Ade Malsasa Akbar contact
Senior author, Open Source enthusiast.
Tuesday, February 7, 2017 at 00:20

# LibreOffice Math: Square Root Equation Examples

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## Example 1

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nroot{n}{a} = a^{ 1 over n }
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nroot{ n }{ a^m } = a^{ m over n }
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sqrt{ 9 } = 3 ...
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## Example 2

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nroot{ n }{ a } times nroot{ n }{ b }  = nroot{ n }{ a times b }
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nroot{ n }{ a } over { nroot{n}{b} } = nroot{ n }{ a over b }
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p nroot{ n }{ a } +- q nroot{ n }{ a } = ( p +- q ) nroot{ n }{ a }
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## Example 3

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sqrt{ 54 } = sqrt{ 9 cdot 6 } = sqrt{ 9 } cdot sqrt{ 6 } = 3 sqrt{ 6 }
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sqrt{ 72 } = sqrt{ 36 cdot 6 } = sqrt{ 36 } cdot sqrt{ 2 } = 6 sqrt{ 2 }
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sqrt{ 2 over 25 } = sqrt{ 2 } over sqrt{ 25 } = sqrt{ 2 } over 5
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nroot{ 3 }{ 128 } = nroot{ 3 }{ 64 } = nroot{ 3 }{ 64 } cdot nroot{ 3 }{ 2 } = 4 nroot{ 3 }{ 2 }
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## Example 4

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sqrt{ 45 } + 3 sqrt{ 20 } - 5 sqrt{ 5 } = 3 sqrt{ 5 } + 3( 2 sqrt{5} ) - 5 sqrt{ 5 }
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( 2 sqrt{3} + sqrt{2} )( 3 sqrt{3} - 5 sqrt{2} ) = 6 cdot 3 - 10 sqrt{ 6 } + 3 sqrt{ 6 } - 5 cdot 2
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## Example 5

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sqrt{ x } = x^{ 1 over 2 }
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nroot{ 3 }{ 5 } = 5^{ 1 over 3 }
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nroot{ 4 }{ p^4 } = p^{ 3 over 4 }
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nroot{ 5 }{ a^10 } = a^{ 10 over 5 } = a^2
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## Example 6

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( x^2 ){ 1 over 3 } = x{ 2 over 3 } = nroot{ 3 }{ x^2 }
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( 6p ){ 3 over 4 } = nroot{ 4 }{ (6p)^3 } = nroot{ 4 }{ 6^3 p^3 }
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3x^{ 2 over 5 }y^{ 3 over 5 } = 3 nroot{ 5 }{ x^2 y^3 }
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## Example 7

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{ a }over{ sqrt{b} } = { a }over{ sqrt{b} } times {{ sqrt{b} }over{ sqrt{b} }} = a over b { sqrt{b} }
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3 over sqrt{ 6 } = 3 over sqrt{ 6 } times { {sqrt{6}}over{sqrt{6}} } = 3 over 6 { sqrt{6} } = 1 over 2 { sqrt{6} }
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## Example 8

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a over { b + sqrt{c} } = a over { b + sqrt{c} } times { {b - sqrt{c}}over{b - sqrt{c}} } = a( b - sqrt{7} ) over { b^2 - c}
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a over { b - sqrt{c} } = a over { b - sqrt{c} } times { {b + sqrt{c}}over{b + sqrt{c}} } = a( b + sqrt{c} ) over { b^2 - c }
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## Example 9

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sqrt{ 1 + x } = sum from{ n=0 } to{ infinity } {{ (-1)^n (2n)! }over{ (1-2n)(n!)^2(4^n) } x^n} = 1 + 1 over 2 { x } - 1 over 8 { x^2 } + 1 over 16 { x^3 } - 5 over 128 { x^4 } + ...,
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sqrt{ i } = 1 over 2 { sqrt{2} } + i { 1 over 2 }{ sqrt{2} } = sqrt{ 2 } over 2 { (1 + i) }
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i = cos( %pi over 2 ) + i sin( %pi over 2 )
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## Example 10

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sqrt{ z } = sqrt{ r } e^{ i %ivarphi/2 } .
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sqrt{ r(cos %ivarphi + i sin %ivarphi)  } = sqrt{ r } [ cos %ivarphi over 2 + i sin {%ivarphi over 2} ]
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