Hello! We have a system of equations here, so there are a few ways to solve it, but I'll use the substitution method. We have
[tex]x^2+y^2=13[/tex]
[tex]y=x^2-1[/tex]
The second equation gives us y in terms of x, so we'll plug that into the first equation and solve:
[tex]x^2+(x^2-1)^2=13[/tex]
[tex]x^2+x^4-2x^2+1=13[/tex]
[tex]x^4-x^2+1=13[/tex]
[tex]x^4-x^2-12=0[/tex]
[tex](x^2+3)(x^2-4)=0[/tex]
[tex](x^2+3)(x+2)(x-2)=0[/tex]
Now we have the tricky part. Initially, you may think that there are three x values that are relevant, but in reality we can never make the [tex]x^2+3[/tex] term equal to 0 because [tex]x^2[/tex] can only go as small as zero itself. This means we only care about the other two factors, so:
[tex]x+2=0[/tex] and [tex]x-2=0[/tex]. Then [tex]x=-2[/tex] and [tex]x=2[/tex].
Now we need to find out which y-values correspond to these x-values. I'll use the second equation to do this:
[tex]y=(-2)^2-1=4-1=3[/tex]
[tex]y=(2)^2-1=4-1=3[/tex]
So our points are (-2,3) and (2,3).
