$16\ge x^2\ge 25$ contradiction no solutions

Try to find complex solutions.

No complex solutions either. Eliminating radicals is routine and gives $x^8-98x^6-2x^5+3609x^4+102x^3-59171x^2-1272x+364496=0$. We can't solve this in radicals (Galois Group is $S_8$), but the numerically solving gives eight complex roots. $x\approx 5.27341\pm 0.332764 i$ $x\approx 4.63336\pm 0.252035 i$ $x\approx -4.90722\pm 0.329298 i$ $x\approx -4.99955\pm 0.26558 i$ All eight of these are extraneous when evaluating square-roots on the principal branch.

Yes, and we don't need to actually solve the equation for $x \in \text{R}$. Just checking the domain does the trick. For complex solutions, we solve to obtain the equation: $$x^{4}+4x^{3}-29x^{2}-65x+284 = 0$$.

无解吧，如果不考虑虚根的情况下，，，

This is really easy. Just notice that if x is real, $\sqrt{16-x^2}$ is a real number, so the expression $16-x^2$ is positive. Then also we have $x^2-25>0$(otherwise that wouldn't make sense, since the radicals cannot be negative).Then if we consider complex roots, finish like rchoker.

$\sqrt{}$ is defined only for positive real number, so $16 - x^2$ is non-negative or $\geq 0$. This means $x^2 \leq 16$ ...(1) Because the LHS is positive, so is the RHS. This means that $x^2 \geq 25$ ...(2) Now from (1) and (2) we get a contradiction, so there are no solotions for this equation

This, was pretty easy. Note, that $x^2 \le 16$, because then, $\sqrt{16-x^2}$, is imaginary. Now, since, the LHS, is positive, we need to have $x^2 \ge 25$, now, there are no solutions.