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Diketahui polinomial  habis dibagi  dan bersisa apabila dibagi . Tentukan akar-akar persamaan polinomial .

Pertanyaan

Diketahui polinomial begin mathsize 14px style f left parenthesis x right parenthesis equals x to the power of 4 minus x cubed plus a x squared plus x plus b end style habis dibagi begin mathsize 14px style left parenthesis x minus 3 right parenthesis end style dan bersisa begin mathsize 14px style negative 12 space end styleapabila dibagi begin mathsize 14px style left parenthesis x minus 2 right parenthesis end style. Tentukan akar-akar persamaan polinomial begin mathsize 14px style f left parenthesis x right parenthesis equals 0 end style.

  1. ...space  

  2. ...space 

Pembahasan Video:

Pembahasan Soal:

Polinomial begin mathsize 14px style straight f left parenthesis straight x right parenthesis equals straight x to the power of 4 minus straight x cubed plus ax squared plus straight x plus straight b end style 

Polinomial begin mathsize 14px style straight f left parenthesis straight x right parenthesis end style  habis dibagi begin mathsize 14px style left parenthesis x minus 3 right parenthesis end style, berarti:

 begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell straight f left parenthesis 3 right parenthesis end cell equals 0 row cell 3 to the power of 4 minus 3 cubed plus straight a times 3 squared plus 3 plus straight b end cell equals 0 row cell 57 plus 9 straight a plus straight b end cell equals 0 row cell 9 straight a plus straight b end cell equals cell negative 57 space horizontal ellipsis left parenthesis persamaan space 1 right parenthesis end cell end table end style 

Polinomial undefined dibagi begin mathsize 14px style open parentheses x minus 2 close parentheses end style bersisa begin mathsize 14px style negative 12 end style, berarti:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell straight f left parenthesis 2 right parenthesis end cell equals cell negative 12 end cell row cell 2 to the power of 4 minus 2 cubed plus straight a times 2 squared plus 2 plus straight b end cell equals cell negative 12 end cell row cell 10 plus 4 a plus straight b end cell equals cell negative 12 end cell row cell 4 straight a plus straight b end cell equals cell negative 22 space horizontal ellipsis left parenthesis persamaan space 2 right parenthesis end cell end table end style 

Eliminasi b dari  persamaan 1 dan 2 untuk menentukan nilai a

begin mathsize 14px style stack attributes charalign center stackalign right end attributes row 9 a plus b equals negative 57 end row row 4 a plus b equals negative 22 end row horizontal line row 5 a equals negative 35 end row row a equals negative 7 end row end stack minus end style 

Substitusi nilai begin mathsize 14px style a space end stylepada salah satu persamaan untuk menentukan nilai begin mathsize 14px style b end style

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 9 a plus b end cell equals cell negative 57 end cell row cell 9 open parentheses negative 7 close parentheses plus b end cell equals cell negative 57 end cell row cell negative 63 plus b end cell equals cell negative 57 end cell row b equals 6 end table end style 

Sehingga persamaan polinomialnya menjadi begin mathsize 14px style straight f left parenthesis straight x right parenthesis equals x to the power of 4 minus x cubed minus 7 x squared plus x plus 6 end style.
Karena begin mathsize 14px style left parenthesis x minus 3 right parenthesis end style habis membagi polinom tersebut artinya dengan begin mathsize 14px style x equals 3 end style dapat dicari akar lainnya menggunakan skema Horner.

begin mathsize 14px style long division with stack on the left by table row 1 row vertical ellipsis end table number space table row cell negative 1 end cell row cell space space 3 end cell end table number space table row cell negative 7 end cell row cell space 6 end cell end table number space table row 1 row cell negative 3 end cell end table number space table row 6 row cell negative 6 end cell end table yields space 1 number space space space space space space 2 number space space space minus 1 number space space minus 2 number space space space space space space 0 pile 3 end pile end long division plus end style 

Diperoleh hasil bagi begin mathsize 14px style x cubed plus 2 x squared minus x minus 2 end style, sehingga:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses x cubed plus 2 x squared minus x minus 2 close parentheses end cell equals 0 row cell x squared open parentheses x plus 2 close parentheses minus 1 open parentheses x plus 2 close parentheses end cell equals 0 row cell open parentheses x plus 2 close parentheses open parentheses x squared minus 1 close parentheses end cell equals 0 row cell open parentheses x plus 2 close parentheses open parentheses x plus 1 close parentheses open parentheses x minus 1 close parentheses end cell equals 0 row cell x equals negative 2 comma space x equals minus 1 comma space x end cell equals 1 end table end style  

Jadi akar-akar persamaan polinomialnya adalah begin mathsize 14px style left curly bracket negative 2 comma negative 1 comma 1 comma 3 right curly bracket end style.

