Himpunan bilangan real x yang merupakan penyelesaian dari persamaan adalah ....

Pertanyaan

Himpunan bilangan real $x$ yang merupakan penyelesaian dari persamaan $begin mathsize 14px style open vertical bar fraction numerator x squared plus 2 x minus 1 over denominator x plus 1 end fraction close vertical bar equals 3 comma space x not equal to negative 1 end style$ adalah ....

D. Natalia

Master Teacher

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah C.

Pembahasan

Berdasarkan definisi nilai mutlak, diperoleh nilai-nilai sebagai berikut.

$begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar fraction numerator x squared plus 2 x minus 1 over denominator x plus 1 end fraction close vertical bar end cell equals cell 3 open curly brackets table row cell fraction numerator x squared plus 2 x minus 1 over denominator x plus 1 end fraction equals 3 comma end cell cell jika blank fraction numerator x squared plus 2 x minus 1 over denominator x plus 1 end fraction greater or equal than 0 end cell row cell fraction numerator x squared plus 2 x minus 1 over denominator x plus 1 end fraction equals negative 3 comma end cell cell jika blank fraction numerator x squared plus 2 x minus 1 over denominator x plus 1 end fraction less than 0 end cell end table close end cell end table end style$

Selanjutnya, cari masing-masing penyelesaian dari bentuk di atas.

Perhatikan perhitungan berikut!

$table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator x squared plus 2 x minus 1 over denominator x plus 1 end fraction end cell equals cell 3 semicolon space x not equal to negative 1 end cell row cell fraction numerator x squared plus 2 x minus 1 over denominator x plus 1 end fraction times open parentheses x plus 1 close parentheses end cell equals cell 3 times open parentheses x plus 1 close parentheses end cell row cell x squared plus 2 x minus 1 end cell equals cell 3 x plus 3 end cell row cell x squared minus x end cell equals 4 row cell open parentheses x minus 1 half close parentheses squared minus 1 fourth end cell equals 4 row cell open parentheses x minus 1 half close parentheses squared end cell equals cell 4 plus 1 fourth end cell row cell open parentheses x minus 1 half close parentheses squared end cell equals cell 17 over 4 end cell row cell x minus 1 half end cell equals cell plus-or-minus square root of 17 over 4 end root end cell row x equals cell 1 half plus-or-minus 1 half square root of 17 end cell row x equals cell 1 half open parentheses 1 plus-or-minus square root of 17 close parentheses end cell end table$

Dari persamaan $begin mathsize 14px style fraction numerator x squared plus 2 x minus 1 over denominator x plus 1 end fraction equals 3 end style$ , diperoleh penyelesaian dan , serta keduanya memenuhi syarat x not equal to negative 1 .

Berikutnya, perhatikan pula perhitungan berikut!

$table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator x squared plus 2 x minus 1 over denominator x plus 1 end fraction end cell equals cell negative 3 semicolon space x not equal to negative 1 end cell row cell fraction numerator x squared plus 2 x minus 1 over denominator x plus 1 end fraction times open parentheses x plus 1 close parentheses end cell equals cell negative 3 times open parentheses x plus 1 close parentheses end cell row cell x squared plus 2 x minus 1 end cell equals cell negative 3 x minus 3 end cell row cell x squared plus 5 x end cell equals cell negative 2 end cell row cell open parentheses x plus 5 over 2 close parentheses squared minus 25 over 4 end cell equals cell negative 2 end cell row cell open parentheses x plus 5 over 2 close parentheses squared end cell equals cell negative 2 plus 25 over 4 end cell row cell open parentheses x plus 5 over 2 close parentheses squared end cell equals cell 17 over 4 end cell row cell x plus 5 over 2 end cell equals cell plus-or-minus square root of 17 over 4 end root end cell row x equals cell negative 5 over 2 plus-or-minus 1 half square root of 17 end cell row x equals cell 1 half open parentheses negative 5 plus-or-minus square root of 17 close parentheses end cell end table$

Dari persamaan $begin mathsize 14px style fraction numerator x squared plus 2 x minus 1 over denominator x plus 1 end fraction equals negative 3 end style$ , diperoleh penyelesaian dan , serta keduanya memenuhi syarat x not equal to negative 1 .

Jika kita gambarkan dalam koordinat kartesius, antara fungsi $begin mathsize 14px style f open parentheses x close parentheses equals open vertical bar fraction numerator x squared plus 2 x minus 1 over denominator x plus 1 end fraction close vertical bar end style$ dan akan berpotongan di empat titik seperti pada gambar berikut ini.

Dengan demikian, himpunan penyelesaian dari persamaan $begin mathsize 14px style open vertical bar fraction numerator x squared plus 2 x minus 1 over denominator x plus 1 end fraction close vertical bar equals 3 end style$ adalah begin mathsize 12px style open curly brackets 1 half open parentheses negative 5 minus square root of 17 close parentheses comma blank 1 half open parentheses 1 minus square root of 17 close parentheses comma blank 1 half open parentheses 1 plus square root of 17 close parentheses comma blank 1 half open parentheses negative 5 plus square root of 17 close parentheses close curly brackets end style .

Jadi, jawaban yang tepat adalah C.