Question

Find points of intersections. x^2+y^2-4=0 and 3x-y^2=0

Answer 1

I solved for y^2 and set them equal, factored and found the zeros but that isn't right.

Answer 2

You'll want to start by solving those both in terms on on variable, then you can find out when they are equal to each other.

Alternately you could try recognizing the graphs and seeing where they intersect. The first one is a circle for instance...

x^2 + y^2 = (radius)^2

Answer 3

Should the method I used have worked correctly?

Answer 4

y^2 =x^2-4 and y^2=3x. x^2-4-3x=0. (x+1)(x-4)=0

Answer 5

Do you have the solutions?
I get x=3 and x=-4
I obtained that by adding them both together so that the y^2 cancels

Answer 6

Yes. http://www.wolframalpha.com/input/?i=x%5E2%2By%5E2-4%3D0+and+3x-y%5E2%3D0

Answer 7

The back of the book agrees.

Answer 8

Thank you, then I would do it this way to be honest.

Answer 9

Seems to be the most efficient. Add equation 1 to equation two, and solve the quadratic for x

Answer 10

Why did the method we both used not work?

Answer 11

@DanielHendrycks , your method should work if you solved for y^2 correctly..
in the first equation, y^2=4+x^2 and in the second, y^2=3x