Pembahasan terverifikasi oleh Roboguru

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Terakhir diupdate 01 Mei 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Diketahui polinomial  habis dibagi  dan bersisa apabila dibagi . Tentukan nilai dan .

Pembahasan Soal:

Polinomial begin mathsize 14px style straight f left parenthesis straight x right parenthesis equals straight x to the power of 4 minus straight x cubed plus ax squared plus straight x plus straight b end style 

Polinomial begin mathsize 14px style straight f left parenthesis straight x right parenthesis end style  habis dibagi begin mathsize 14px style left parenthesis x minus 3 right parenthesis end style, berarti:

 begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell straight f left parenthesis 3 right parenthesis end cell equals 0 row cell 3 to the power of 4 minus 3 cubed plus straight a times 3 squared plus 3 plus straight b end cell equals 0 row cell 57 plus 9 straight a plus straight b end cell equals 0 row cell 9 straight a plus straight b end cell equals cell negative 57 space horizontal ellipsis left parenthesis persamaan space 1 right parenthesis end cell end table end style 

Polinomial undefined dibagi begin mathsize 14px style open parentheses x minus 2 close parentheses end style bersisa begin mathsize 14px style negative 12 end style, berarti:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell straight f left parenthesis 2 right parenthesis end cell equals cell negative 12 end cell row cell 2 to the power of 4 minus 2 cubed plus straight a times 2 squared plus 2 plus straight b end cell equals cell negative 12 end cell row cell 10 plus 4 a plus straight b end cell equals cell negative 12 end cell row cell 4 straight a plus straight b end cell equals cell negative 22 space horizontal ellipsis left parenthesis persamaan space 2 right parenthesis end cell end table end style 

Eliminasi b dari  persamaan 1 dan 2 untuk menentukan nilai a

begin mathsize 14px style stack attributes charalign center stackalign right end attributes row 9 a plus b equals negative 57 end row row 4 a plus b equals negative 22 end row horizontal line row 5 a equals negative 35 end row row a equals negative 7 end row end stack minus end style 

Substitusi nilai begin mathsize 14px style a space end stylepada salah satu persamaan untuk menentukan nilai begin mathsize 14px style b end style

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 9 a plus b end cell equals cell negative 57 end cell row cell 9 open parentheses negative 7 close parentheses plus b end cell equals cell negative 57 end cell row cell negative 63 plus b end cell equals cell negative 57 end cell row b equals 6 end table end style 

Jadi nilai begin mathsize 14px style a equals negative 7 end style dan begin mathsize 14px style b equals 6 end style 

0

Roboguru

Tentukan akar-akar dari persamaan polinomial .

Pembahasan Soal:

Terdapat kesalahan pada soal. Persamaan yang dimaksud bukan begin mathsize 14px style 2 x to the power of 4 plus 3 x cubed minus 14 x squared minus 3 x plus 2 equals 0 end style, melainkan begin mathsize 14px style 2 x to the power of 4 plus 3 x cubed minus 4 x squared minus 3 x plus 2 equals 0 end style.

Persamaan: begin mathsize 14px style 2 x to the power of 4 plus 3 x cubed minus 4 x squared minus 3 x plus 2 equals 0 end style

Error converting from MathML to accessible text.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank Suku end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank konstan end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank adalah end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank p end table table attributes columnalign right center left columnspacing 0px end attributes row blank equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell 2 space left parenthesis faktor space dari space 2 space adalah space plus-or-minus 1 comma space dan space plus-or-minus 2 right parenthesis end cell end table end style
begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank Akar end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank yang end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank mungkin end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank adalah end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell straight p over straight q end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank plus-or-minus end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell 1 over 1 end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank comma end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank plus-or-minus end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell 1 half end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank comma end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank plus-or-minus end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell 2 over 1 end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank comma end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank dan end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank plus-or-minus end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell 2 over 2 end cell end table end style  

Uji untuk begin mathsize 14px style x equals 1 end style

 begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 2 x to the power of 4 plus 3 x cubed minus 4 x squared minus 3 x plus 2 end cell equals cell 2 times 1 to the power of 4 plus 3 times 1 cubed minus 4 times 1 squared minus 3 times 1 plus 2 end cell row blank equals cell 2 plus 3 minus 4 minus 3 plus 2 end cell row blank equals 0 end table end style 

Karena sisa = 0, maka diperoleh begin mathsize 14px style x equals 1 end style merupakan salah satu akarnya.

