Question 169252
Solve the following system of nonlinear equations.
x^2 + y^2 = 3
x^2 + y = 0
---------------Subtraction eliminate x^2
y^2 - y = 3
:
y^2 - y - 3 = 0
:
Solve the quadratic equation using the quadratic formula:
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
In this equation a=1, b=-1, c=-3
{{{y = (-(-1) +- sqrt(-1^2 - 4*1*-3 ))/(2*1) }}}
;
{{{y = (1 +- sqrt(1 - (-12) ))/(2) }}}
:
{{{y = (1 +- sqrt(13 ))/(2) }}}
Two solutions
{{{y = (1 + 3.6)/(2) }}}
{{{y = 4.6/2}}}
y = 2.3
and
{{{y = (1 - 3.6)/(2) }}}
{{{y = -2.6/2}}}
y = -1.3
:
Find x using the 2nd equation using y = +2.3
x^2 + 2.3 = 0
x^2 = -2.3
x = Sqrt(-2.3) not a real solution
:
Using y = -1.3
x^2 - 1.3 = 0
x^2 = +1.3
x = sqrt(1.3)
x = 1.14
:
Check solution in the 1st equation
1.14^2 + (-1.3^2) = 
1.3 + 1.69 = 2.99 ~ 3
:
Solutions: x = 1.14, y = -1.3
;
You can check in the 2nd equation: