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Pertanyaan

Akar-akar persamaan x squared plus left parenthesis straight p plus 1 right parenthesis x plus 3 straight p equals 0 adalah x subscript 1 dan x subscript 2. Jika nilai x subscript 1 squared plus x subscript 2 squared equals 85, tentukan nilai straight p!

Pembahasan Soal:

Apabila persamaan kuadrat straight a x squared plus straight b x plus straight c equals 0 mempunyai penyelesaian x subscript 1 dan x subscript 2, maka nilai:


x subscript 1 plus x subscript 2 equals negative straight b over straight a


Dan


x subscript 1 cross times x subscript 2 equals straight c over straight a


Dari persamaan kuadrat x squared plus left parenthesis straight p plus 1 right parenthesis x plus 3 straight p equals 0 yang mempunyai nilai straight a equals 1straight b equals left parenthesis straight p plus 1 right parenthesis, dan straight c equals 3 straight p,, maka:


table attributes columnalign right center left columnspacing 0px end attributes row cell x subscript 1 plus x subscript 2 end cell equals cell negative straight b over straight a end cell row blank equals cell negative fraction numerator left parenthesis straight p plus 1 right parenthesis over denominator 1 end fraction end cell row blank equals cell negative open parentheses straight p plus 1 close parentheses end cell end table


Dan


table attributes columnalign right center left columnspacing 0px end attributes row cell x subscript 1 cross times x subscript 2 end cell equals cell straight c over straight a end cell row blank equals cell fraction numerator 3 straight p over denominator 1 end fraction end cell row blank equals cell 3 straight p end cell end table


Diketahui bahwa x subscript 1 squared plus x subscript 2 squared equals 85.

Ingat bahwa open parentheses x subscript 1 plus x subscript 2 close parentheses squared equals x subscript 1 squared plus 2 x subscript 1 times x subscript 2 plus x subscript 2 squared, sehingga:


table attributes columnalign right center left columnspacing 0px end attributes row cell x subscript 1 squared plus x subscript 2 squared end cell equals 85 row cell open parentheses x subscript 1 plus x subscript 2 close parentheses squared minus 2 x subscript 1 times x subscript 2 end cell equals 85 row cell open parentheses negative open parentheses straight p plus 1 close parentheses close parentheses squared minus 2 times 3 straight p end cell equals 85 row cell straight p squared plus 2 straight p plus 1 minus 6 straight p minus 85 end cell equals 0 row cell straight p squared minus 4 straight p minus 84 end cell equals 0 end table


Dengan menggunakan rumus kuadratik, diperoleh akar-akar dari persamaan kuadrat tersebut adalah:


table attributes columnalign right center left columnspacing 0px end attributes row straight p equals cell fraction numerator negative straight b plus-or-minus square root of straight b squared minus 4 ac end root over denominator 2 straight a end fraction end cell row blank equals cell fraction numerator negative left parenthesis negative 4 right parenthesis plus-or-minus square root of left parenthesis negative 4 right parenthesis squared minus 4 times 1 times left parenthesis negative 84 right parenthesis end root over denominator 2 times 1 end fraction end cell row blank equals cell fraction numerator 4 plus-or-minus square root of 16 plus 336 end root over denominator 2 end fraction end cell row blank equals cell fraction numerator 4 plus-or-minus square root of 352 over denominator 2 end fraction end cell row blank equals cell fraction numerator 4 plus-or-minus 4 square root of 22 over denominator 2 end fraction end cell row blank equals cell 2 plus-or-minus 2 square root of 22 end cell end table


Jadi, nilai straight p yang dapat membuat nilai x subscript 1 squared plus x subscript 2 squared equals 85 adalah straight p equals 2 plus 2 square root of 22 atau straight p equals 2 minus 2 square root of 22.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

M. Rochmat

Terakhir diupdate 17 September 2021

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Pertanyaan yang serupa

Akar-akar persamaan  adalah  dan . Susunlah persamaan kuadrat baru dalam  yang akar-akarnya masing-masing kuadrat dari   dan !

