<s>[INST] Finding all real values of $a$ for which $\sqrt{9-a^2+2ax-x^2}>\sqrt{16-x^2}$ Find all real values of $a$ for which $\sqrt{9-a^2+2ax-x^2}>\sqrt{16-x^2}$ for all $x>0$
Attempt: Let $$\sqrt{9-(x-a)^2}> \sqrt{16-x^2}$$
So $$x^2-(x-a)^2>7$$
So $$a(2x-a)>7\Rightarrow a^2-2ax+7<0$$
could some help me how to find range of $a,$ Thanks
 [/INST] For the square roots to be defined, we need $|x|\leq 4$ and $|x-a|\leq 3$.
To solve a quadratic inequality, it's useful to find its roots. We have
$$a^2-2ax+7=0\iff a=\frac{2x\pm\sqrt{4x^2-28}}{2}=x\pm\sqrt{x^2-7}$$
In particular, if $|x|\leq\sqrt{7}$ the polynomial is always nonnegative in $a$.
If $\sqrt{7}<|x|\leq 4$, then the solution set will be all $a$ such that $|x-a|\leq 3$ and
\begin{align}&x-\sqrt{x^2-7}<a<x+\sqrt{x^2-7}\\
\iff &-\sqrt{x^2-7}<a-x<\sqrt{x^2-7}\\
\iff &|a-x|=|x-a|<\sqrt{x^2-7}\end{align}
Now, observe that if $|x|\leq 4$, then $x^2-7\leq 9$, so that $\sqrt{x^2-7}\leq 3$. In the end, all we need is:

$$\sqrt{7}<|x|\leq 4\,\,\text{ and }\,\,|x-a|<\sqrt{x^2-7}.$$

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