How do you solve the system $x^2 + y^2 = 13$ , $y=x^2 - 1$ ?

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Answer 1 · 2015-11-18T00:07:26

Use the second equation to substitute for $x^2$ in the first equation to get a quadratic in $y$ . Solve for $y$ and hence for $x$ to find solutions:

$(+-2, 3)$ and $(+-sqrt(3)i, -4)$

Add $1$ to both sides of the second equation to get:

$x^2 = y+1$

Substitute this expression for $x^2$ into the first equation to get:

$(y+1)+y^2=13$

Subtract $13$ from both sides and rearrange to get:

$y^2+y-12 = 0$

Note that $4xx3 = 12$ and $4-3 = 1$

Hence:

$0 = y^2+y-12 = (y+4)(y-3)$

So $y = 3$ or $y = -4$

So $x^2 = y+1 = 4$ or $x^2 = y+1 = -3$

So $x = +-sqrt(4) = +-2$ or $x = +-sqrt(-3) = +-sqrt(3)i$

So the possible Real solutions are $(+-2, 3)$ and the possible Complex solutions are $(+-sqrt(3)i, -4)$

How do you solve the system x^2 + y^2 = 13x^2 + y^2 = 13, y=x^2 - 1 y=x^2 - 1?