Pembahasan Soal:

Kita gunakan sifat jumlah dari dan hasil kali akar-akar persamaan kuadrat. Persamaan 2 x squared plus 12 x plus 9 equals 0 mempunyai nilai straight a equals 2straight b equals 12 dan straight c equals 9, sehingga diperoleh:


table attributes columnalign right center left columnspacing 0px end attributes row cell x subscript 1 plus x subscript 2 end cell equals cell negative straight b over straight a end cell row blank equals cell negative 12 over 2 end cell row blank equals cell negative 6 end cell end table


Dan


x1x2==ac29


Ditanyakan persamaan kuadrat dalam y yang akar-akarnya adalah y subscript 1 equals x subscript 1 squared dan y subscript 2 equals x subscript 2 squared. Langkah berikutnya adalah hitung jumlah dan hasil kali dari akar-akar tersebut. Diperoleh:


table attributes columnalign right center left columnspacing 0px end attributes row cell y subscript 1 plus y subscript 2 end cell equals cell x subscript 1 squared plus x subscript 2 squared end cell row blank equals cell open parentheses x subscript 1 plus x subscript 2 close parentheses squared minus 2 x subscript 1 times x subscript 2 end cell row blank equals cell open parentheses negative 6 close parentheses squared minus 2 times open parentheses 9 over 2 close parentheses end cell row blank equals cell 36 minus 9 end cell row blank equals 27 end table


Dan


y1y2====x12x22(x1x2)2(29)2481


Persamaan kuadrat baru yang akar-akarnya sudah diketahui adalah:


table attributes columnalign right center left columnspacing 0px end attributes row cell y squared minus open parentheses y subscript 1 plus y subscript 2 close parentheses y plus y subscript 1 times y subscript 2 end cell equals 0 row cell y squared minus 27 y plus 81 over 4 end cell equals 0 row blank blank cell Kedua space ruas space dikali space 4 end cell row cell 4 y squared minus 108 y plus 81 end cell equals 0 end table


Jadi, persamaan kuadrat yang baru adalah Error converting from MathML to accessible text..

Roboguru

Pada persamaan , tentukan nilai akar-akar berikut!

Pembahasan Soal:

Apabila persamaan kuadrat straight a x squared plus straight b x plus straight c equals 0 mempunyai penyelesaian x subscript 1 dan x subscript 2, maka nilai:


x subscript 1 plus x subscript 2 equals negative straight b over straight a


Dan


x subscript 1 cross times x subscript 2 equals straight c over straight a


Maka, dari persamaan kuadrat 2 x squared minus 4 x minus 6 equals 0 yang mempunyai nilai straight a equals 2straight b equals negative 4, dan straight c equals negative 6, diperoleh:


table attributes columnalign right center left columnspacing 0px end attributes row cell x subscript 1 plus x subscript 2 end cell equals cell negative straight b over straight a end cell row blank equals cell negative fraction numerator left parenthesis negative 4 right parenthesis over denominator 2 end fraction end cell row blank equals 2 end table


Dan


table attributes columnalign right center left columnspacing 0px end attributes row cell x subscript 1 cross times x subscript 2 end cell equals cell straight c over straight a end cell row blank equals cell fraction numerator negative 6 over denominator 2 end fraction end cell row blank equals cell negative 3 end cell end table


Sehingga diperoleh:


table attributes columnalign right center left columnspacing 0px end attributes row cell 1 over x subscript 1 squared plus 1 over x subscript 2 squared end cell equals cell fraction numerator x subscript 1 squared over denominator x subscript 1 squared times x subscript 2 squared end fraction plus fraction numerator x subscript 2 squared over denominator x subscript 1 squared times x subscript 2 squared end fraction end cell row blank equals cell fraction numerator x subscript 1 squared plus x subscript 2 squared over denominator open parentheses x subscript 1 times x subscript 2 close parentheses squared end fraction end cell end table


