<s>[INST] Roots of x inside square root In finding the roots of
$ \sqrt{6-4x-x^2}=x+4$
I get that the roots of $x$ are $-5$ and $-1$.
However first I need to take into account that:


*

*$x+4 \ge 0$
and

*$6-4x-x^2 \ge 0$.


Considering the second equation gives me:
$x \ge -2+ \sqrt{10}$,
would the roots of $x$ I found in the first equation be considered invalid?
 [/INST] Let $x+4 \ge 0~~~(1)$  and for tje reality of the roots $$6-4x-x^2 \ge 0~~~~~~(2) \Rightarrow x^2+4x-6 \le 0 \Rightarrow -2-\sqrt{10} \le x \le -2+\sqrt{10}.$$ So that we can square the given equation, then we have
$$x^2+6x+5 \Rightarrow x=-5, -1$$. As per the conditiona (1,2), $x=-1$ is the correct root.
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