Uji untuk undefined

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 2 x to the power of 4 plus 3 x cubed minus 4 x squared minus 3 x plus 2 end cell equals cell 2 times left parenthesis negative 2 right parenthesis to the power of 4 plus 3 times left parenthesis negative 2 right parenthesis cubed minus 4 times left parenthesis negative 2 right parenthesis squared minus 3 times left parenthesis negative 2 right parenthesis plus 2 end cell row blank equals cell 32 minus 24 minus 16 plus 6 plus 2 end cell row blank equals 0 end table end style

Karena sisa = 0, maka diperoleh begin mathsize 14px style x equals negative 2 end style merupakan salah satu akarnya.

Persamaan menjadi

 begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 2 x to the power of 4 plus 3 x cubed minus 4 x squared minus 3 x plus 2 end cell equals 0 row cell left parenthesis x minus 1 right parenthesis left parenthesis x plus 2 right parenthesis times h left parenthesis x right parenthesis end cell equals 0 row cell left parenthesis x squared plus x minus 2 right parenthesis left parenthesis 2 x squared plus x minus 1 right parenthesis end cell equals 0 row cell left parenthesis x squared plus x minus 2 right parenthesis left parenthesis 2 x minus 1 right parenthesis left parenthesis x plus 1 right parenthesis end cell equals 0 end table end style 

Diperoleh akar lainnya yaitu begin mathsize 14px style x equals 1 half space dan space x equals negative 1 end style

Jadi, akar-akar persamaan tersebut adalahbegin mathsize 14px style 1 comma space minus 1 comma space minus 2 comma space dan space 1 half end style.

 

2

Roboguru

Diketahui persamaan suku banyak dengan akar-akarnya  Jika , maka nilai  yang mungkin adalah...

Pembahasan Soal:

Perhatikan penghitungan berikut!

begin mathsize 14px style x cubed plus m x squared minus 4 x minus 16 equals 0 x subscript 1 times x subscript 2 equals x subscript 3 Vieta space formula colon x subscript 1 plus x subscript 2 plus x subscript 3 equals negative m x subscript 1 x subscript 2 plus x subscript 2 x subscript 3 plus x subscript 1 x subscript 3 equals negative 4 x subscript 1 x subscript 2 x subscript 3 equals 16  Berdasarkan space formula space di space atas comma x subscript 1 x subscript 2 plus x subscript 2 x subscript 3 plus x subscript 1 x subscript 3 equals negative 4 x subscript 3 plus x subscript 2 x subscript 3 plus x subscript 1 x subscript 3 equals negative 4 x subscript 3 open parentheses 1 plus x subscript 2 plus x subscript 1 close parentheses equals negative 4 1 plus x subscript 2 plus x subscript 1 equals fraction numerator negative 4 over denominator x subscript 3 end fraction x subscript 2 plus x subscript 1 equals fraction numerator negative 4 over denominator x subscript 3 end fraction minus 1  Berdasarkan space formula space di space atas comma x subscript 1 x subscript 2 x subscript 3 equals 16 x subscript 3 x subscript 3 equals 16 x subscript 3 squared equals 16 x subscript 3 equals plus-or-minus square root of 16 x subscript 3 equals plus-or-minus 4  Berdasarkan space formula space di space atas comma x subscript 1 plus x subscript 2 plus x subscript 3 equals negative m fraction numerator negative 4 over denominator x subscript 3 end fraction minus 1 plus x subscript 3 equals negative m m equals 4 over x subscript 3 plus 1 minus x subscript 3 untuk space x subscript 3 equals 4 comma space maka m equals 4 over 4 plus 1 minus 4 equals negative 2 untuk space x subscript 3 equals negative 4 comma space maka m equals fraction numerator 4 over denominator negative 4 end fraction plus 1 minus open parentheses negative 4 close parentheses equals 4 sehingga comma space nilai space m space yang space mungkin space yaitu space minus 2 space atau space 4. end style 

Oleh karena itu, jawaban yang benar adalah D. space 

0

Roboguru

Suku banyak  berderajat  habis dibagi . Jika  dibagi  bersisa  dan dibagi   bersisa , tentukan hasil bagi dan sisa pembagian jika  dibagi .

