Answer
$(3,\sqrt 3)$
Work Step by Step
$y=\sqrt x$ Equation $(1)$
$x^{2}+y^{2} = 12$ Equation $(2)$
Substitute $y=\sqrt x$ in Equation $(2)$
$x^{2}+y^{2} = 12$
$x^{2}+(\sqrt x)^{2} = 12$
$x^{2}+x = 12$
$x^{2}+x- 12=0$
By factoring,
$(x-3)(x+4)=0$
$x=3$ or $x=-4$
Substitute $x$ values in Equation $(1)$ to get $y$
Let $x=3$
$y=\sqrt x$
$y=\sqrt 3$
Let $x=-4$
$y=\sqrt x$
$y=\sqrt -4$ is not a real number.
$(3,\sqrt 3)$ satisfies both equations. $(3,\sqrt 3)$ is the only solution.
