Roboguru

Tentukan akar-akar dari persamaan kuadrat berikut ini dengan rumus abc. 3.

Pertanyaan

Tentukan akar-akar dari persamaan kuadrat berikut ini dengan rumus abc.

3. a squared minus 3 a minus 10 equals 0 

Pembahasan Soal:

Diketahui persamaan kuadrat:

 a squared minus 3 a minus 10 equals 0
a equals 1 comma space b equals negative 3 comma space c equals negative 10

Rumus kuadratik:

table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row x cell 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 end cell end table

Sehingga didapat akar-akar:

table attributes columnalign right center left columnspacing 0px end attributes row cell a subscript 1 comma thin space 2 end subscript end cell equals cell fraction numerator negative open parentheses negative 3 close parentheses plus-or-minus square root of open parentheses negative 3 close parentheses squared minus 4 times 1 times open parentheses negative 10 close parentheses end root over denominator 2 times 1 end fraction end cell row cell a subscript 1 comma thin space 2 end subscript end cell equals cell fraction numerator negative open parentheses negative 3 close parentheses plus-or-minus square root of 9 plus 40 end root over denominator 2 times 1 end fraction end cell row cell a subscript 1 comma thin space 2 end subscript end cell equals cell fraction numerator negative open parentheses negative 3 close parentheses plus-or-minus square root of 49 over denominator 2 times 1 end fraction end cell row cell a subscript 1 comma thin space 2 end subscript end cell equals cell fraction numerator negative open parentheses negative 3 close parentheses plus-or-minus thin space 7 over denominator 2 times 1 end fraction end cell row cell a subscript 1 end cell equals cell fraction numerator negative open parentheses negative 3 close parentheses plus 7 over denominator 2 times 1 end fraction comma thin space a subscript 2 equals fraction numerator negative open parentheses negative 3 close parentheses minus 7 over denominator 2 times 1 end fraction end cell row cell a subscript 1 end cell equals cell 10 over 2 comma thin space a subscript 2 equals fraction numerator negative 4 over denominator 2 end fraction end cell row a equals cell 5 comma thin space a equals negative 2 end cell end table    

Jadi, akar-akarnya adalah a equals 5 comma thin space a equals negative 2.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

S. Yoga

Mahasiswa/Alumni Universitas Pendidikan Indonesia

Terakhir diupdate 05 Juni 2021

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Pertanyaan yang serupa

Tentukan himpunan penyelesaian persamaan kuadrat berikut dengan cara memfaktorkan ! 14.

Pembahasan Soal:

Bentuk a x squared plus b x plus c equals 0 comma space a not equal to 0 dapat diselesaikan dengan rumus AL-khawarizmi:

x subscript 1 comma 2 space 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

Pada persamaan kuadrat 49 x squared minus 56 x plus 1 equals 0, diketahui a equals 49 comma space b equals negative 56 comma space c equals 1, maka

table attributes columnalign right center left columnspacing 0px end attributes row cell x subscript 1 comma space 2 end subscript end cell equals cell fraction numerator negative left parenthesis negative 56 right parenthesis plus-or-minus square root of left parenthesis negative 56 right parenthesis squared minus 4 left parenthesis 49 right parenthesis left parenthesis 1 right parenthesis end root over denominator 2 left parenthesis 49 right parenthesis end fraction end cell row blank equals cell fraction numerator 56 plus-or-minus square root of 3.136 minus 196 end root over denominator 98 end fraction end cell row blank equals cell fraction numerator 56 plus-or-minus square root of 2.940 end root over denominator 98 end fraction end cell row blank equals cell fraction numerator 56 plus-or-minus square root of 4 cross times 735 end root over denominator 98 end fraction end cell row blank equals cell fraction numerator 56 plus-or-minus 2 square root of 735 over denominator 98 end fraction end cell row blank equals cell fraction numerator 28 plus-or-minus square root of 735 over denominator 49 end fraction end cell end table

Jadi, himpunan penyelesaian dari persamaan 49 x squared minus 56 x plus 1 equals 0 adalah open curly brackets table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator 28 minus square root of 735 over denominator 49 end fraction end cell end table comma space table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator 28 plus square root of 735 over denominator 49 end fraction end cell end table close curly brackets.

Roboguru

Tentukan himpunan penyelesaian dari persamaan .

