SOLUTION: Solve the following system of equations and state answer in coordinate pairs?x^(2)+y^(2)=15 and x^(2)+y=-15 answers...(21,6)(10,-5),(21,-6)(-10,-5),(21,6)(-10,-5), no solution

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Solve the following system of equations and state answer in coordinate pairs?x^(2)+y^(2)=15 and x^(2)+y=-15 answers...(21,6)(10,-5),(21,-6)(-10,-5),(21,6)(-10,-5), no solution      Log On


   

Question 441886: Solve the following system of equations and state answer in coordinate pairs?x^(2)+y^(2)=15 and x^(2)+y=-15
answers...(21,6)(10,-5),(21,-6)(-10,-5),(21,6)(-10,-5), no solution
Found 2 solutions by htmentor, rwm:
Answer by htmentor(1343) About Me  (Show Source):
You can put this solution on YOUR website!
x%5E2+%2B+y%5E2+=+15 [1]
x%5E2+%2B+y+=+-15 [2]
Subtract equation 2 from equation 1:
y%5E2+-+y+=+30+-%3E+y%5E2+-+y+-+30+=+0
Use the quadratic formula to solve for y:
y+=+%281+%2B-+sqrt%281%2B120%29%29%2F2
This gives y+=+%281+%2B-+11%29%2F2
y = -5,6
If we use equation [2] to find x, we see that no solution exists for either value of y:
y=-5: x%5E2+-+5+=+-15+-%3E+x%5E2+%3C+0
y=6: x%5E2+%2B+6+=+-15+-%3E+x%5E2+%3C+0
So there is no solution.
This can also be seen by the graphs of the two functions. Notice there are no points of intersection:

Answer by rwm(914) About Me  (Show Source):
You can put this solution on YOUR website!
the first equation is a circle centered at the origin (0,0) radius sqrt(15) which is slightly less than 4
the second is a parabola with its vertex at y=-15
the two equations never cross so all solutions are complex.