Hvordan løser man dette likningsettet?
[tex]-4x+2y^2+x^2=15[/tex]
[tex]y-x=2[/tex]
Likningssett
Moderators: Aleks855, Gustav, Nebuchadnezzar, Janhaa, DennisChristensen, Emilga
Som alle andre likningssett.
[tex]I: x^2 - 4x + 2y^2 - 15 = 0 \\ II: y-x = 2[/tex]
[tex]II: y = 2+x[/tex]
Setter inn i I
[tex]I: x^2 - 4x + 2(2+x)^2 - 15 = 0[/tex]
[tex]I: x^2 - 4x + 2(4 + 4x + x^2) - 15 = 0[/tex]
[tex]I: x^2 -4x + 8 + 8x + 2x^2 - 15 = 0[/tex]
[tex]I: 3x^2 + 4x - 7 = 0[/tex]
abc-formelen.
[tex]x = 1 \ \vee \ x = -\frac{7}{3}[/tex]
Setter inn i II
[tex]II: y = 2 + 1 = 3 \ \vee \ y = 2 - \frac{7}{3} = -\frac{1}{3}[/tex]
Løsning:
[tex]x = 1 \ \vee \ x = -\frac{7}{3} \\ y = 3 \ \vee \ y = -\frac{1}{3}[/tex]
[tex]I: x^2 - 4x + 2y^2 - 15 = 0 \\ II: y-x = 2[/tex]
[tex]II: y = 2+x[/tex]
Setter inn i I
[tex]I: x^2 - 4x + 2(2+x)^2 - 15 = 0[/tex]
[tex]I: x^2 - 4x + 2(4 + 4x + x^2) - 15 = 0[/tex]
[tex]I: x^2 -4x + 8 + 8x + 2x^2 - 15 = 0[/tex]
[tex]I: 3x^2 + 4x - 7 = 0[/tex]
abc-formelen.
[tex]x = 1 \ \vee \ x = -\frac{7}{3}[/tex]
Setter inn i II
[tex]II: y = 2 + 1 = 3 \ \vee \ y = 2 - \frac{7}{3} = -\frac{1}{3}[/tex]
Løsning:
[tex]x = 1 \ \vee \ x = -\frac{7}{3} \\ y = 3 \ \vee \ y = -\frac{1}{3}[/tex]
