Systems of Non-Linear equations

[datafiend]

Hi all. I'm really at a loss on how to find the "other" points of intersection for this system of equations: x^2-y=4
x^2+y^2=4
Obviously we have a parabola and circle.
I have solved for x and got "2", and plugged that back into get "0". However the answer has 4 points, the others being +/- \sqrt{3} , +/-1.

How do you get the other 2 points? I tried to use Cramers, but I got a zero in the denominator.

Any help?

 

[Dethrone]

Hi (Wave),

I don't know Cramers, but I think the reason why you're not getting all the solutions is because you've isolated for $x$ or $y$ and did not check the $\pm$ when you took the square root.

$ x^2-y=4$
$x^2=4+y$
$|x| = \sqrt{4+y}$
$x =\pm\sqrt{4+y}$

Personally, I would keep the $x$ as $x^2$ and solve:

I have these two equations:
$x^2=4-y^2$
$x^2-y=4$

What happens when I substitute the first equation into the second?

 

[datafiend]

So,
I get 4-y^2-y=4?

Do I factor that out? Maybe use the quadratic formula?

 

[Dethrone]

There is no need, just simple factoring!

Well, we can subtract both sides to get rid of the $4$.
$-y^2-y=0$

Now factor,
$-y(y+1)=0$

Either $y=0$ or $y+1=0$. What are the solutions? (Wondering)

 

[datafiend]

yes, I'm struggling with this...

ok. i got y=0 or y=-1.

at this point, do I just put the -1 into x and solve?

 

[Dethrone]

So, given $-y^2-y=0$, we want to find the values of $y$ such that the equation equals 0. If we factor the equation, this becomes clear:
$-y(y+1)=0$

If either $y=0$ or $y+1=0$, the whole expression will be equal to 0. Which are those $y$ values? Do you understand what I have said so far?

EDIT:

Yes, those are correct! Just put them back into the equation and solve for the corresponding $x$ values.

 

[datafiend]

Yeah. I got the other coordinates by substituting the y=-1 into the other equation.

Thank you for your patience. I'm done for the day.

 

[Amad27]

Alternatively,

$x^2 - y = 4$ and $x^2 + y^2 = 4$

$4 = 4$ therefore,

$x^2 - y = x^2 + y^2$

$-y = y^2$ [subtracting x^2 from both sides]

$0 = y^2 + y$

$y(y + 1) = 0$

$y = 0$ and $y = -1$ now substitute these in.