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Diketahui x1​ dan x2​ adalah akar-akar persamaan (m+3)x2−(2m−1)x+1=0. Jika x1−2​+x2−2​=49, maka nilai m haruslah...

Pertanyaan

Diketahui x subscript 1 space dan space x subscript 2 adalah akar-akar persamaan open parentheses straight m plus 3 close parentheses x squared minus open parentheses 2 straight m minus 1 close parentheses x plus 1 equals 0. Jika x subscript 1 to the power of negative 2 end exponent plus x subscript 2 to the power of negative 2 end exponent equals 49, maka nilai m haruslah...

N. Puspita

Master Teacher

Jawaban terverifikasi

Jawaban

diperoleh nilai straight m equals 9 over 2 space atau space straight m equals negative 3.

Pembahasan

Jika x subscript 1 space dan space x subscript 2 adalah akar-persamaan kuadrat a x squared plus b x plus c equals 0, maka

x subscript 1 plus x subscript 2 equals negative b over a x subscript 1 times x subscript 2 equals c over a

Pada persamaan kuadrat open parentheses straight m plus 3 close parentheses x squared minus open parentheses 2 straight m minus 1 close parentheses x plus 1 equals 0, diketahui nilai

straight a equals straight m plus 3 straight b equals negative left parenthesis 2 straight m minus 1 right parenthesis straight c equals 1

Maka,

x subscript 1 plus x subscript 2 equals fraction numerator 2 straight m minus 1 over denominator straight m plus 3 end fraction x subscript 1 times x subscript 2 equals fraction numerator 1 over denominator straight m plus 3 end fraction

Selanjutnya, diketahui x subscript 1 to the power of negative 2 end exponent plus x subscript 2 to the power of negative 2 end exponent equals 49,

table attributes columnalign right center left columnspacing 0px end attributes row cell x subscript 1 to the power of negative 2 end exponent plus x subscript 2 to the power of negative 2 end exponent end cell equals 49 row cell 1 over x subscript 1 squared plus 1 over x subscript 2 squared end cell equals 49 row cell fraction numerator x subscript 2 squared plus x subscript 1 squared over denominator x subscript 1 squared x subscript 2 squared end fraction end cell equals 49 row cell fraction numerator open parentheses x subscript 1 plus x subscript 2 close parentheses squared minus 2 x subscript 1 x subscript 2 over denominator x subscript 1 squared x subscript 2 squared end fraction end cell equals 49 row cell fraction numerator open parentheses begin display style fraction numerator 2 straight m minus 1 over denominator straight m plus 3 end fraction end style close parentheses squared minus 2 open parentheses begin display style fraction numerator 1 over denominator straight m plus 3 end fraction end style close parentheses over denominator open parentheses begin display style fraction numerator 1 over denominator straight m plus 3 end fraction end style close parentheses squared end fraction end cell equals 49 row cell 4 straight m squared minus 4 straight m plus 1 minus 2 left parenthesis straight m plus 3 right parenthesis end cell equals 49 row cell 4 straight m squared minus 6 straight m minus 54 end cell equals 0 row cell 2 straight m squared minus 3 straight m minus 27 end cell equals 0 row cell open parentheses 2 straight m minus 9 close parentheses open parentheses straight m plus 3 close parentheses end cell equals 0 row straight m equals cell 9 over 2 space atau end cell row straight m equals cell negative 3 end cell end table

Jadi, diperoleh nilai straight m equals 9 over 2 space atau space straight m equals negative 3.

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