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

Note by Sakib Nazmus 3 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?

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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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