Persamaan kuadrat   mempunyai akar-akar x1​ dan x2​. Jika x1​>1 dan x2​< 1 maka ….

Pertanyaan

Persamaan kuadrat begin mathsize 14px style straight x squared plus 2 straight x plus open parentheses straight c plus 2 close parentheses equals 0 end style  mempunyai akar-akar begin mathsize 14px style straight x subscript 1 end style dan begin mathsize 14px style straight x subscript 2 end style. Jika begin mathsize 14px style straight x subscript 1 greater than 1 end style dan begin mathsize 14px style text x end text subscript 2 less than space 1 end style maka ….

  1. begin mathsize 14px style straight c less than negative 1 end style

  2. begin mathsize 14px style straight c less than negative 5 end style

  3. begin mathsize 14px style straight c greater than negative 5 end style

  4. begin mathsize 14px style negative 5 less than straight c less than negative 1 end style

  5. begin mathsize 14px style straight c less than negative 3 end style atau begin mathsize 14px style straight c greater than negative 1 end style

M. Haidar.

Master Teacher

Jawaban terverifikasi

Jawaban

jawabannya adalah B.

Pembahasan

Ingat kembali:

Jika undefined  memiliki akar-akar begin mathsize 14px style straight x subscript 1 end style dan begin mathsize 14px style straight x subscript 2 end style, maka berlaku:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell straight x subscript 1 plus straight x subscript 2 end cell equals cell negative straight b over straight a end cell row cell straight x subscript 1 straight x subscript 2 end cell equals cell straight c over straight a end cell end table end style 

Sehingga, pada persamaan begin mathsize 14px style straight x squared plus 2 straight x plus straight c plus 2 equals 0 end style  berlaku:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell straight x subscript 1 plus straight x subscript 2 end cell equals cell negative 2 end cell row cell straight x subscript 1 straight x subscript 2 end cell equals cell straight c plus 2 end cell end table end style 


Diketahui pula bahwa,

begin mathsize 14px style straight x subscript 1 greater than 1 space maka space straight x subscript 1 minus 1 greater than 0 straight x subscript 2 less than 1 space maka space straight x subscript 2 minus 1 less than 0 end style 


Jika keduanya dikalikan, maka akan menjadi:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses straight x subscript 1 minus 1 close parentheses open parentheses straight x subscript 2 minus 1 close parentheses end cell less than 0 row cell straight x subscript 1 straight x subscript 2 minus straight x subscript 1 minus straight x subscript 2 plus 1 end cell less than 0 row cell straight x subscript 1 straight x subscript 2 minus left parenthesis straight x subscript 1 plus straight x subscript 2 right parenthesis plus 1 end cell less than 0 row cell straight c plus 2 plus 2 plus 1 end cell less than 0 row straight c less than cell negative 5 end cell end table end style 

Jadi, jawabannya adalah B.

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