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Can you show how you get the value of \(\large\sqrt{868}\) without a calculator?

Note by Sakib Nazmus 8 months, 2 weeks ago

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You could try expressing it as a continued fraction

let,

\(\begin{align} x&= \sqrt{868}\\ \lfloor{x\rfloor}&=29\\ x^2-29^2&=27\\ (x+29)(x-29)&=27\\ x&=29+\dfrac{27}{(x+29)}\end{align}\)

Substituting for \(x\) repeatedly we get the continued fraction of \(x\),i.e.

\(x=29+\dfrac{27}{58+\dfrac{27}{58+\dfrac{27}{58+\dfrac{27}{58+_\ddots}}}} \)

You could truncate the expression at about 2 or 3 levels and end up with a fairly decent approximation.

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I'm really confused... how can you simplify it more? Square roots are irrational, right? Do you mean the actual value or just an approximation?

You may use differential calculus to find approximate value

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Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

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TopNewestYou could try expressing it as a continued fraction

let,

\(\begin{align} x&= \sqrt{868}\\ \lfloor{x\rfloor}&=29\\ x^2-29^2&=27\\ (x+29)(x-29)&=27\\ x&=29+\dfrac{27}{(x+29)}\end{align}\)

Substituting for \(x\) repeatedly we get the continued fraction of \(x\),i.e.

\(x=29+\dfrac{27}{58+\dfrac{27}{58+\dfrac{27}{58+\dfrac{27}{58+_\ddots}}}} \)

You could truncate the expression at about 2 or 3 levels and end up with a fairly decent approximation.

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I'm really confused... how can you simplify it more? Square roots are irrational, right? Do you mean the actual value or just an approximation?

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You may use differential calculus to find approximate value

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