SOLUTION: 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. x^2 + y^2 = 3 x^2 + y = 0

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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.
x^2 + y^2 = 3
x^2 + y = 0
Found 3 solutions by Earlsdon, ankor@dixie-net.com, Edwin McCravy:
Answer by Earlsdon(6294)   (Show Source): You can put this solution on YOUR website!
Solve the system of equations:
1) This is a circle with its center at the origin and radius of .
2) This is a parabola that opens downard with its vertex at the origin.
Subtract equation 1) from equation 2) to get:
Subtract 3 from both sides.
Solve using the quadratic formula: where: a = 1, b = -1, and c = -3



or 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)
Subtract y from both sides.
Substitute
Take the square root of both sides.

So one solution is:
(,)
Answer by ankor@dixie-net.com(22740)   (Show Source): You can put this solution on YOUR website!
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:

In this equation a=1, b=-1, c=-3

;

:

Two solutions


y = 2.3
and


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:
Answer by Edwin McCravy(20056)   (Show Source): You can put this solution on YOUR website!
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. 



Solve the second for y:




Substitute in the first original equation:









Let .  Then 

Substitute those:



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



where 

 





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

since we let , we have

  

Using the +, and the principle of
square roots:

 = ±, approximately

Substituting in 



, 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:

= ±= ±1.517489914i, approximately.

Substituting in 



, approximately.

So we have two imaginary solutions:

(x,y) = (±1.517489914i,2.302775638)

Edwin
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