Question

X squared+ y squared = 2 y = 2x squared – 3 Which of the following describes the system?

Answer 1

`x=-1,1,-\sqrt{(7)/(4) },\sqrt{(7)/(4)}` and
`y=-1,(1)/(2)`

The ordered pair solutions are
`(-\sqrt{(7)/(4)},0.5)`,
`(\sqrt{(7)/(4)},0.5)`,
`(-1,-1)`, and
`(1,-1)`.

Explanation:

I'm assuming the system is
`\left \{ {x^2+y^2=2} \atop {y=2x^2-3}} \right.`:

`x^2+y^2=2`

`x^2+(2x^2-3)^2=2`

`x^2+(4x^4-12x^2+9)=2`

`x^2+4x^4-12x^2+9=2`

`4x^4-11x^2+9=2`

`4x^4-11x^2+7=0`

`x^4-11x^2+28=0`

`(x^2-7)(x^2-4)=0`

`(4x^2-7)(x^2-1)=0`

`4x^2-7=0`

`4x^2=7`

`x^2=(7)/(4)`

`x=\pm\sqrt{(7)/(4)}`

`x^2-1=0`

`x^2=1`

`x=\pm1`

`y=2x^2-3`

`y=2(\pm\sqrt{(7)/(4)})^2-3`

`y=2({(7)/(4)})-3`

`y=(7)/(2)-3`

`y=(1)/(2)`

`y=2x^2-3`

`y=2(\pm1)^2-3`

`y=2(1)-3`

`y=2-3`

`y=-1`

Therefore,
`x=-1,1,-\sqrt{(7)/(4) },\sqrt{(7)/(4)}` and
`y=-1,(1)/(2)`

The ordered pair solutions are
`(-\sqrt{(7)/(4)},0.5)`,
`(\sqrt{(7)/(4)},0.5)`,
`(-1,-1)`, and
`(1,-1)`.