.5^2 - 3 = -0.75$

Since f(c) is negative, the root is in the interval [1.5, 2]. We set $b = c$ and repeat the process.

The new midpoint of the interval is:

$c = (a + b) \div 2 = (1.5 + 2) \div 2 = 1.75$

We evaluate the function at the midpoint:

$f(c) = c^2 - 3 = 1.75^2 - 3 = 0.0625$

Since f(c) is positive, the root is in the interval [1.5, 1.75]. We set $a = c$ and repeat the process.

The new midpoint of the interval is:

$c = (a + b) \div2 = (1.5 + 1.75) \div2 = 1.625$

We evaluate the function at the midpoint:

$f(c) = c^2 - 3 = 1.625^2 - 3 = -0.3671875$

Since f(c) is negative, the root is in the interval [1.625, 1.75]. We set $b = c$ and repeat the process.

We can continue this process until the desired level of accuracy is achieved. To obtain an approximation correct to 2 decimal places, we can stop when the length of the interval is less than 0.01.

After a few more iterations (detailed table attached below), we obtain the interval [1.732421875, 1.7333984375], which has a length of 0.0009765625.

Therefore, an approximate value of √3 correct to 2 decimal places using the bisection method is: $\sqrt3 \approx 1.73 .$

Learn more about the bisection method at:'

https://brainly.in/question/16110641

https://brainly.in/question/24494459

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