Pembahasan Soal:

Suku banyak undefined berderajat undefined , misalkan begin mathsize 14px style f left parenthesis x right parenthesis equals a x squared plus b x plus c end style,

undefined habis dibagi begin mathsize 14px style 2 x plus 3 end style,

begin mathsize 14px style f open parentheses negative begin inline style 3 over 2 end style close parentheses equals a open parentheses negative begin inline style 3 over 2 end style close parentheses squared plus b open parentheses negative begin inline style 3 over 2 end style close parentheses plus c equals 0 space space space space space space space space space left right double arrow begin inline style 9 over 4 end style a minus begin inline style 3 over 2 end style b plus c equals 0 space left parenthesis cross times 4 right parenthesis space space space space space space space space space left right double arrow 9 a minus 6 b plus 4 c equals 0 space... left parenthesis i right parenthesis end style

  undefined dibagi undefined bersisa begin mathsize 14px style 36 end style,

begin mathsize 14px style f left parenthesis 3 right parenthesis equals a left parenthesis 3 right parenthesis squared plus b left parenthesis 3 right parenthesis plus c equals 36 space space space space space left right double arrow 9 a plus 3 b plus c equals 36 space... left parenthesis i i right parenthesis end style

undefined dibagi begin mathsize 14px style x plus 2 end style  bersisa begin mathsize 14px style 1 end style,

begin mathsize 14px style f left parenthesis negative 2 right parenthesis equals a left parenthesis negative 2 right parenthesis squared plus b left parenthesis negative 2 right parenthesis plus c equals 1 space space space space space space space space space left right double arrow 4 a minus 2 b plus c equals 1 space... left parenthesis i i i right parenthesis end style

Eliminasi c:

selanjutnya akan mengeliminasi c dengan cara :

kalikan persamaan (ii) yaitu begin mathsize 14px style 9 a plus 3 b plus c equals 36 end style dengan 4 sehingga didapat begin mathsize 14px style 36 a plus 12 b plus 4 c equals 144 end style, kemudian kurangkan persamaan tersebut dengan persaman (i) yaitu begin mathsize 14px style 9 a minus 6 b plus 4 c equals 0 end style sehingga didapat persamaan baru yaitu :

begin mathsize 14px style 27 a plus 18 b equals 144... left parenthesis i v right parenthesis end style 

kalikan persamaan (iii) yaitu begin mathsize 14px style 4 a minus 2 b plus c equals 1 end style  dengan 4 sehingga didapat begin mathsize 14px style 16 a minus 8 b plus 4 c equals 4 end style , kemudian kurangkan persamaan tersebut dengan persaman (i) yaitu begin mathsize 14px style 9 a minus 6 b plus 4 c equals 0 end style sehingga didapat persamaan baru yaitu :

begin mathsize 14px style 7 a minus 2 b equals 4... left parenthesis v right parenthesis end style 

Eliminasi b:

selanjutnya akan mengeliminasi b dengan cara :

kalikan persamaan (v) yaitu begin mathsize 14px style 7 a minus 2 b equals 4 end style   dengan 9 sehingga didapat begin mathsize 14px style 63 a minus 18 b equals 36 end style  , kemudian jumlahkan persamaan tersebut dengan persaman (iv) yaitu begin mathsize 14px style 27 a plus 18 b equals 144 end style sehingga didapat :

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 90 a end cell equals 180 row a equals 2 end table end style 

Substitusi begin mathsize 14px style a equals 2 end style ke persamaan (v):

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 7 a minus 2 b end cell equals 4 row cell 7 left parenthesis 2 right parenthesis minus 2 b end cell equals 4 row cell 14 minus 2 b end cell equals 4 row cell negative 2 b end cell equals cell negative 10 end cell row b equals 5 end table end style 

 

Substitusi begin mathsize 14px style a equals 2 comma space b equals 5 end style ke persamaan (i):

 begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 9 a minus 6 b plus 4 c end cell equals 0 row cell 9 left parenthesis 2 right parenthesis minus 6 left parenthesis 5 right parenthesis plus 4 c end cell equals 0 row cell 18 minus 30 plus 4 c end cell equals 0 row cell negative 12 plus 4 c end cell equals 0 row cell 4 c end cell equals 12 row c equals 3 end table end style 

Diperoleh suku banyak begin mathsize 14px style f left parenthesis x right parenthesis equals 2 x squared plus 5 x plus 3 end style.

Hasil bagi dan sisa bagi begin mathsize 14px style f left parenthesis x right parenthesis end style dibagi begin mathsize 14px style 3 x minus 1 end style,

  

Hasil bagi: begin mathsize 14px style begin inline style 2 over 3 end style x plus begin inline style 17 over 9 end style end style, Sisa pembagian: begin mathsize 14px style begin inline style 44 over 9 end style end style.

