Pertanyaan

Carilah solusi ( x , y ) bilangan real untuk setiap persamaan kuadrat berikut dan lukiska sketsa grafik penyelesaiannya! { 2 x 2 + x y + y 2 = 16 2 x 2 + 7 x y + 6 y 2 = 0

Carilah solusi $(x, y)$ bilangan real untuk setiap persamaan kuadrat berikut dan lukiska sketsa grafik penyelesaiannya!

${2 x^{2} + x y + y^{2} = 16 2 x^{2} + 7 x y + 6 y^{2} = 0$

Ikuti Tryout SNBT & Menangkan E-Wallet 100rb

Habis dalam

Klaim

D. Wahyu

Master Teacher

Mahasiswa/Alumni Universitas Negeri Semarang

Jawaban terverifikasi

Pembahasan

Diketahui: Dari persamaan (2) dapat difaktorkan, sehingga Nilai-nilai disubtitusikan ke persamaan (2), sehingga Selanjutnya, subtitusikan nilai Jadi, solusi adalah . Grafik

Diketahui:

2 x squared plus x y plus y squared equals 16... left parenthesis 1 right parenthesis 2 x squared plus 7 x y plus 6 y squared equals 0... left parenthesis 2 right parenthesis

Dari persamaan (2) dapat difaktorkan, sehingga

2 x squared plus 7 x y plus 6 y squared equals 0 left parenthesis 2 x plus 3 y right parenthesis left parenthesis x plus 2 y right parenthesis equals 0 2 x plus 3 y equals 0 3 y equals negative 2 x y equals negative 2 over 3 x x plus 2 y equals 0 2 y equals negative x y equals negative x over 2

Nilai-nilai disubtitusikan ke persamaan (2), sehingga

Ketika space y equals negative 2 over 3 x comma space maka

$2 x squared plus 7 x times open parentheses negative 2 over 3 x close parentheses plus 6 times open parentheses negative 2 over 3 x close parentheses squared equals 0 2 x squared minus 14 over 3 x squared space plus up diagonal strike 6 squared end strike times open parentheses fraction numerator 4 over denominator up diagonal strike 9 cubed end strike end fraction x squared close parentheses equals 0 2 x squared minus 14 over 3 x squared space plus 8 over 3 x squared equals 0 space left parenthesis Kedua space ruas space dikali space 3 right parenthesis 6 x squared minus 14 x squared plus 8 x squared equals 0 minus 8 x squared plus 8 x squared equals 0 x space tidak space punya space solusi$

Ketika space y equals negative x over 2 comma space maka

$2 x squared plus 7 x times open parentheses fraction numerator negative x over denominator 2 end fraction close parentheses plus 6 times open parentheses negative x over 2 close parentheses squared equals 0 2 x squared minus fraction numerator 7 x squared over denominator 3 end fraction space plus up diagonal strike 6 cubed end strike times open parentheses fraction numerator x squared over denominator up diagonal strike 4 squared end strike end fraction close parentheses equals 0 2 x squared minus 14 over 3 x squared space plus fraction numerator 3 x squared over denominator 2 end fraction equals 0 space left parenthesis Kedua space ruas space dikali space 3 right parenthesis fraction numerator 6 x squared over denominator 3 end fraction minus fraction numerator 14 x squared over denominator 3 end fraction plus fraction numerator 3 x squared over denominator 2 end fraction equals 0 fraction numerator negative 8 x squared over denominator 3 end fraction plus fraction numerator 3 x squared over denominator 2 end fraction equals 0 fraction numerator negative 16 x squared plus 9 x squared over denominator 6 end fraction equals negative 8 space fraction numerator negative 7 x squared over denominator 6 end fraction equals negative 8 left parenthesis Kedua space ruas space dikali space minus 6 right parenthesis 7 x squared equals 48 x squared equals 48 over 7 x equals plus-or-minus square root of 48 over 7 end root x subscript 1 equals square root of space 48 over 7 end root comma space x subscript 2 equals negative square root of 48 over 7 end root$

Selanjutnya, subtitusikan nilai

y equals negative x over 2 space ke space x subscript 1 space space dan space space x subscript 2 comma space sehingga

$y subscript 1 equals negative fraction numerator square root of begin display style 48 over 7 end style end root over denominator 2 end fraction y subscript 2 equals negative fraction numerator open parentheses negative square root of begin display style 48 over 7 end style end root close parentheses over denominator 2 end fraction equals fraction numerator square root of 48 over 7 end root over denominator 7 end fraction$

Jadi, solusi adalah $open parentheses square root of 48 over 7 end root comma space fraction numerator negative square root of begin display style 48 over 7 end style end root over denominator 2 end fraction close parentheses space space dan space space open parentheses negative square root of 48 over 7 end root comma space fraction numerator square root of begin display style 48 over 7 end style end root over denominator 2 end fraction close parentheses$ .

Grafik

Perdalam pemahamanmu bersama Master Teacher
di sesi Live Teaching, GRATIS!

0.0 (0 rating)

Pertanyaan serupa

Let a , b dan c constans (with a  = 0 ) and consider the system. { y = a x 2 + b x + c y = k for which value of k (In terms will the system have exactly one solution? What is that solution ? ...

0.0

Jawaban terverifikasi

Banyaknya anggota dari himpunan penyelesaian SPLK di bawah ini ialah .... y = x + 7 y = x 2 + 4 x − 12 .

5.0

Jawaban terverifikasi

Himpunan penyelesaian dari sistem persamaan linear-kuadrat { y = x 2 y = 2 x + 3 adalah ....

5.0

Jawaban terverifikasi

Diketahui ( − 1 , − 2 ) merupakan salah satu solusi real dari sistem persamaan berikut. { 2 a x 2 − 7 x y − 2 a y 2 = − 20 − a x 2 + 4 x y + a y 2 = 11 Carilahsolusi !

168

0.0

Jawaban terverifikasi

The straight line 2 x + y = 2 meets the curve x 2 + x y + 3 = 0 at the points of A and B . Find: a. The mid -point coordinates of A B .

0.0

Jawaban terverifikasi