Consider this problem:
$$\sqrt{(x^2-16)^2}=9$$
When solving algebraically I get the following:
$$x^2-16=9$$ $$x^2=25$$ $$x=\pm5$$
However I noticed that $\pm\sqrt7$ is also a solution to the equation. I'm trying to understand how you would arrive at this solution algebraically. I suspect I have assumed or done something wrong in my solution.
On
$$\sqrt{(x^2-16)^2}=9
$$
Squaring both sides we get $$(x^2-16)^2=9^2$$
Again taking square roots we get $$(x^2-16)=\pm9\\x^2=16\pm9\\x=\pm\sqrt{16\pm9}\\x=\pm5,\pm\sqrt{7}.$$ This is the solution as desired.
My method is indicating towards the fact that $\sqrt{x^2}=|x|$. I get there from the definition.
Bear in mind that
$$\sqrt{x^2} = |x|, \text{ not just } x$$
To see why, plug in $x=-1$; certainly $\sqrt{(-1)^2} = \sqrt{1} = 1 \ne -1$.
Thus, you have a second equation to solve when you take the root:
$$x^2 - 16 = -9$$
in addition to
$$x^2 - 16 = +9$$
which you already solved. That is, you really are solving
$$|x^2-16| = 9$$