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?
 A: 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 $x$ = $\sqrt{3}, -\sqrt{3}, 2, -2$
A: The solutions to $\sqrt y = y$ are $0$ and $1$.
Observe that we're looking at $\sqrt{4-x^2}=f(x)=g(x)=4-x^2$.
A: Notice: $$-x^4+7x^2-12=0 \ \ \ \ \rlap{\rlap{\rlap==}}\Rightarrow \ \ \ \ -(\color{red}{x^2})^2 +7(\color{red}{x})-12=0.$$
If we let $t=x^2$ then our expression becomes a quadratic: $$-t^2+7t-12=0.\tag{$\star$}$$ To solve it we can use the quadratic formula, which says that if we have an equation of the form $ax^2+bx+c=0$ where $a\neq0$, then its solutions are given by the formula: $$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}.$$ So using it we found that $(\star)$ has two roots, which are: $t=3$ and $t=4.$ Now to return back to our original equation and to solve it therein we need to get back our $x$ values from the $t$ values we obtained, that is solving: $$t=x^2 \ \ \ \ \rlap{\rlap{\rlap==}}\Rightarrow \ \ \ \ t=3=x^2 \quad\text{and}\quad t=4=x^2.\rlap{\phantom{X}}\phantom{XX}$$
 And I'm sure you can take it from there. ;-)
Side note: This $t$-substitution isn't necessary at all, when you noticed that you had an expression of the form $a(\text{something})^2+b(\text{something})+c=0$ you can directly use the quadratic formula to solve for that $\text{something}$.

We can get some graphical sense of what this means by looking at the graphs of $\color{darkblue}{f(x)=}$$\color{darkblue}{-x^2+4}$ and of $\color{darkmagenta}{g(x)=\sqrt{4-x^2}}$: (The $x$-coordinates of the points of intersection of the two curves are the solutions to our equation)
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