Pembahasan Soal:

Diketahui persamaan 5 x squared plus 4 x minus 12 equals 0 maka a equals 5 comma space b equals 4 comma space c equals negative 12.

table attributes columnalign right center left columnspacing 0px end attributes row cell x subscript 1 comma 2 end subscript end cell 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 4 plus-or-minus square root of 4 squared minus 4 open parentheses 5 close parentheses open parentheses negative 12 close parentheses end root over denominator 2 open parentheses 5 close parentheses end fraction end cell row blank equals cell fraction numerator negative 4 plus-or-minus square root of 16 plus 240 end root over denominator 10 end fraction end cell row blank equals cell fraction numerator negative 4 plus-or-minus square root of 256 over denominator 10 end fraction end cell row x equals cell fraction numerator negative 4 plus-or-minus 16 over denominator 10 end fraction end cell row cell x subscript 1 end cell equals cell fraction numerator negative 4 plus 16 over denominator 10 end fraction equals 6 over 5 end cell row blank blank cell a t a u end cell row cell x subscript 2 end cell equals cell fraction numerator negative 4 minus 16 over denominator 10 end fraction equals negative 2 end cell end table

Jadi, diperoleh himpunan penyelesaian dari persamaan 5 x squared plus 4 x minus 12 equals 0 adalah open curly brackets negative 2 comma space 6 over 5 close curly brackets.

Roboguru

Akar-akar persamaan  adalah

Pembahasan Soal:

Perhatikan bahwa persamaan kuadrat begin mathsize 14px style 1 half x squared minus x minus 3 equals 0 end style memiliki nilai begin mathsize 14px style a equals 1 half end stylesize 14px b size 14px equals size 14px minus size 14px 1, dan size 14px c size 14px equals size 14px minus size 14px 3.  Maka nilai diskriminan dari persamaan kuadrat tersebut adalah

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row D equals cell b squared minus 4 a c end cell row blank equals cell open parentheses negative 1 close parentheses squared minus up diagonal strike 4 squared open parentheses 1 over up diagonal strike 2 to the power of 1 close parentheses open parentheses negative 3 close parentheses end cell row blank equals cell 1 plus 6 end cell row blank equals 7 end table end style

Dengan menggunakan rumus kuadratik

begin mathsize 14px style x subscript 1 comma 2 end subscript equals fraction numerator negative b plus-or-minus square root of D over denominator 2 a end fraction end style

diperoleh akar-akar

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell x subscript 1 comma 2 end subscript end cell equals cell fraction numerator negative open parentheses negative 1 close parentheses plus-or-minus square root of 7 over denominator up diagonal strike 2 to the power of 1 open parentheses begin display style 1 over up diagonal strike 2 to the power of 1 end style close parentheses end fraction end cell row blank equals cell 1 plus-or-minus square root of 7 end cell end table end style

Jadi, jawaban yang tepat adalah B.

Roboguru

Bagaimana cara mendapatkan rumus abc?

Pembahasan Soal:

Diketahui persamaan kuadrat:

a x squared plus b x plus c equals 0 

Kemudian didapat:

table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell a x squared plus b x plus c end cell cell equals 0 blank text  (bagi  end text a text  kedua ruas) end text end cell row cell x squared plus b over a x plus c over a end cell cell equals 0 end cell row cell x squared plus b over a x end cell cell equals negative c over a end cell row cell text  gunakan  end text x squared plus p x end cell cell equals open parentheses x plus p over 2 close parentheses squared minus open parentheses p over 2 close parentheses squared end cell row cell open parentheses x plus fraction numerator b over denominator 2 a end fraction close parentheses squared minus open parentheses fraction numerator b over denominator 2 a end fraction close parentheses squared end cell cell equals negative c over a end cell row cell open parentheses x plus fraction numerator b over denominator 2 a end fraction close parentheses squared minus fraction numerator b squared over denominator 4 a squared end fraction end cell cell equals negative c over a end cell row cell open parentheses x plus fraction numerator b over denominator 2 a end fraction close parentheses squared end cell cell equals fraction numerator b squared over denominator 4 a squared end fraction minus c over a end cell row cell open parentheses x plus fraction numerator b over denominator 2 a end fraction close parentheses squared end cell cell equals fraction numerator b squared over denominator 4 a squared end fraction minus fraction numerator 4 a c over denominator 4 a squared end fraction end cell row cell open parentheses x plus fraction numerator b over denominator 2 a end fraction close parentheses squared end cell cell equals fraction numerator b squared minus 4 a c over denominator 4 a squared end fraction end cell row cell open parentheses x plus fraction numerator b over denominator 2 a end fraction close parentheses end cell cell equals plus-or-minus square root of fraction numerator b squared minus 4 a c over denominator 4 a squared end fraction end root end cell row cell open parentheses x plus fraction numerator b over denominator 2 a end fraction close parentheses end cell cell equals plus-or-minus fraction numerator square root of b squared minus 4 a c end root over denominator 2 a end fraction end cell row x cell equals negative fraction numerator b over denominator 2 a end fraction plus-or-minus fraction numerator square root of b squared minus 4 a c end root over denominator 2 a end fraction end cell row x cell 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 end cell end table 

