Answer
$(2, -\sqrt 8)$, $(2,\sqrt 8)$
Work Step by Step
$4x-y^2=0$
$2x^2+y^2=16$
$4x-y^2=0$
$4x-y^2+y^2=0+y^2$
$4x=y^2$
$2x^2+y^2=16$
$2x^2+4x=16$
$2x^2+4x-16=16-16$
$2x^2+4x-16=0$
$a=2$, $b=4$, $c=-16$
$x=(−b±\sqrt {b^2−4ac})/2a$
$x=(−4±\sqrt {4^2−4*2*-16})/2*2$
$x=(-4±\sqrt {16+128})/4$
$x=(-4±\sqrt {144})/4$
$x=(-4±12)/4$
$x=(-4+12)/4$
$x=8/4$
$x=2$
$x=(-4-12)/4$
$x=-16/4$
$x=-4$
$x=-4$
$4x-y^2=0$
$4*-4-y^2=0$
$-16-y^2=0$
$-16-y^2+y^2=0+y^2$
$-16=y^2$
Since we are working on the real numbers, we cannot have the square root of a negative number. Thus, $x \ne -4$.
$x=2$
$4x-y^2=0$
$4*2-y^2=0$
$8-y^2=0$
$8-y^2+y^2=0+y^2$
$8=y^2$
$\sqrt 8 =\sqrt {y^2}$
$±\sqrt 8 =y$
