Answer
$(4,2),(4,-2)$
Work Step by Step
$y=\sqrt x$ Equation $(1)$
$x^{2}+y^{2} = 20$ Equation $(2)$
Substitute $y=\sqrt x$ in Equation $(2)$
$x^{2}+y^{2} = 20$
$x^{2}+(\sqrt x)^{2} = 20$
$x^{2}+x = 20$
$x^{2}+x-20=0$
By factoring,
$(x-4)(x+5)=0$
$x=4$ or $x=-5$
Substitute $x$ values in Equation $(1)$ to get corresponding $y$ values.
Let $x=4$
$y=\sqrt x$
$y=\sqrt 4$
$y=±2$
Let $x=-5$
$y=\sqrt x$
$y=\sqrt -5$ is not a real number.
$(4,2),(4,-2)$ satisfy the given equations.
The solution set is $\{(4,2),(4,-2)\}$
