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Tentukan penyelesaian dari persamaan berikut! 10. 2 m 4 + m 3 − 11 m 2 + m + 2 = 0

Tentukan penyelesaian dari persamaan berikut!

10.  

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N. Puspita

Master Teacher

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Menentukan penyelesaian dari persamaan Selesaikan persamaan berikut! Gunakan rumus abc untuk menyelesaikan persamaan . Jadi, penyelesaian persamaan di atas adalah m 1 ​ m 2 ​ m 3 ​ m 4 ​ ​ = = = = ​ 2 − 3 + 5 ​ ​ 2 − 3 − 5 ​ ​ 2 1 ​ 2 ​

Menentukan penyelesaian dari persamaan 2 m to the power of 4 plus m cubed minus 11 m squared plus m plus 2 equals 0

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 m to the power of 4 plus m cubed minus 11 m squared plus m plus 2 end cell equals 0 row cell 2 m to the power of 4 plus open square brackets negative m cubed plus 2 m cubed close square brackets plus open square brackets negative m squared minus 10 m squared close square brackets plus open square brackets 5 m minus 4 m close square brackets plus 2 end cell equals 0 row cell open parentheses 2 m to the power of 4 minus m cubed close parentheses plus open parentheses 2 m cubed minus m squared close parentheses minus open parentheses 10 m squared plus 5 m close parentheses minus open parentheses 4 m plus 2 close parentheses end cell equals 0 row cell m cubed open parentheses 2 m minus 1 close parentheses plus m squared open parentheses 2 m minus 1 close parentheses minus 5 m open parentheses 2 m minus 1 close parentheses minus 2 open parentheses 2 m minus 1 close parentheses end cell equals 0 row cell open parentheses 2 m minus 1 close parentheses cross times open parentheses m cubed plus m squared minus 5 m minus 2 close parentheses end cell equals 0 row cell open parentheses 2 m minus 1 close parentheses cross times open square brackets m cubed plus open parentheses negative 2 m squared plus 3 m squared close parentheses plus open parentheses negative 6 m plus m close parentheses minus 2 close square brackets end cell equals 0 row cell open parentheses 2 m minus 1 close parentheses cross times open square brackets open parentheses m cubed minus 2 m squared close parentheses plus open parentheses 3 m squared minus 6 m close parentheses plus open parentheses m minus 2 close parentheses close square brackets end cell equals 0 row cell open parentheses 2 m minus 1 close parentheses cross times open square brackets m squared open parentheses m minus 2 close parentheses plus 3 m open parentheses m minus 2 close parentheses plus open parentheses m minus 2 close parentheses close square brackets end cell equals 0 row cell open parentheses 2 m minus 1 close parentheses open parentheses m minus 2 close parentheses open parentheses m squared plus 3 m plus 1 close parentheses end cell equals 0 end table 

Selesaikan persamaan berikut!

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 m minus 1 end cell equals 0 row cell 2 m end cell equals 1 row m equals cell 1 half end cell end table 

table attributes columnalign right center left columnspacing 0px end attributes row cell m minus 2 end cell equals 0 row m equals 2 end table 

Gunakan rumus abc untuk menyelesaikan persamaan m squared plus 3 m plus 1 equals 0.

table attributes columnalign right center left columnspacing 0px end attributes row cell m squared plus 3 m plus 1 end cell equals 0 row a equals cell 1 comma space b equals 3 comma space c equals 1 end cell row m equals cell fraction numerator negative b plus-or-minus square root of b squared minus 4 a c end root over denominator 2 a end fraction end cell row blank equals cell fraction numerator negative 3 plus-or-minus square root of open parentheses 3 close parentheses squared minus 4 open parentheses 1 close parentheses open parentheses 1 close parentheses end root over denominator 2 open parentheses 1 close parentheses end fraction end cell row blank equals cell fraction numerator negative 3 plus-or-minus square root of 9 minus 4 end root over denominator 2 end fraction end cell row blank equals cell fraction numerator negative 3 plus-or-minus square root of 5 over denominator 2 end fraction end cell end table 

Jadi, penyelesaian persamaan di atas adalah

 

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