Pertanyaan

Himpunan penyelesaian x yang memenuhi pertidaksamaan 4 − 3 x ≤ x 2 − 4 x ≤ 2 + 6 x adalah

Himpunan penyelesaian x yang memenuhi pertidaksamaan $4 - 3 x \leq x^{2} - 4 x \leq 2 + 6 x$ adalah

$open curly brackets x element of R vertical line x less or equal than fraction numerator 1 minus square root of 17 over denominator 2 end fraction space atau space x greater or equal than fraction numerator 1 plus square root of 17 over denominator 2 end fraction close curly brackets$
$open curly brackets x element of R vertical line fraction numerator 1 minus square root of 17 over denominator 2 end fraction space less or equal than x less or equal than fraction numerator 1 plus square root of 17 over denominator 2 end fraction close curly brackets$
$open curly brackets x element of R vertical line 5 minus 3 square root of 3 space less or equal than x less or equal than fraction numerator 1 plus square root of 17 over denominator 2 end fraction close curly brackets$
$open curly brackets x element of R vertical line fraction numerator 1 plus square root of 17 over denominator 2 end fraction space less or equal than x less or equal than 5 plus 3 square root of 3 close curly brackets$

M. Nasrullah

Master Teacher

Mahasiswa/Alumni Universitas Negeri Makassar

Jawaban terverifikasi

Pembahasan

Ingat bahwa pentuk pertidaksamaan Sehingga diperoleh perhitungan: Untuk Dengan menggunakan rumus abc, maka akar-akar dari persamaan di atas titik uji Sehingga himpunan penyelesainya: Untuk engan menggunakan rumus abc, maka akar-akar dari persamaan di atas titik uji Sehingga himpunan penyelesainya: Sehingga irisan dari kedua himpunan tersebut adalah Dengan demikian, Himpunan penyelesaian x yang memenuhi pertidaksamaan tersebut adalah Jadi, jawaban yang tepat adalah E

Ingat bahwa pentuk pertidaksamaan

Sehingga diperoleh perhitungan:

4 minus 3 x less or equal than x squared minus 4 x space dan space x squared minus 4 x less or equal than 2 plus 6 x

Untuk 4 minus 3 x less or equal than x squared minus 4 x

table attributes columnalign right center left columnspacing 0px end attributes row cell 4 minus 3 x end cell less or equal than cell x squared minus 4 x end cell row 0 less or equal than cell x squared minus 4 x plus 3 x minus 4 end cell row 0 less or equal than cell x squared minus x minus 4 end cell row cell x squared minus x minus 4 end cell greater or equal than 0 end table

Dengan menggunakan rumus abc, maka akar-akar dari persamaan di atas

$table attributes columnalign right center left columnspacing 0px end attributes row cell x subscript 1 comma 2 end subscript end cell equals cell fraction numerator negative b plus-or-minus square root of b squared minus 4 a c end root over denominator 2 a end fraction end cell row blank equals cell fraction numerator negative open parentheses negative 1 close parentheses plus-or-minus square root of open parentheses negative 1 close parentheses minus 4 open parentheses 1 close parentheses open parentheses negative 4 close parentheses end root over denominator 2 open parentheses 1 close parentheses end fraction end cell row blank equals cell fraction numerator 1 plus-or-minus square root of 7 over denominator 2 end fraction end cell row x equals cell fraction numerator 1 plus square root of 17 over denominator 2 end fraction space atau space x equals fraction numerator 1 minus square root of 17 over denominator 2 end fraction end cell end table$

titik uji x equals 0

Sehingga himpunan penyelesainya:

$H p subscript 1 equals open curly brackets x vertical line x less or equal than fraction numerator 1 minus square root of 17 over denominator 2 end fraction space atau space x greater or equal than fraction numerator 1 plus square root of 17 over denominator 2 end fraction close curly brackets$

Untuk x squared minus 4 x less or equal than 2 plus 6 x

table attributes columnalign right center left columnspacing 0px end attributes row cell x squared minus 4 x end cell less or equal than cell 2 plus 6 x end cell row cell x squared minus 4 x minus 6 x minus 2 end cell less or equal than 0 row cell x squared minus 10 x minus 2 end cell less or equal than 0 end table

engan menggunakan rumus abc, maka akar-akar dari persamaan di atas

$table attributes columnalign right center left columnspacing 0px end attributes row cell x subscript 1 comma 2 end subscript end cell equals cell fraction numerator negative b plus-or-minus square root of b squared minus 4 a c end root over denominator 2 a end fraction end cell row blank equals cell fraction numerator negative open parentheses negative 10 close parentheses plus-or-minus square root of open parentheses negative 1 close parentheses minus 4 open parentheses 1 close parentheses open parentheses negative 2 close parentheses end root over denominator 2 open parentheses 1 close parentheses end fraction end cell row blank equals cell fraction numerator 10 plus-or-minus square root of 108 over denominator 2 end fraction end cell row x equals cell 5 plus 3 square root of 3 space atau space x equals 5 minus 3 square root of 3 end cell end table$

titik uji x equals 0

table attributes columnalign right center left columnspacing 0px end attributes row cell x squared minus 10 x minus 2 end cell less or equal than 0 row cell open parentheses 0 close parentheses minus open parentheses 0 close parentheses minus 2 end cell less or equal than 0 row cell negative 2 end cell less or equal than cell 0 space open parentheses Benar close parentheses end cell end table

Sehingga himpunan penyelesainya:

Sehingga irisan dari kedua himpunan tersebut adalah $H p equals open curly brackets x vertical line fraction numerator 1 plus square root of 17 over denominator 2 end fraction table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table table attributes columnalign right center left columnspacing 0px end attributes row blank less or equal than blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank x end table table attributes columnalign right center left columnspacing 0px end attributes row blank less or equal than blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 5 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell plus 3 end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell square root of 3 end cell end table close curly brackets$

Dengan demikian, Himpunan penyelesaian x yang memenuhi pertidaksamaan tersebut adalah $H p equals open curly brackets x vertical line fraction numerator 1 plus square root of 17 over denominator 2 end fraction table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table table attributes columnalign right center left columnspacing 0px end attributes row blank less or equal than blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank x end table table attributes columnalign right center left columnspacing 0px end attributes row blank less or equal than blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 5 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell plus 3 end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell square root of 3 end cell end table close curly brackets$

Jadi, jawaban yang tepat adalah E