SOLUTION: I have a system of nonlinear equations, x^2+y^2=49 y^2-3x=49 I need to find the real solutions... Please help!

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Question 1050549: I have a system of nonlinear equations,
x^2+y^2=49
y^2-3x=49
I need to find the real solutions... Please help!
Found 2 solutions by advanced_Learner, jorel555:
Answer by advanced_Learner(501) About Me  (Show Source):
You can put this solution on YOUR website!
I have a system of nonlinear equations,
x^2+y^2=49
y^2-3x=49
I need to find the real solutions... Please help!
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it is straighforward
49-x^2-3x=49
x^2+3x=0
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B3x%2B0+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%283%29%5E2-4%2A1%2A0=9.

Discriminant d=9 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-3%2B-sqrt%28+9+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%283%29%2Bsqrt%28+9+%29%29%2F2%5C1+=+0
x%5B2%5D+=+%28-%283%29-sqrt%28+9+%29%29%2F2%5C1+=+-3

Quadratic expression 1x%5E2%2B3x%2B0 can be factored:
1x%5E2%2B3x%2B0+=+1%28x-0%29%2A%28x--3%29
Again, the answer is: 0, -3. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B3%2Ax%2B0+%29

Answer by jorel555(1290) About Me  (Show Source):
You can put this solution on YOUR website!
x²+y²=49
y²-3x=49;
Subtracting the second equation from the first, we get:
x²+3x=0
x²=-3x
x=-3
49-9=y²
y²=40
y=√40=2√10. ☺☺☺☺