Empma u2 ecuaciones cubicas

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π‘₯ 3 + 𝑏π‘₯ 2 + 𝑐π‘₯ + 𝑑 = 0 ,

π‘₯ =π‘¦βˆ’

𝑦 3 + 𝑝𝑦 + π‘ž = 0 , 𝑦

𝑝=π‘βˆ’

𝑏2 , 3

π‘ž=π‘‘βˆ’

𝑝

π‘ž

𝑏𝑐 2𝑏 3 + . 3 27

𝑦1 , 𝑦2 , 𝑦3

π‘₯1 = 𝑦1 βˆ’

𝑏 , 3

π‘₯2 = 𝑦2 βˆ’

𝑏 , 3

𝑦 3 + 𝑝𝑦 + π‘ž = 0

𝑦=π‘§βˆ’

𝑝 , 3𝑧

𝑏 π‘₯3 = 𝑦3 βˆ’ . 3

𝑏 3


𝑧3 βˆ’

𝑝3 27𝑧 3

+π‘ž =0

𝑧3 𝑧 6 + π‘žπ‘§ 3 βˆ’

𝑝3 = 0. 27 𝑑 = 𝑧3

𝑑 2 + π‘žπ‘‘ βˆ’

π‘ž 𝑑 = 𝑧 3 = βˆ’ Β± βˆšπ‘…, 2

𝑝3 =0, 27

π‘ž 2 𝑝 3 𝑅 =( ) +( ) . 2 3

1 1 πœ” = βˆ’ + π‘–βˆš3, 2 2

1 1 πœ”2 = βˆ’ βˆ’ π‘–βˆš3 , 2 2

π‘ž 3 𝐴 = βˆšβˆ’ + βˆšπ‘… , 2

π‘ž 3 𝐡 = βˆšβˆ’ βˆ’ βˆšπ‘… , 2

𝑝

𝐴𝐡 = βˆ’ 3 𝑝 3

(βˆ’ 3)

𝑧 𝐴, πœ”π΄, πœ”2 𝐴, 𝐡, πœ”π΅, πœ”2 𝐡. 𝑝

βˆ’3


𝑝 𝐴𝐡 = βˆ’ , 3

𝑝 πœ”π΄ βˆ™ πœ”2 𝐡 = βˆ’ , 3

𝑝 πœ”2 𝐴 βˆ™ πœ”π΅ = βˆ’ . 3

𝑧

βˆ’ 𝑦.

𝑝 3𝑧

𝑦=π‘§βˆ’

𝑝 3𝑧

𝑦 𝑦2 = πœ”π΄ + πœ”2 𝐡,

𝑦1 = 𝐴 + 𝐡,

𝑦3 = πœ”2 𝐴 + πœ”π΅ ,

π‘ž 2 𝑝3 Ξ” =√ , βˆšπ‘… = √ + 4 27 108 Ξ” = 4𝑝3 + 27π‘ž2

Ξ”>0

Ξ”

βˆšπ‘… 𝑦1 𝑦2 , 𝑦3 𝑦1 = 𝐴 + 𝐡 , 𝑦2 = πœ”π΄ + πœ”2 𝐡 , 𝑦2 = πœ”2 𝐴 + πœ”π΅ .

Ξ”<0

βˆšπ‘…

𝑦1 = 2π‘Ž , 𝑦2 = βˆ’π‘Ž βˆ’ π‘βˆš3 , 𝑦2 = βˆ’π‘Ž + π‘βˆš3 . π‘Ž

𝑏 3

√𝐴 = π‘Ž + 𝑖𝑏

Ξ”=0

π‘ž 𝐴 = 𝐡 = βˆ’ ⁄2

3

√𝐡 = π‘Ž βˆ’ 𝑖𝑏


3 π‘ž 𝑦1 = 2βˆšβˆ’ ⁄2 , 3 π‘ž 𝑦2 = 𝑦3 = 2√ ⁄2 .

π‘ž=0

π‘₯ 3 + 6π‘₯ 2 + 2π‘₯ + 12 = 0 . 𝑏 = 6, 𝑐 = 2

𝑑 = 12

𝑏2 62 𝑝=π‘βˆ’ =2βˆ’ = βˆ’10, 3 3

𝑝

π‘ž

𝑏𝑐 2𝑏 3 6 βˆ™ 2 2 βˆ™ 63 π‘ž=π‘‘βˆ’ + = 12 βˆ’ + = 24, 3 27 3 27

𝑦 3 βˆ’ 10𝑦 + 24 = 0 , Ξ” Ξ” = 4𝑝3 + 27π‘ž2 = 4 βˆ™ (βˆ’10)3 + 27 βˆ™ 242 = 11552 , Ξ”

βˆšπ‘… = √108 𝑦1 𝑦2 , 𝑦3

3 π‘ž 24 2888 3 βˆ’108 + 38√6 3 𝐴 = βˆšβˆ’ + βˆšπ‘… = βˆšβˆ’ + √ =√ , 2 2 27 9


3 π‘ž 24 2888 3 βˆ’108 βˆ’ 38√6 𝐡 = βˆšβˆ’ βˆ’ βˆšπ‘… = βˆšβˆ’ βˆ’ √ =√ . 2 2 27 9 3

3 βˆ’108 + 38√6 3 βˆ’108 βˆ’ 38√6 𝑦1 = 𝐴 + 𝐡 = √ +√ = βˆ’4 . 9 9

3 3 1 1 βˆ’108 + 38√6 1 1 βˆ’108 βˆ’ 38√6 𝑦2 = πœ”π΄ + πœ”2 𝐡 = (βˆ’ + π‘–βˆš3) √ + (βˆ’ βˆ’ π‘–βˆš3) √ , 2 2 9 2 2 9 1 1 = (1 βˆ’ π‘–βˆš3)(6 βˆ’ √6) + (1 + π‘–βˆš3)(6 + √6) , 6 6 = 2 + π‘–βˆš2 . 3 3 1 1 βˆ’108 + 38√6 1 1 βˆ’108 βˆ’ 38√6 𝑦2 = πœ”2 𝐴 + πœ”π΅ = (βˆ’ βˆ’ π‘–βˆš3) √ + (βˆ’ + π‘–βˆš3) √ , 2 2 9 2 2 9 1 1 = (1 + π‘–βˆš3)(6 βˆ’ √6) + (1 βˆ’ π‘–βˆš3)(6 + √6) , 6 6 = 2 βˆ’ π‘–βˆš2.

π‘₯ 3 + 6π‘₯ 2 + 2π‘₯ + 12 = 0 π‘₯1 = 𝑦1 βˆ’ 2 = βˆ’6 , π‘₯2 = 𝑦2 βˆ’ 2 = π‘–βˆš2 , π‘₯3 = 𝑦3 βˆ’ 2 = βˆ’π‘–βˆš2 .


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