# Thread: [SOLVED] Halving the interval (bisection)?

1. ## [SOLVED] Halving the interval (bisection)?

can someone help me with halving the interval, i know the theory, method, etc but i dont know how to arrive at the final answer when it asks to solve to two decimal places.
e.g Question:
Use the method of bisection to solve the following equation to 2 decimal places given that:
x^3 - 3x - 20 = 0 has a root between 3 and 3.1

thanks

2. Originally Posted by noobonastick
can someone help me with halving the interval, i know the theory, method, etc but i dont know how to arrive at the final answer when it asks to solve to two decimal places.
e.g Question:
Use the method of bisection to solve the following equation to 2 decimal places given that:
x^3 - 3x - 20 = 0 has a root between 3 and 3.1

thanks
Use a table: $f(x)=x^3-3x-20$

$1. f(3)=-2~~~~f(3.1)=0.491~~~~\longrightarrow~3

$2. f(3.05)=-0.777...~~~~f(3.1)=0.491~~~~\longrightarrow~3.05

$3. f(3.075)=-0.1489... ~~~~f(3.1)=0.491~~~~ \longrightarrow~ 3.075

Because $f(3.0875)=\bold{+0.165...}$ you now have to examine:

$f(3.08125)=\bold{+0.0099...}$ It is still positive. Therefore you have to examine now:

$4. f(3.078125)=-0.695...~~~~f(3.08125)=0.0099~~~~ \longrightarrow~3.078125

Thus the result to 2 dp is 3.08

By the way: A linear interpolation yields the same result in one step.

3. Thanks, so basically I have to keep halving until the upper and lower x values round off to the same two decimal places?

4. Originally Posted by noobonastick
Thanks, so basically I have to keep halving until the upper and lower x values round off to the same two decimal places?
Correct!