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Carilah solusi ( x , y ) bilangan real untuk setiap persamaan kuadrat berikut dan lukiska sketsa grafik penyelesaiannya! { y = − x ​ ( x − 3 ) 2 + y 2 = 4 ​

Carilah solusi  bilangan real untuk setiap persamaan kuadrat berikut dan lukiska sketsa grafik penyelesaiannya!

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D. Wahyu

Master Teacher

Mahasiswa/Alumni Universitas Negeri Semarang

Jawaban terverifikasi

Pembahasan

Diketahui: subtitusikan persamaan (1) ke persamaan (2) , sehingga Menentukan nilai dengan rumus yaitu Dari persamaan diketahui: , sehingga Selanjutnya, subtitusikan nilai ke persamaan (1), sehingga Jadi, solusi adalah . Grafik. Terlebih dahulu menentukan titik-titik nya yaitu Grafik

Diketahui:

y equals negative square root of x space.. left parenthesis 1 right parenthesis left parenthesis x minus 3 right parenthesis squared plus y to the power of 2 end exponent equals 4 space... left parenthesis 2 right parenthesis

subtitusikan persamaan (1) ke persamaan (2) , sehingga

table attributes columnalign right center left columnspacing 0px end attributes row cell left parenthesis x minus 3 right parenthesis squared plus left parenthesis negative square root of x right parenthesis squared end cell equals 4 row cell 4 space x squared minus 6 x plus 9 plus x minus 4 end cell equals 0 row cell x minus 5 x plus 5 end cell equals cell 0... left parenthesis 3 right parenthesis end cell end table

Menentukan nilai x dengan rumus a comma b comma c yaitu

x subscript 1 comma 2 end subscript equals fraction numerator negative b plus-or-minus square root of b squared minus 4 a c end root over denominator 2 a end fraction

Dari persamaan diketahui:   a equals 1 comma space b equals negative 5 comma space c equals 5, sehingga

x subscript 1 comma 2 end subscript equals fraction numerator negative left parenthesis negative 5 right parenthesis plus-or-minus square root of left parenthesis negative 5 right parenthesis squared minus 4 times 1 times 5 end root over denominator 2 times 5 end fraction x subscript 1 comma 2 end subscript equals fraction numerator 5 plus-or-minus square root of 25 minus 20 end root over denominator 10 end fraction x subscript 1 comma 2 end subscript equals fraction numerator 5 plus-or-minus square root of 5 over denominator 10 end fraction x subscript 1 equals fraction numerator 5 plus square root of 5 over denominator 10 end fraction space space comma space x subscript 2 equals fraction numerator 5 minus square root of 5 over denominator 10 end fraction

Selanjutnya, subtitusikan nilai x subscript 1 space dan space x subscript 2 space spaceke persamaan (1), sehingga

y subscript 1 equals fraction numerator negative square root of 5 plus square root of 5 end root over denominator 10 end fraction y subscript 2 equals fraction numerator square root of 5 plus square root of 5 end root over denominator 10 end fraction

Jadi, solusi adalah open parentheses fraction numerator 5 plus square root of 5 over denominator 10 end fraction comma fraction numerator square root of 5 plus square root of 5 end root over denominator 10 end fraction close parentheses space space dan space space open parentheses fraction numerator 5 plus square root of 5 over denominator 10 end fraction comma negative fraction numerator square root of 5 plus square root of 5 end root over denominator 10 end fraction close parentheses.

Grafik.

Terlebih dahulu menentukan titik-titik nya yaitu 

table attributes columnalign right center left columnspacing 0px end attributes row y equals cell negative square root of x end cell row cell negative Sumbu space x comma space y end cell equals cell 0 colon end cell row 0 equals cell negative square root of x end cell row cell 0 squared end cell equals cell negative x end cell row 0 equals x row cell negative sumbu space y comma space x end cell equals cell 0 colon end cell row y equals cell negative square root of 0 end cell row y equals 0 row blank blank blank row cell left parenthesis x minus 3 right parenthesis squared plus y squared end cell equals 4 row cell negative Sumbu space x comma space y end cell equals cell 0 colon end cell row cell left parenthesis x minus 3 right parenthesis squared plus 0 end cell equals 4 row cell left parenthesis x minus 3 right parenthesis squared end cell equals cell 4 space left parenthesis x minus 3 right parenthesis equals plus-or-minus square root of 4 end cell row cell left parenthesis x minus 3 right parenthesis end cell equals cell plus-or-minus 2 end cell row blank blank blank row cell left parenthesis x subscript 1 minus 3 right parenthesis end cell equals 2 row cell x subscript 1 end cell equals cell 2 plus 3 end cell row cell x subscript 1 end cell equals 5 row blank blank blank row cell left parenthesis x subscript 2 minus 3 right parenthesis end cell equals cell negative 2 end cell row cell left parenthesis x subscript 2 minus 3 right parenthesis end cell equals cell negative 2 end cell row cell x subscript 2 end cell equals cell negative 2 plus 3 end cell row cell x subscript 2 end cell equals 1 row cell negative Sumbu space y comma space x end cell equals cell 0 colon end cell row cell left parenthesis 0 minus 3 right parenthesis squared plus y to the power of 2 end exponent end cell equals 4 row cell 0 plus y squared end cell equals 4 row cell y squared end cell equals 4 row y equals cell plus-or-minus square root of 4 end cell row y equals cell plus-or-minus 2 end cell row cell y subscript 1 end cell equals cell 2 space comma space y subscript 2 equals negative 2 end cell end table

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Carilah solusi ( x , y ) bilangan real untuk setiap persamaan kuadrat berikut dan lukiska sketsa grafik penyelesaiannya! { y 2 = x − 1 ( x − 3 ) 2 + y 2 = 4 ​

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