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Jika  akar-akar persamaan polinomial   nilai  adalah ....

Pertanyaan

Jika begin mathsize 14px style p comma space q comma space r comma space dan space s end style akar-akar persamaan polinomial

begin mathsize 14px style x to the power of 4 minus 2 x cubed minus 7 x squared plus 8 x plus 12 equals 0 end style 

nilai begin mathsize 14px style fraction numerator 1 over denominator p q r end fraction plus fraction numerator 1 over denominator p q s end fraction plus fraction numerator 1 over denominator p r s end fraction plus fraction numerator 1 over denominator q r s end fraction end style adalah ....

  1. begin mathsize 14px style 6 end style 

  2. undefined 

  3. begin mathsize 14px style 1 third end style 

  4. begin mathsize 14px style 1 over 6 end style 

  5. begin mathsize 14px style negative 1 half end style 

Pembahasan Video:

Pembahasan Soal:

begin mathsize 14px style x to the power of 4 minus 2 x cubed minus 7 x squared plus 8 x plus 12 equals 0 end style 

begin mathsize 14px style a equals 1 comma space b equals negative 2 comma space c equals negative 7 comma space d equals 8 comma space e equals 12 end style

begin mathsize 14px style p comma space q comma space r comma space dan space s end style akar-akar persamaan polinomial, maka

begin mathsize 14px style begin inline style fraction numerator 1 over denominator p q r end fraction end style plus begin inline style fraction numerator 1 over denominator p q s end fraction end style plus begin inline style fraction numerator 1 over denominator p r s end fraction end style plus begin inline style fraction numerator 1 over denominator q r s end fraction end style equals begin inline style fraction numerator s plus r plus q plus p over denominator p q r s end fraction end style equals begin inline style fraction numerator p plus q plus r plus s over denominator p q r s end fraction end style end style

begin mathsize 14px style fraction numerator p plus q plus r plus s over denominator p q r s end fraction equals fraction numerator negative b over a over denominator e over a end fraction equals negative begin inline style b over a end style cross times begin inline style a over e end style equals negative begin inline style b over e end style equals negative begin inline style fraction numerator left parenthesis negative 2 right parenthesis over denominator 12 end fraction end style equals begin inline style 1 over 6 end style end style

Jadi, jawaban yang tepat adalah D.

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 12 April 2021

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