Solve the equation $4\sqrt{2-x^2}=-x^3-x^2+3x+3$ Solve the equation in $\Bbb R$:
$$4\sqrt{2-x^2}=-x^3-x^2+3x+3$$
Is there a unique solution $x=1$? I have trouble when I try to prove it. 
I really appreciate if some one can help me. Thanks!
 A: Let's rewrite it and try to solve it:
$$4\sqrt{2-x^2}=-(x+1)(x^2-3)$$
Squaring both sides (this will keep originial solutions but introduce new ones) gives
$$16(2-x^2)=(x+1)^2(x^2-3)^2$$
Moving everything over, expanding, and then factoring by the root we already had (when $x=1$) yields
$$(x-1)(x^5+3x^4-2x^3-14x^2+5x+23)=0$$
Now as the second factor is a degree $5$ polynomial over $\mathbb{R}$, it has at least one real root. (In fact, it has only one root since the derivative $f'(x)$ of the polynomial $f(x):=x^5+3x^4-2x^3-14x^2+5x+23$ has only two roots but I'll try to think of a better argument than this).
Making a translation $x=-1+y$ simplifies the right factor to
$$f(y)=(y^5-2y^4-4y^3+20y+8)=0$$
To see there is a solution $y$ in $[-1,0)$ note that $[f(-1)=-11,f(0)=8)$ so by the intermediate value theorem there is a solution in $y\in [-1,0)$.
So if this was a solution to our original relation, we would have $x\in [-2,-1)$ but this can't be the case since we may restrict solutions of our original equation to $[-1,\sqrt{2}]$ following the argument of Paul Vergeev.
A: You can rewrite the expression as 
$$ 16(2-x^2) = (x+1)^2(x^2-3)^2 $$
Note that the right hand side is non-negative so the left hand side should be too. Does it help?
EDIT: As Eoin points out, keep in mind that both sides of the original equation have the same sign only on $[-1;\sqrt2]$.
A: It can be seen that $4 \sqrt{2 - x^2} = (1+x)(3 - x^2)$ leads to the expanded form, after squaring both sides and combining terms,
$$ x^6 + 2 x^5 - 5 x^4 - 12 x^3 + 19 x^2 + 18 x - 23 = 0 .$$
It is readily identified that $x=1$ is a solution and can then be factored out leading to
$$ (x-1)( x^5 + 3 x^4 - 2 x^3 - 14 x^2 + 5x + 23) = 0.$$
The polynomial of order $5$ has one real solution and $4$ complex solutions. 
