Pertanyaan

Banyak pasangan dengan x , y ∈ R yang memenuhi persamaan 3 x 2 − ∣ x y ∣ + 2 = 0 dan ( 6 x − y ) 2 + y 2 = 24 adalah ....

Banyak pasangan dengan $x, y \in R$ yang memenuhi persamaan $3 x^{2} - ∣ x y ∣ + 2 = 0$ dan $(6 x - y)^{2} + y^{2} = 24$ adalah ....

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F. Ayudhita

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Jawaban terverifikasi

Pembahasan

Dapat diperhatikan bahwa terdapat dua kemungkinan nilai , yaitu sebagai berikut. Kemungkinan pertamayaitu saat , maka . Kemudian, substitusi nilai ke . Saat , didapat Kemudaian, saat , didapat Didapat 2 pasang ,yaitu dan . Kita cek dulu kedua pasang tersebut memenuhi syarat atau tidak. Untuk , didapat Nilai . Akibatnya, pasangan memenuhi syarat . Untuk , didapat Nilai . Akibatnya, pasangan memenuhi syarat . Oleh karena itu, didapat2 pasang dari kemungkinan pertama. Kemungkinan kedua yaitu saat ,maka . Kemudian, substitusi ke . Kita cari nilai diskriminan dari . Dari persamaan , didapat , , dan . Didapat sehingga . Artinya, persamaan memiliki akar-akar tidak real. Oleh karena itu, saat atau ,tidak ada pasangan dengan yang memenuhi persamaan . Dengan demikian, banyak pasangan dengan yang memenuhi adalah 2. Jadi, jawaban yang tepat adalah D. Untuk mempelajarinya lebih jelas, tonton video selanjutnya.

Dapat diperhatikan bahwa terdapat dua kemungkinan nilai , yaitu sebagai berikut.

Kemungkinan pertama yaitu saat , maka .

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 3 x squared minus open vertical bar x y close vertical bar plus 2 end cell equals 0 row cell 3 x squared minus x y plus 2 end cell equals 0 row cell x y end cell equals cell 3 x squared plus 2 end cell row y equals cell fraction numerator 3 x squared plus 2 over denominator x end fraction end cell end table end style$

Kemudian, substitusi nilai $begin mathsize 14px style y equals fraction numerator 3 x squared plus 2 over denominator x end fraction end style$ ke .

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell left parenthesis 6 x minus y right parenthesis squared plus y squared end cell equals 24 row cell open parentheses 6 x minus open parentheses fraction numerator 3 x squared plus 2 over denominator x end fraction close parentheses close parentheses squared plus open parentheses fraction numerator 3 x squared plus 2 over denominator x end fraction close parentheses squared end cell equals 24 row cell open parentheses fraction numerator 6 x squared over denominator x end fraction minus open parentheses fraction numerator 3 x squared plus 2 over denominator x end fraction close parentheses close parentheses squared plus open parentheses fraction numerator 3 x squared plus 2 over denominator x end fraction close parentheses squared end cell equals 24 row cell open parentheses fraction numerator 6 x squared minus 3 x squared minus 2 over denominator x end fraction close parentheses squared plus open parentheses fraction numerator 3 x squared plus 2 over denominator x end fraction close parentheses squared end cell equals 24 row cell open parentheses fraction numerator 3 x squared minus 2 over denominator x end fraction close parentheses squared plus open parentheses fraction numerator 3 x squared plus 2 over denominator x end fraction close parentheses squared end cell equals 24 row cell fraction numerator 9 x to the power of 4 minus 12 x squared plus 4 over denominator x squared end fraction plus fraction numerator 9 x to the power of 4 plus 12 x squared plus 4 over denominator x squared end fraction end cell equals 24 row cell fraction numerator 18 x to the power of 4 plus 8 over denominator x squared end fraction end cell equals 24 row cell 18 x to the power of 4 plus 8 end cell equals cell 24 x squared end cell row cell 18 x to the power of 4 minus 24 x squared plus 8 end cell equals cell 0 space space space space space left parenthesis divided by 2 right parenthesis end cell row cell 9 x to the power of 4 minus 12 x squared plus 4 end cell equals 0 row cell left parenthesis 3 x squared minus 2 right parenthesis squared end cell equals 0 row cell 3 x squared minus 2 end cell equals 0 row cell left parenthesis square root of 3 x minus square root of 2 right parenthesis left parenthesis square root of 3 x plus square root of 2 right parenthesis end cell equals 0 row x equals cell fraction numerator square root of 2 over denominator square root of 3 end fraction space atau space x equals negative fraction numerator square root of 2 over denominator square root of 3 end fraction end cell end table end style$

Saat $begin mathsize 14px style x equals fraction numerator square root of 2 over denominator square root of 3 end fraction end style$ , didapat

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row y equals cell fraction numerator 3 x squared plus 2 over denominator x end fraction end cell row blank equals cell fraction numerator 3 open parentheses begin display style fraction numerator square root of 2 over denominator square root of 3 end fraction end style close parentheses squared plus 2 over denominator begin display style fraction numerator square root of 2 over denominator square root of 3 end fraction end style end fraction end cell row blank equals cell open parentheses 3 open parentheses 2 over 3 close parentheses plus 2 close parentheses times fraction numerator square root of 3 over denominator square root of 2 end fraction end cell row blank equals cell fraction numerator 4 square root of 3 over denominator square root of 2 end fraction end cell row blank equals cell 2 square root of 6 end cell end table end style$

Kemudaian, saat $begin mathsize 14px style x equals negative fraction numerator square root of 2 over denominator square root of 3 end fraction end style$ , didapat