Jadi, terbukti rumus ABC nya adalah

table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row x cell 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 end cell end table 

Roboguru

Tentukan himpunan penyelesaian persamaan-persamaan berikut dengan menggunakan rumus!

Pembahasan Soal:

Ingat kembali rumus penyelesaian persamaan kuadrat atau rumus kuadratik abc dengan bentuk umum a p squared plus b p plus c equals 0.

table attributes columnalign right center left columnspacing 2px end attributes row p equals cell fraction numerator negative b plus-or-minus square root of open parentheses b squared minus 4 a c close parentheses end root over denominator 2 a end fraction end cell end table

Akan ditentukan akar (penyelesaian) dari 5 open parentheses p minus 3 close parentheses open parentheses p minus 2 close parentheses equals 4 open parentheses p minus 2 close parentheses plus 9 open parentheses p minus 3 close parentheses. Terlebih dahulu ubah persamaan tersebut ke bentuk umum persamaan kuadrat.

table attributes columnalign right center left columnspacing 2px end attributes row cell 5 open parentheses p minus 3 close parentheses open parentheses p minus 2 close parentheses end cell equals cell 4 open parentheses p minus 2 close parentheses plus 9 open parentheses p minus 3 close parentheses end cell row cell 5 open parentheses p squared minus 2 p minus 3 p plus 6 close parentheses end cell equals cell 4 p minus 8 plus 9 p minus 27 end cell row cell 5 open parentheses p squared minus 5 p plus 6 close parentheses end cell equals cell 13 p minus 35 end cell row cell 5 p squared minus 25 p plus 30 end cell equals cell 13 p minus 35 end cell row cell 5 p squared minus 25 p plus 30 minus 13 p plus 35 end cell equals 0 row cell 5 p squared minus 38 p plus 65 end cell equals 0 end table

Berdasarkan bentuk umum a p squared plus b p plus c equals 0, maka diperoleh a equals 5b equals negative 38, dan c equals 65.

Perhatikan perhitungan berikut.

table attributes columnalign right center left columnspacing 2px end attributes row p equals cell fraction numerator negative b plus-or-minus square root of open parentheses b squared minus 4 a c close parentheses end root over denominator 2 a end fraction end cell row blank equals cell fraction numerator negative open parentheses negative 38 close parentheses plus-or-minus square root of open parentheses negative 38 close parentheses squared minus 4 open parentheses 5 close parentheses open parentheses 65 close parentheses end root over denominator 2 open parentheses 5 close parentheses end fraction end cell row blank equals cell fraction numerator 38 plus-or-minus square root of 1444 minus 1300 end root over denominator 10 end fraction end cell row blank equals cell fraction numerator 38 plus-or-minus square root of 144 over denominator 10 end fraction end cell row p equals cell fraction numerator 38 plus-or-minus 12 over denominator 10 end fraction end cell end table table attributes columnalign right center left columnspacing 2px end attributes row cell p subscript 1 end cell equals cell fraction numerator 38 plus 12 over denominator 10 end fraction end cell row cell p subscript 1 end cell equals cell 50 over 10 end cell row cell p subscript 1 end cell equals 5 end table space space atau space space table attributes columnalign right center left columnspacing 2px end attributes row cell p subscript 2 end cell equals cell fraction numerator 38 minus 12 over denominator 10 end fraction end cell row cell p subscript 2 end cell equals cell 26 over 10 end cell row cell p subscript 2 end cell equals cell 13 over 5 end cell end table

Diperoleh penyelesaiannya adalah p subscript 1 equals 5 atau p subscript 2 equals 13 over 5.

Jadi, himpunan penyelesaiannya adalah open curly brackets 13 over 5 comma space 5 close curly brackets.

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