Question

Using the Babylonian method find √600 ???????????? √235.

how?

Answer 1

With the Babylonian method
`√(600) \approx 24.494897` and
`√(235) \approx 15.329709`.

Explanation:

• To find the square root of 600, do the following:

1. Make an initial guess: Because
`√(576)=24` and
`√(625) =25` you can start with
`x_(0)=24` as your initial guess.

2. Apply the formula:

`x_(1)=((x_(0)+(x)/(x_(0))))/(2)` where
`x=600`

`x_(1)=((24+(600)/(24)))/(2)\\x_(1)=24.5`

The number
`x_(1)` is a better approximation to
`√(600)`

3. Iterate until convergence:

For this apply the formula:

`x_(n+1)=((x_(n)+(x)/(x_(n))))/(2)`

Convergence is achieved when the digits of
`x_(n+1)` and
`x_(n)` agree to as many decimal places as you desire.

`x_(2)=((x_(1)+(x)/(x_(1))))/(2)\\x_(2)=((24.5+(600)/(24.5)))/(2)\\x_(2)=24.494897`

`x_(3)=((x_(2)+(x)/(x_(2))))/(2)\\x_(3)=((24.494897+(600)/(24.494897)))/(2)\\x_(3)=24.494897\\`

Because
`x_(2)` and
`x_(3)` agree to six decimal places. we can say that an estimate for
`√(600) \approx 24.494897`.

You can compare with the value that WolframAlpha gives you which is
`√(600) \approx 24.49489742` you can see that it agrees to six decimal places.

• To find the square root of 235, do the following:

1. Make an initial guess: Because
`√(225)=15` and
`√(256) =16` you can start with
`x_(0)=15` as your initial guess.

2. Apply the formula:

`x_(1)=((x_(0)+(x)/(x_(0))))/(2)` where
`x=235`

`x_(1)=((15+(235)/(15)))/(2)\\x_(1)=(46)/(3)`

3. Iterate until convergence:

Apply the formula:

`x_(n+1)=((x_(n)+(x)/(x_(n))))/(2)`

`x_(2)=((x_(1)+(235)/(x_(1))))/(2)\\x_(2)=(((46)/(3)+(235)/((46)/(3))))/(2) \\x_(2)=15.329710`

`x_(3)=((x_(2)+(235)/(x_(2))))/(2)\\x_(3)=((15.329710+(235)/(15.329710)))/(2) \\x_(3)=15.329709`

`x_(4)=((x_(3)+(235)/(x_(3))))/(2)\\x_(4)=((15.329709+(235)/(15.329709)))/(2) \\x_(3)=15.329709`

Because
`x_(3)` and
`x_(4)` agree to six decimal places. We can say that an estimate for
`√(235) \approx 15.329709`.

WolframAlpha gives you
`√(235) \approx 15.329709716`