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Jika akar-akar persamaan kuadrat ,maka persamaan kuadrat yang akar-akarnya x 1 ​ + x 1 ​ 1 ​ d an x 2 ​ + x 2 ​ 1 ​ adalah...

Jika x subscript 1 space d a n space x subscript 2 akar-akar persamaan kuadrat x squared minus 3 x plus 1 equals 0, maka persamaan kuadrat yang akar-akarnya  adalah...

  1. x squared plus 9 x minus 6 equals 0

  2. x squared minus 6 x minus 6 equals 0

  3. x squared minus 6 x plus 9 equals 0

  4. x squared plus 6 x plus 9 equals 0

  5. x squared minus 6 x minus 9 equals 0

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N. Rahayu

Master Teacher

Mahasiswa/Alumni Universitas Negeri Jakarta

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Pembahasan

Tentukan hasil jumlah dan hasil kali akar-akar yang baru Maka persamaan barunya:

U n t u k space p e r s a m a a n space x squared minus 3 x plus 1 equals 0 x subscript 1 plus x subscript 2 equals negative b over a equals negative fraction numerator open parentheses negative 3 close parentheses over denominator 1 end fraction equals 3 x subscript 1 space. space x subscript 2 equals c over a equals 1 over 1 equals 1 x subscript 1 superscript space space 2 end superscript plus x subscript 2 superscript space space 2 end superscript equals open parentheses x subscript 1 plus x subscript 2 close parentheses squared minus 2 open parentheses x subscript 1 plus x subscript 2 close parentheses equals open parentheses 3 close parentheses squared minus 2 open parentheses 1 close parentheses equals 9 minus 2 equals 7

Tentukan hasil jumlah dan hasil kali akar-akar yang baru

open parentheses x subscript 1 plus 1 over x subscript 1 close parentheses plus open parentheses x subscript 2 plus 1 over x subscript 2 close parentheses equals open parentheses x subscript 1 plus x subscript 2 close parentheses plus open parentheses 1 over x subscript 1 plus 1 over x subscript 2 close parentheses equals open parentheses x subscript 1 plus x subscript 2 close parentheses plus open parentheses fraction numerator x subscript 1 plus x subscript 2 over denominator x subscript 1 space x subscript 2 end fraction close parentheses equals 3 plus open parentheses 3 over 1 close parentheses equals 6  open parentheses x subscript 1 plus 1 over x subscript 1 close parentheses open parentheses x subscript 2 plus 1 over x subscript 2 close parentheses equals x subscript 1 space. space x subscript 2 plus open parentheses x subscript 1 over x subscript 2 plus x subscript 2 over x subscript 1 close parentheses plus fraction numerator 1 over denominator x subscript 1 space. space x subscript 2 end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals x subscript 1 space. space x subscript 2 plus open parentheses fraction numerator x subscript 1 superscript space space 2 end superscript plus x subscript 2 superscript space space 2 end superscript over denominator x subscript 1 x subscript 2 end fraction close parentheses plus fraction numerator 1 over denominator x subscript 1 space. space x subscript 2 end fraction equals 1 plus open parentheses 7 over 1 close parentheses plus 1 over 1 equals 9Maka persamaan barunya:

x squared minus open square brackets open parentheses x subscript 1 plus 1 over x subscript 1 close parentheses plus open parentheses x subscript 2 plus 1 over x subscript 2 close parentheses close square brackets x plus open square brackets open parentheses x subscript 1 plus 1 over x subscript 1 close parentheses open parentheses x subscript 2 plus 1 over x subscript 2 close parentheses close square brackets equals 0 x squared minus 6 x plus 9 equals 0 J a d i space p e r s a m a a n space k u a d r a t space y a n g space b a r u space a d a l a h space x squared minus 6 x plus 9 equals 0

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