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row y equals cell fraction numerator 3 x squared plus 2 over denominator x end fraction end cell row blank equals cell fraction numerator 3 begin display style open parentheses negative fraction numerator square root of 2 over denominator square root of 3 end fraction close parentheses end style squared plus 2 over denominator begin display style negative fraction numerator square root of 2 over denominator square root of 3 end fraction end style end fraction end cell row blank equals cell open parentheses 3 open parentheses 2 over 3 close parentheses plus 2 close parentheses times open parentheses negative fraction numerator square root of 3 over denominator square root of 2 end fraction close parentheses end cell row blank equals cell negative fraction numerator 4 square root of 3 over denominator square root of 2 end fraction end cell row blank equals cell negative 2 square root of 6 end cell end table end style$

Didapat 2 pasang , yaitu $begin mathsize 14px style open parentheses fraction numerator square root of 2 over denominator square root of 3 end fraction comma space 2 square root of 6 close parentheses end style$ dan $begin mathsize 14px style open parentheses negative fraction numerator square root of 2 over denominator square root of 3 end fraction comma space minus 2 square root of 6 close parentheses end style$ .

Kita cek dulu kedua pasang tersebut memenuhi syarat atau tidak.

Untuk $begin mathsize 14px style open parentheses fraction numerator square root of 2 over denominator square root of 3 end fraction comma space 2 square root of 6 close parentheses end style$ , didapat

$begin mathsize 14px style x y equals fraction numerator square root of 2 over denominator square root of 3 end fraction times 2 square root of 6 equals fraction numerator square root of 2 over denominator square root of 3 end fraction times 2 square root of 2 times square root of 3 equals 4 end style$

Nilai . Akibatnya, pasangan $begin mathsize 14px style open parentheses fraction numerator square root of 2 over denominator square root of 3 end fraction comma space 2 square root of 6 close parentheses end style$ memenuhi syarat .

Untuk $begin mathsize 14px style open parentheses negative fraction numerator square root of 2 over denominator square root of 3 end fraction comma space minus 2 square root of 6 close parentheses end style$ , didapat

$begin mathsize 14px style x y equals open parentheses negative fraction numerator square root of 2 over denominator square root of 3 end fraction close parentheses times open parentheses negative 2 square root of 6 close parentheses equals fraction numerator square root of 2 over denominator square root of 3 end fraction times 2 square root of 6 equals fraction numerator square root of 2 over denominator square root of 3 end fraction times 2 square root of 2 times square root of 3 equals 4 end style$

Oleh karena itu, didapat 2 pasang dari kemungkinan pertama.

Kemungkinan kedua yaitu saat , maka .

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 3 x squared minus open vertical bar x y close vertical bar plus 2 end cell equals 0 row cell 3 x squared minus open parentheses negative x y close parentheses plus 2 end cell equals 0 row cell 3 x squared plus x y plus 2 end cell equals 0 row cell 3 x squared plus 2 end cell equals cell negative x y end cell row y equals cell fraction numerator 3 x squared plus 2 over denominator negative x end fraction end cell end table end style$

Kemudian, substitusi $begin mathsize 14px style y equals fraction numerator 3 x squared plus 2 over denominator negative x end fraction end style$ ke .

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses 6 x minus open parentheses fraction numerator 3 x squared plus 2 over denominator negative x end fraction close parentheses close parentheses squared plus open parentheses fraction numerator 3 x squared plus 2 over denominator negative x end fraction close parentheses squared end cell equals 24 row cell open parentheses 6 x plus open parentheses fraction numerator 3 x squared plus 2 over denominator x end fraction close parentheses close parentheses squared plus open parentheses fraction numerator 3 x squared plus 2 over denominator negative x end fraction close parentheses squared end cell equals 24 row cell open parentheses fraction numerator 6 x squared plus 3 x squared plus 2 over denominator x end fraction close parentheses squared plus open parentheses fraction numerator 3 x squared plus 2 over denominator negative x end fraction close parentheses squared end cell equals 24 row cell open parentheses fraction numerator 9 x squared plus 2 over denominator x end fraction close parentheses squared plus open parentheses fraction numerator 3 x squared plus 2 over denominator negative x end fraction close parentheses squared end cell equals 24 row cell fraction numerator 81 x to the power of 4 plus 36 x squared plus 4 over denominator x squared end fraction plus fraction numerator 9 x to the power of 4 plus 12 x squared plus 4 over denominator x squared end fraction end cell equals 24 row cell fraction numerator 90 x to the power of 4 plus 48 x squared plus 8 over denominator x squared end fraction end cell equals 24 row cell 90 x to the power of 4 plus 48 x squared plus 8 end cell equals cell 24 x squared end cell row cell 90 x to the power of 4 plus 24 x squared plus 8 end cell equals cell 0 space space space space space space space left parenthesis divided by 2 right parenthesis end cell row cell 45 x to the power of 4 plus 12 x squared plus 4 end cell equals 0 end table end style$

Kita cari nilai diskriminan dari . Dari persamaan , didapat , , dan .

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 12 squared minus 4 open parentheses 45 close parentheses open parentheses 4 close parentheses end cell row blank equals cell 144 minus 720 end cell row blank equals cell negative 576 end cell end table end style

Didapat sehingga . Artinya, persamaan memiliki akar-akar tidak real.

Oleh karena itu, saat atau , tidak ada pasangan dengan yang memenuhi persamaan .

Dengan demikian, banyak pasangan dengan yang memenuhi adalah 2.

Jadi, jawaban yang tepat adalah D.

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