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Diketahui persamaan kuadrat begin mathsize 14px style x squared minus 4 x plus open parentheses m minus 4 close parentheses equals 0 end style mempunyai akar-akar real begin mathsize 14px style p end style dan begin mathsize 14px style q end style.

Jika begin mathsize 14px style p greater than negative 2 end style dan begin mathsize 14px style q greater than negative 2 end style, maka nilai begin mathsize 14px style m end style adalah ....space 

  1. begin mathsize 14px style m greater or equal than 8 end stylespace 

  2. undefinedspace 

  3. begin mathsize 14px style negative 8 less than m less or equal than 8 end stylespace 

  4. undefinedspace 

  5. begin mathsize 14px style m less than negative 8 end stylespace 

A. Acfreelance

Master Teacher

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah C.space 

Pembahasan

Perhatikan bahwa pada persamaan kuadrat undefined, didapatkan nilai begin mathsize 14px style a equals 1 end stylebegin mathsize 14px style b equals negative 4 end styledan  begin mathsize 14px style c equals m minus 4 end style.

Karena akar-akarnya adalah undefined dan undefined, maka didapat hubungan 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell p plus q end cell equals cell negative b over a equals negative fraction numerator negative 4 over denominator 1 end fraction equals 4 end cell row cell p q end cell equals cell c over a equals fraction numerator m minus 4 over denominator 1 end fraction equals m minus 4 end cell end table end style

Karena persamaan kuadrat ini memiliki akar-akar real undefined dan undefined, maka haruslah begin mathsize 14px style D greater or equal than 0 end style. Dengan demikian, didapat

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell left parenthesis negative 4 right parenthesis squared minus 4 left parenthesis 1 right parenthesis open parentheses m minus 4 close parentheses end cell greater or equal than 0 row cell 16 minus 4 m plus 16 end cell greater or equal than 0 row cell negative 4 m plus 32 end cell greater or equal than 0 row cell negative 4 m end cell greater or equal than cell negative 32 end cell row m less or equal than 8 end table end style

 

Karena begin mathsize 14px style p greater than negative 2 end style dan begin mathsize 14px style q greater than negative 2 end style, maka didapat begin mathsize 14px style p plus 2 greater than 0 end style dan begin mathsize 14px style q plus 2 greater than 0 end style. Oleh karena itu, didapat

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell left parenthesis p plus 2 right parenthesis left parenthesis q plus 2 right parenthesis end cell greater than cell 0 space end cell row cell p q plus 2 left parenthesis p plus q right parenthesis plus 4 end cell greater than cell 0 space end cell row cell left parenthesis m minus 4 right parenthesis plus 2 left parenthesis 4 right parenthesis plus 4 end cell greater than cell 0 space end cell row cell m minus 4 plus 8 plus 4 end cell greater than 0 row cell m plus 8 end cell greater than cell 0 space end cell row m greater than cell negative 8 end cell end table end style

 

Karena begin mathsize 14px style m less or equal than 8 end style dan begin mathsize 14px style m greater than negative 8 end style, maka didapat begin mathsize 14px style negative 8 less than m less or equal than 8 end style.

Jadi, jawaban yang tepat adalah C.space 

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