Tentukan semua penyelesaian (x,y) bilangan real dari sistem persamaan berikut. {y=x−6​(x−3)2+y2=4​

Pertanyaan

Tentukan semua penyelesaian open parentheses x comma y close parentheses bilangan real dari sistem persamaan berikut.

open curly brackets table attributes columnalign left end attributes row cell y equals square root of x minus 6 end root end cell row cell left parenthesis x minus 3 right parenthesis squared plus y to the power of 2 end exponent equals 4 end cell end table close

D. Wahyu

Master Teacher

Mahasiswa/Alumni Universitas Negeri Semarang

Jawaban terverifikasi

Jawaban

solusi dari sistem persamaan ini adalah open parentheses square root of fraction numerator 5 plus square root of 29 over denominator 2 end fraction end root comma square root of fraction numerator negative 7 plus square root of 29 over denominator 2 end fraction end root close parentheses space space dan space space open parentheses square root of fraction numerator 5 minus square root of 29 over denominator 2 end fraction end root comma square root of fraction numerator negative 7 minus square root of 29 over denominator 2 end fraction end root close parentheses. 

Pembahasan

Diketahui:

y equals square root of x minus 6 end root y squared equals x minus 6 space... left parenthesis 1 right parenthesis  left parenthesis x minus 3 right parenthesis squared plus y to the power of 2 end exponent equals 4 space... left parenthesis 2 right parenthesis

Terlebih dahulu, subtitusikan persamaan (1) ke persamaan (2), sehingga

left parenthesis x minus 3 right parenthesis squared plus left parenthesis x minus 6 right parenthesis equals 4 x squared minus 6 x plus 9 plus x minus 6 minus 4 equals 0 x squared minus 5 x minus 1 equals 0 space... left parenthesis 3 right parenthesis

Untuk mendapatkan akar-akar x , dengan menggunakan rumus a comma b comma c sehingga

x subscript 1 comma 2 end subscript equals fraction numerator negative b plus-or-minus square root of b squared minus 4 a c end root over denominator 2 a end fraction

Dari persamaan (3), diperoleh :a equals 1 comma space b equals negative 5 comma space c equals negative 1, sehingga

table attributes columnalign right center left columnspacing 0px end attributes row cell x subscript 1 comma 2 end subscript end cell equals cell fraction numerator negative left parenthesis negative 5 right parenthesis plus-or-minus square root of left parenthesis negative 5 right parenthesis squared minus 4 left parenthesis 1 right parenthesis left parenthesis negative 1 right parenthesis end root over denominator 2 left parenthesis 1 right parenthesis end fraction end cell row cell x subscript 1 comma 2 end subscript end cell equals cell fraction numerator 5 plus-or-minus square root of 25 plus 4 end root over denominator 2 end fraction end cell row cell x subscript 1 comma 2 end subscript end cell equals cell fraction numerator 5 plus-or-minus square root of 29 over denominator 2 end fraction end cell row cell x subscript 1 end cell equals cell fraction numerator 5 plus square root of 29 over denominator 2 end fraction space comma space x subscript 2 equals fraction numerator 5 minus square root of 29 over denominator 2 end fraction end cell end table

Selanjutnya, subtitusikan nilai  x subscript 1 space dan space space x subscript 2 ke persamaan (1), sehingga

table attributes columnalign right center left columnspacing 0px end attributes row cell y subscript 1 end cell equals cell square root of fraction numerator 5 plus square root of 29 over denominator 2 end fraction minus 6 end root end cell row blank equals cell square root of fraction numerator 5 minus 12 plus square root of 29 over denominator 2 end fraction end root end cell row blank equals cell square root of fraction numerator negative 7 plus square root of 29 over denominator 2 end fraction end root end cell row blank blank blank row blank blank blank row cell y subscript 2 end cell equals cell square root of fraction numerator 5 minus square root of 29 over denominator 2 end fraction minus 6 end root end cell row blank equals cell square root of fraction numerator 5 minus 12 minus square root of 29 over denominator 2 end fraction end root end cell row blank equals cell square root of fraction numerator negative 7 minus square root of 29 over denominator 2 end fraction end root end cell end table

Jadi, solusi dari sistem persamaan ini adalah open parentheses square root of fraction numerator 5 plus square root of 29 over denominator 2 end fraction end root comma square root of fraction numerator negative 7 plus square root of 29 over denominator 2 end fraction end root close parentheses space space dan space space open parentheses square root of fraction numerator 5 minus square root of 29 over denominator 2 end fraction end root comma square root of fraction numerator negative 7 minus square root of 29 over denominator 2 end fraction end root close parentheses. 

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