How do you solve x^2 + y^2 = 4 and y^2 = 3x?

Aug 4, 2015

$\left(x , y\right) = \left(1 , - \sqrt{3}\right) , \left(1 , \sqrt{3}\right) , \left(4 , i \sqrt{12}\right) , \left(4 , - i \sqrt{12}\right)$

Explanation:

Substitute the second equation into the first to obtain a quadratic equation for $x$:

${x}^{2} + {y}^{2} = {x}^{2} + 3 x = 4$ => ${x}^{2} + 3 x - 4 = \left(x + 4\right) \left(x - 1\right) = 0$

This has solutions $x = - 4 , 1$, substituting this into the second equation we have $y = \pm \sqrt{3} , \pm i \sqrt{12}$.

Therefore we have:

$\left(x , y\right) = \left(1 , - \sqrt{3}\right) , \left(1 , \sqrt{3}\right) , \left(4 , i \sqrt{12}\right) , \left(4 , - i \sqrt{12}\right)$

Aug 4, 2015

Substitute the second equation into the first to get a quadratic in $x$, the positive root of which gives two possible Real values for $y$ in the second equation.

$\left(x , y\right) = \left(1 , \pm \sqrt{3}\right)$

Explanation:

Substitute ${y}^{2} = 3 x$ into the first equation to get:

${x}^{2} + 3 x = 4$

Subtract $4$ from both sides to get:

$0 = {x}^{2} + 3 x - 4 = \left(x + 4\right) \left(x - 1\right)$

So $x = 1$ or $x = - 4$.

If $x = - 4$ then the second equation becomes ${y}^{2} = - 12$, which has no Real valued solutions.

If $x = 1$ then the second equation becomes ${y}^{2} = 3$, so $y = \pm \sqrt{3}$