Ingat bahwa open parentheses x subscript 1 plus x subscript 2 close parentheses squared equals x subscript 1 squared plus 2 x subscript 1 times x subscript 2 plus x subscript 2 squared, sehingga:


table attributes columnalign right center left columnspacing 0px end attributes row cell 1 over x subscript 1 squared plus 1 over x subscript 2 squared end cell equals cell fraction numerator x subscript 1 squared over denominator x subscript 1 squared times x subscript 2 squared end fraction plus fraction numerator x subscript 2 squared over denominator x subscript 1 squared times x subscript 2 squared end fraction end cell row blank equals cell fraction numerator x subscript 1 squared plus x subscript 2 squared over denominator open parentheses x subscript 1 times x subscript 2 close parentheses squared end fraction end cell row blank equals cell fraction numerator open parentheses x subscript 1 plus x subscript 2 close parentheses squared minus 2 x subscript 1 times x subscript 2 over denominator open parentheses x subscript 1 times x subscript 2 close parentheses squared end fraction end cell row blank equals cell fraction numerator open parentheses 2 close parentheses squared minus 2 times left parenthesis negative 3 right parenthesis over denominator left parenthesis negative 3 right parenthesis squared end fraction end cell row blank equals cell fraction numerator 4 plus 6 over denominator 9 end fraction end cell row blank equals cell 10 over 9 end cell end table


Jadi, nilai 1 over x subscript 1 squared plus 1 over x subscript 2 squared equals 10 over 9.

Roboguru

Susunlah persamaan kuadrat dalam  yang akar-akarnya adalah kuadrat dari masing-masing akar persamaan berikut!

Pembahasan Soal:

Kita gunakan sifat jumlah dari dan hasil kali akar-akar persamaan kuadrat. Persamaan 3 x squared minus 6 x minus 4 equals 0 mempunyai nilai straight a equals 3straight b equals negative 6 dan straight c equals negative 4, sehingga diperoleh:


table attributes columnalign right center left columnspacing 0px end attributes row cell x subscript 1 plus x subscript 2 end cell equals cell negative straight b over straight a end cell row blank equals cell negative fraction numerator negative 6 over denominator 3 end fraction end cell row blank equals 2 end table


Dan


table attributes columnalign right center left columnspacing 0px end attributes row cell x subscript 1 cross times x subscript 2 end cell equals cell straight c over straight a end cell row blank equals cell negative 4 over 3 end cell end table


Ditanyakan persamaan kuadrat dalam y yang akar-akarnya adalah y subscript 1 equals x subscript 1 squared dan y subscript 2 equals x subscript 2 squared. Langkah berikutnya adalah hitung jumlah dan hasil kali dari akar-akar tersebut. Diperoleh:


table attributes columnalign right center left columnspacing 0px end attributes row cell y subscript 1 plus y subscript 2 end cell equals cell x subscript 1 squared plus x subscript 2 squared end cell row blank equals cell open parentheses x subscript 1 plus x subscript 2 close parentheses squared minus 2 x subscript 1 times x subscript 2 end cell row blank equals cell open parentheses 2 close parentheses squared minus 2 times open parentheses negative 4 over 3 close parentheses end cell row blank equals cell 4 plus 8 over 3 end cell row blank equals cell 20 over 3 end cell end table


Dan


table attributes columnalign right center left columnspacing 0px end attributes row cell y subscript 1 cross times y subscript 2 end cell equals cell x subscript 1 squared times x subscript 2 squared end cell row blank equals cell open parentheses x subscript 1 times x subscript 2 close parentheses squared end cell row blank equals cell open parentheses negative 4 over 3 close parentheses squared end cell row blank equals cell 16 over 9 end cell end table


Persamaan kuadrat baru yang akar-akarnya sudah diketahui adalah:


table attributes columnalign right center left columnspacing 0px end attributes row cell y squared minus open parentheses y subscript 1 plus y subscript 2 close parentheses y plus y subscript 1 times y subscript 2 end cell equals 0 row cell y squared minus 20 over 3 y plus 16 over 9 end cell equals 0 row blank blank cell Kedua space ruas space dikali space 9 end cell row cell 9 y squared minus 60 y plus 16 end cell equals 0 end table


Jadi, persamaan kuadrat yang baru adalah Error converting from MathML to accessible text..

Roboguru

dan  adalah akar-akar dari persamaan . Nilai  = ...

