<s>[INST] Find the intersections of the functions I have $f(x)=-x^2+4$  a parabola and $g(x)=\sqrt{(4-x^2})$ a semi circle with a raduis of $2$
if I say  $g(x)=f(x)$ and solve for $x$. I should find the points at which $x$ intercepts
$\sqrt{4-x^2}=-x^2+4$
 then
$4-x^2=x^4-8x^2+16\Rightarrow
-x^4+7x^2-12=0$
this is as far as i got, how do I continue from here?
 [/INST] I would use the quadratic formula. 
if you set $f(x) = g(x)$,
you get $x^4 - 7x^2 + 12 = 0$
factoring, we get $(x^2-4)*(x^2-3) = 0$
So $x^2 -4 = 0$,
$x = 2, -2$
Then we have $x^2 - 3$ $= 0$,
$x$ = $\sqrt{3},$ $-\sqrt{3}$
But now we must check for extraneous solutions.
Seeing as there are none, our answers are $x$ = $\sqrt{3}, -\sqrt{3}, 2, -2$
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