1

Roboguru

Suku banyak  berderajat  habis dibagi . Jika  dibagi  bersisa  dan dibagi   bersisa , tentukan rumus suku banyak .

Pembahasan Soal:

Suku banyak undefined berderajat undefined , misalkan begin mathsize 14px style f left parenthesis x right parenthesis equals a x squared plus b x plus c end style,

undefined habis dibagi begin mathsize 14px style 2 x plus 3 end style,

begin mathsize 14px style f open parentheses negative begin inline style 3 over 2 end style close parentheses equals a open parentheses negative begin inline style 3 over 2 end style close parentheses squared plus b open parentheses negative begin inline style 3 over 2 end style close parentheses plus c equals 0 space space space space space space space space space left right double arrow begin inline style 9 over 4 end style a minus begin inline style 3 over 2 end style b plus c equals 0 space left parenthesis cross times 4 right parenthesis space space space space space space space space space left right double arrow 9 a minus 6 b plus 4 c equals 0 space... left parenthesis i right parenthesis end style

  undefined dibagi undefined bersisa begin mathsize 14px style 36 end style,

begin mathsize 14px style f left parenthesis 3 right parenthesis equals a left parenthesis 3 right parenthesis squared plus b left parenthesis 3 right parenthesis plus c equals 36 space space space space space left right double arrow 9 a plus 3 b plus c equals 36 space... left parenthesis i i right parenthesis end style

undefined dibagi begin mathsize 14px style x plus 2 end style  bersisa begin mathsize 14px style 1 end style,

begin mathsize 14px style f left parenthesis negative 2 right parenthesis equals a left parenthesis negative 2 right parenthesis squared plus b left parenthesis negative 2 right parenthesis plus c equals 1 space space space space space space space space space left right double arrow 4 a minus 2 b plus c equals 1 space... left parenthesis i i i right parenthesis end style

Eliminasi c:

Selanjutnya akan mengeliminasi c dengan cara  kalikan persamaan (ii) yaitu begin mathsize 14px style 9 a plus 3 b plus c equals 36 end style  dengan 4 sehingga didapat begin mathsize 14px style 36 a plus 9 b plus 4 c equals 144 end style , kemudian kurangkan persamaan tersebut dengan persaman (i) yaitu begin mathsize 14px style 9 a minus 6 b plus 4 c equals 0 end style sehingga didapat persamaan baru yaitu :

begin mathsize 14px style 27 a plus 18 b equals 144... left parenthesis i v right parenthesis end style 

Kemudian, kalikan persamaan (iii) yaitu begin mathsize 14px style 4 a minus 2 b plus c equals 1 end style  dengan 4 sehingga didapat begin mathsize 14px style 16 a minus 8 b plus 4 c equals 4 end style , kemudian kurangkan persamaan tersebut dengan persaman (i) yaitu begin mathsize 14px style 9 a minus 6 b plus 4 c equals 0 end style sehingga didapat persamaan baru yaitu :

begin mathsize 14px style 7 a minus 2 b equals 4... left parenthesis v right parenthesis end style 

Eliminasi b:

Selanjutnya akan mengeliminasi b dengan cara : kalikan persamaan (v) yaitu begin mathsize 14px style 7 a minus 2 b equals 4 end style  dengan 9 sehingga didapat begin mathsize 14px style 63 a minus 18 b equals 36 end style , kemudian jumlahkan persamaan tersebut dengan persaman (iv) yaitu begin mathsize 14px style 27 a plus 18 b equals 144 end style sehingga didapat :

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 90 a end cell equals 180 row a equals 2 end table end style 

Substitusi begin mathsize 14px style a equals 2 end style  ke persamaan (v):

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 7 a minus 2 b end cell equals 4 row cell 7 left parenthesis 2 right parenthesis minus 2 b end cell equals 4 row cell 14 minus 2 b end cell equals 4 row cell negative 2 b end cell equals cell negative 10 end cell row b equals 5 end table end style 

Substitusi begin mathsize 14px style a equals 2 space dan space b equals 5 end style ke persamaan (i):

 begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 9 a minus 6 b plus 4 c end cell equals 0 row cell 9 left parenthesis 2 right parenthesis minus 6 left parenthesis 5 right parenthesis plus 4 c end cell equals 0 row cell 18 minus 30 plus 4 c end cell equals 0 row cell negative 12 plus 4 c end cell equals 0 row cell 4 c end cell equals 12 row c equals 3 end table end style 

Diperoleh suku banyak undefined.

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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