Answer:
[tex]x=-1,1,-\sqrt{\frac{7}{4} },\sqrt{\frac{7}{4}}[/tex] and [tex]y=-1,\frac{1}{2}[/tex]
The ordered pair solutions are [tex](-\sqrt{\frac{7}{4}},0.5)[/tex], [tex](\sqrt{\frac{7}{4}},0.5)[/tex], [tex](-1,-1)[/tex], and [tex](1,-1)[/tex].
Step-by-step explanation:
I'm assuming the system is [tex]\left \{ {x^2+y^2=2} \atop {y=2x^2-3}} \right.[/tex]:
[tex]x^2+y^2=2[/tex]
[tex]x^2+(2x^2-3)^2=2[/tex]
[tex]x^2+(4x^4-12x^2+9)=2[/tex]
[tex]x^2+4x^4-12x^2+9=2[/tex]
[tex]4x^4-11x^2+9=2[/tex]
[tex]4x^4-11x^2+7=0[/tex]
[tex]x^4-11x^2+28=0[/tex]
[tex](x^2-7)(x^2-4)=0[/tex]
[tex](4x^2-7)(x^2-1)=0[/tex]
[tex]4x^2-7=0[/tex]
[tex]4x^2=7[/tex]
[tex]x^2=\frac{7}{4}[/tex]
[tex]x=\pm\sqrt{\frac{7}{4}}[/tex]
[tex]x^2-1=0[/tex]
[tex]x^2=1[/tex]
[tex]x=\pm1[/tex]
[tex]y=2x^2-3[/tex]
[tex]y=2(\pm\sqrt{\frac{7}{4}})^2-3[/tex]
[tex]y=2({\frac{7}{4}})-3[/tex]
[tex]y=\frac{7}{2}-3[/tex]
[tex]y=\frac{1}{2}[/tex]
[tex]y=2x^2-3[/tex]
[tex]y=2(\pm1)^2-3[/tex]
[tex]y=2(1)-3[/tex]
[tex]y=2-3[/tex]
[tex]y=-1[/tex]
Therefore, [tex]x=-1,1,-\sqrt{\frac{7}{4} },\sqrt{\frac{7}{4}}[/tex] and [tex]y=-1,\frac{1}{2}[/tex]
The ordered pair solutions are [tex](-\sqrt{\frac{7}{4}},0.5)[/tex], [tex](\sqrt{\frac{7}{4}},0.5)[/tex], [tex](-1,-1)[/tex], and [tex](1,-1)[/tex].
