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Diketahui (2,1) merupakan salah satu solusi dari sistem persamaan berikut. { x 2 − y 2 = 3 x 2 − a y − y 3 = − 1 ​ Carilah solusi lainnya!

Diketahui (2,1) merupakan salah satu solusi dari sistem persamaan berikut.

Carilah solusi lainnya!

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D. Wahyu

Master Teacher

Mahasiswa/Alumni Universitas Negeri Semarang

Jawaban terverifikasi

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solusi adalah .

solusi adalah left parenthesis square root of 19 comma 2 right parenthesis space space dan space space left parenthesis negative square root of 19 comma 2 right parenthesis.

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Pembahasan

Diketahui: Terlebih dahulu, mensubtitusikan persamaan (3) ke persamaan (2), sehingga Selanjutnya, nilai disubtitusikan persmaan (2), sehingga Subtitusikan persamaan (3) ke persamaan (1), sehingga Selanjutnya, mendapatkan nilai dengan mensubtitusikan nilai ke pesamaan (2) . sehingga Jadi, solusi adalah .

Diketahui:

table attributes columnalign right center left columnspacing 0px end attributes row cell x squared minus y squared end cell equals cell 3 space... left parenthesis 1 right parenthesis end cell row cell x squared minus a y minus y cubed end cell equals cell negative 1 space... left parenthesis 2 right parenthesis end cell row cell left parenthesis x comma y right parenthesis end cell equals cell left parenthesis 2 comma 1 right parenthesis space... left parenthesis 3 right parenthesis end cell end table

Terlebih dahulu, mensubtitusikan persamaan (3) ke persamaan (2), sehingga

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 squared minus a open parentheses 1 close parentheses minus left parenthesis 1 right parenthesis cubed end cell equals cell negative 1 end cell row cell 4 minus a minus 1 end cell equals cell negative 1 end cell row cell 3 minus a end cell equals cell negative 1 end cell row cell negative a end cell equals cell negative 1 minus 3 end cell row cell negative a end cell equals cell negative 4 space left parenthesis Kedua space ruas space dikali space left parenthesis negative right parenthesis right parenthesis end cell row a equals cell 4 space end cell end table

Selanjutnya, nilai a disubtitusikan persmaan (2), sehingga

table attributes columnalign right center left columnspacing 0px end attributes row cell x squared minus 4 times y minus y cubed end cell equals cell negative 1 end cell row cell x squared minus 4 y minus y cubed end cell equals cell negative 1 end cell row cell x squared end cell equals cell negative 1 plus 4 y plus y cubed space... left parenthesis 3 right parenthesis end cell end table

Subtitusikan persamaan (3) ke persamaan (1), sehingga

table attributes columnalign right center left columnspacing 0px end attributes row cell x squared minus y squared end cell equals 3 row cell left parenthesis negative 1 plus 4 y plus y cubed right parenthesis minus y squared end cell equals 3 row cell y cubed minus y squared plus 4 y minus 1 minus 3 end cell equals 0 row cell y cubed minus y squared plus 4 y minus 4 end cell equals cell 0 space left parenthesis Kedua space ruas space dikali space left parenthesis negative right parenthesis right parenthesis end cell row cell left parenthesis negative y cubed right parenthesis plus left parenthesis y squared minus 4 y plus 4 right parenthesis end cell equals 0 row cell left parenthesis negative y right parenthesis cubed end cell equals cell 0 space comma space left parenthesis y minus 2 right parenthesis squared equals 0 space end cell row cell negative y end cell equals cell cube root of 0 end cell row cell negative y end cell equals 0 row y equals 0 row blank blank blank row cell left parenthesis y minus 2 right parenthesis squared end cell equals cell 0 space left parenthesis y minus 2 right parenthesis equals square root of 0 end cell row cell left parenthesis y minus 2 right parenthesis end cell equals 0 row y equals 2 end table

Selanjutnya, mendapatkan nilai x dengan mensubtitusikan nilai  y subscript 1 space space space dan space space space y subscript 2 ke pesamaan (2) . sehingga

table attributes columnalign right center left columnspacing 0px end attributes row cell Ketika space y end cell equals cell 0 comma space maka end cell row cell x squared end cell equals cell negative 1 plus left parenthesis 0 right parenthesis squared plus 4 left parenthesis 0 right parenthesis plus left parenthesis 0 right parenthesis cubed x squared equals negative 1 end cell row x equals cell plus-or-minus square root of negative 1 end root end cell row blank blank cell x space tidak space punya space solusi end cell end table

Ketika space y equals 2 comma space maka x squared equals negative 1 plus left parenthesis 2 right parenthesis squared plus 4 left parenthesis 2 right parenthesis plus 2 cubed x squared equals negative 1 plus 4 plus 8 plus 8 x squared equals 19 x equals plus-or-minus square root of 19 x subscript 1 equals square root of 19 space space comma space x subscript 2 equals negative square root of 19

Jadi, solusi adalah left parenthesis square root of 19 comma 2 right parenthesis space space dan space space left parenthesis negative square root of 19 comma 2 right parenthesis.

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