Pembahasan Soal:

Jika p dan q adalah akar-akar dari persamaan x squared minus 10 x plus 20 equals 0, maka sifat akar-akarnya adalah:


table attributes columnalign right center left columnspacing 0px end attributes row cell p plus q end cell equals cell negative straight b over straight a end cell row blank equals cell negative fraction numerator negative 10 over denominator 1 end fraction end cell row blank equals 10 end table


Dan


table attributes columnalign right center left columnspacing 0px end attributes row cell p cross times q end cell equals cell straight c over straight a end cell row blank equals cell 20 over 1 end cell row blank equals 20 end table


Ditanyakan nilai p squared plus q squared. Ingat bahwa open parentheses p plus q close parentheses squared equals p squared plus 2 p q plus q squared, maka p squared plus q squared equals open parentheses p plus q close parentheses squared minus 2 p q.


table attributes columnalign right center left columnspacing 0px end attributes row cell p squared plus q squared end cell equals cell open parentheses p plus q close parentheses squared minus 2 p q end cell row blank equals cell open parentheses 10 close parentheses squared minus 2 times 20 end cell row blank equals cell 100 minus 40 end cell row blank equals 60 end table


Jadi, nilai table attributes columnalign right center left columnspacing 0px end attributes row cell p squared plus q squared end cell equals 60 end table.

Oleh karena itu, jawaban yang benar adalah A.

Roboguru

Pada persamaan , tentukan nilai akar-akar berikut!

Pembahasan Soal:

Apabila persamaan kuadrat straight a x squared plus straight b x plus straight c equals 0 mempunyai penyelesaian x subscript 1 dan x subscript 2, maka nilai:


x subscript 1 plus x subscript 2 equals negative straight b over straight a


Dan


x subscript 1 cross times x subscript 2 equals straight c over straight a


Maka, dari persamaan kuadrat 2 x squared minus 4 x minus 6 equals 0 yang mempunyai nilai straight a equals 2straight b equals negative 4, dan straight c equals negative 6, diperoleh:


table attributes columnalign right center left columnspacing 0px end attributes row cell x subscript 1 plus x subscript 2 end cell equals cell negative straight b over straight a end cell row blank equals cell negative fraction numerator left parenthesis negative 4 right parenthesis over denominator 2 end fraction end cell row blank equals 2 end table


Dan


table attributes columnalign right center left columnspacing 0px end attributes row cell x subscript 1 cross times x subscript 2 end cell equals cell straight c over straight a end cell row blank equals cell fraction numerator negative 6 over denominator 2 end fraction end cell row blank equals cell negative 3 end cell end table


Sehingga diperoleh:


table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses 2 x subscript 1 close parentheses squared plus open parentheses 2 x subscript 2 close parentheses squared end cell equals cell 4 x subscript 1 squared plus 4 x subscript 2 squared end cell row blank equals cell 4 open parentheses x subscript 1 squared plus x subscript 2 squared close parentheses end cell end table


Ingat bahwa open parentheses x subscript 1 plus x subscript 2 close parentheses squared equals x subscript 1 squared plus 2 x subscript 1 times x subscript 2 plus x subscript 2 squared, sehingga:


table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses 2 x subscript 1 close parentheses squared plus open parentheses 2 x subscript 2 close parentheses squared end cell equals cell 4 x subscript 1 squared plus 4 x subscript 2 squared end cell row blank equals cell 4 open parentheses x subscript 1 squared plus x subscript 2 squared close parentheses end cell row blank equals cell 4 open parentheses open parentheses x subscript 1 plus x subscript 2 close parentheses squared minus 2 x subscript 1 times x subscript 2 close parentheses end cell row blank equals cell 4 open parentheses open parentheses 2 close parentheses squared minus 2 times open parentheses negative 3 close parentheses close parentheses end cell row blank equals cell 4 open parentheses 4 plus 6 close parentheses end cell row blank equals cell 4 times 10 end cell row blank equals 40 end table


Jadi, nilai open parentheses 2 x subscript 1 close parentheses squared plus open parentheses 2 x subscript 2 close parentheses squared equals 40.

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