Question 169252
I solved 48 problems but this one has me stumped and I have been working on it for 2hrs.
Solve the following system of nonlinear equations. 
<pre><font size = 4 color = "indigo"><b>
{{{system(x^2 + y^2 = 3,x^2 + y = 0)}}}

Solve the second for y:

{{{x^2+y=0}}}
{{{y=-x^2}}}

Substitute in the first original equation:

{{{x^2+y^2=3}}}

{{{x^2+(-x^2)^2=3}}}

{{{x^2+(x^4)=3}}}

{{{x^4+x^2-3=0}}}

Let {{{x^2=W}}}.  Then {{{x^4=W^2}}}

Substitute those:

{{{W^2+W-3=0}}}

This does not factor so we have to use
the quadratic formula:

{{{W = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}

where {{{matrix(1,5,   a=1, ",", b=1, ",", c=-3)}}}

{{{W = (-(1) +- sqrt( (1)^2-4*(1)*(-3) ))/(2*(1)) }}} 

{{{W = (-1 +- sqrt(1+12))/2 }}}

{{{W = (-1 +- sqrt(13))/2 }}}

But we do not want W.  We want x, so

since we let {{{x^2=W}}}, we have

{{{x^2 = (-1 +- sqrt(13))/2 }}}  

Using the +, and the principle of
square roots:

{{{x = ""+-sqrt((-1 + sqrt(13))/2)}}} = ±{{{1.141391974}}}, approximately

Substituting in {{{y=-x^2}}}

{{{y=-x^2}}}

{{{y = -(-1 + sqrt(13))/2) = -1.302775638}}}, approximately.

So we have two solutions:

(x, y) = (±1.141391974, -1.302775638)

Those are the only real solutions.  

To find the imaginary solutions,

we use the -, and the principle of
square roots:

{{{x = ""+-sqrt( (-1 - sqrt(13))/2 )  }}}= ±{{{i*sqrt(  (1+sqrt(13))/2 )}}}= ±1.517489914i, approximately.

Substituting in {{{y=-x^2}}}

{{{y=-x^2}}}

{{{y = -(-1 - sqrt(13))/2) = (1+sqrt(13))/2= 2.302775638}}}, approximately.

So we have two imaginary solutions:

(x,y) = (±1.517489914i,2.302775638)

Edwin</pre>