Question 169252
Solve the system of equations:
1) {{{x^2+y^2 = 3}}} This is a circle with its center at the origin and radius of {{{sqrt(3)}}}.
2) {{{x^2+y = 0}}} This is a parabola  that opens downard with its vertex at the origin.
Subtract equation 1) from equation 2) to get:
{{{y^2-y = 3}}} Subtract 3 from both sides.
{{{y^2-y-3 = 0}}} Solve using the quadratic formula: {{{y = (-b+-sqrt(b^2-4ac))/2a}}} where: a = 1, b = -1, and c = -3
{{{y = (-(-1)+-sqrt((-1)^2-4(1)(-3)))/2(1)}}}
{{{y = (1+-sqrt(1-(-12)))/2}}}
{{{y = (1+-sqrt(13))/2}}}
{{{y = (1+sqrt(13))/2}}} or {{{y = (1-sqrt(13))/2}}} Now substitute these values of y, one-at-a-time, into either one of the two given equations to solve for x. Let's use equation 2)
{{{x^2+y = 0}}} Subtract y from both sides.
{{{x^2 = -y}}} Substitute {{{y = (1+sqrt(13))/2}}}
{{{x^2 = -(1+sqrt(13))/2}}} Take the square root of both sides.
{{{x = sqrt(-(1+sqrt(13))/2)}}}
So one solution is:
({{{sqrt(-(1+sqrt(13))/2)}}},{{{(1+sqrt(13))/2}}})