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Jumlah nilai  yang memenuhi sistem persamaan  adalah ...

Pertanyaan

Jumlah nilai begin mathsize 14px style x end style yang memenuhi sistem persamaan begin mathsize 14px style open curly brackets table attributes columnalign left end attributes row cell x squared plus x y minus y squared equals negative 4 end cell row cell x plus 2 y equals 2 end cell end table close end style adalah ... 

Pembahasan Soal:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell x squared plus x y minus y squared end cell equals cell negative 4 space...... open parentheses 1 close parentheses end cell row cell x plus 2 y end cell equals cell 2 space space........ open parentheses 2 close parentheses end cell end table end style 

Dari persamaan (2) diperoleh 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell x plus 2 y end cell equals 2 row cell 2 y end cell equals cell 2 minus x end cell row y equals cell fraction numerator 2 minus x over denominator 2 end fraction space....... space open parentheses 3 close parentheses end cell end table end style 

Substitusi persamanan (3) ke persamaan (1), diperoleh 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell x squared plus x y minus y squared end cell equals cell negative 4 end cell row cell x squared plus x open parentheses fraction numerator 2 minus x over denominator 2 end fraction close parentheses minus open parentheses fraction numerator 2 minus x over denominator 2 end fraction close parentheses squared end cell equals cell negative 4 end cell row cell x squared plus fraction numerator 2 x minus x squared over denominator 2 end fraction minus open parentheses fraction numerator 2 minus x over denominator 2 end fraction close parentheses squared end cell equals cell negative 4 end cell row cell x squared plus fraction numerator 2 x minus x squared over denominator 2 end fraction minus fraction numerator 4 minus 4 x plus x squared over denominator 4 end fraction end cell equals cell negative 4 end cell row cell 4 open parentheses x squared plus fraction numerator 2 x minus x squared over denominator 2 end fraction minus fraction numerator 4 minus 4 x plus x squared over denominator 4 end fraction close parentheses end cell equals cell 4 open parentheses negative 4 close parentheses end cell row cell 4 x squared plus 4 x minus 2 x squared minus 4 plus 4 x minus 4 x squared end cell equals cell negative 16 end cell row cell negative 2 x squared plus 8 x minus 4 end cell equals 0 end table end style 

Dengan menggunakan rumus jumlah akar-akar persamaan kuadrat, maka jumlah nilai undefined yang memenuhi sistem persamaan tersebut adalah 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row straight a equals cell negative 2 end cell row straight b equals 8 row cell x subscript 1 plus x subscript 2 end cell equals cell fraction numerator negative straight b over denominator straight a end fraction end cell row blank equals cell fraction numerator negative 8 over denominator negative 2 end fraction end cell row blank equals 4 end table end style 

Jadi jumlah nilai undefined yang memenuhi sistem persamaan tersebut adalah begin mathsize 14px style x subscript 1 plus x subscript 2 equals 4 end style

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 04 April 2021

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