$$3\sqrt{7x-5}-4=8$$
On my homework, it said, "Solve each of the following radical equations algebraically. State any restrictions on the variable." I already solved the equation algebraically and got an answer of $x = 3$. I'm not exactly sure how to state any restrictions on the variable. Any pointers in the right direction would be greatly appreciated.
On
Whenever you see a square root (or any even $n$-th root), the expression under that root cannot be negative (because even powers are always positive!). This means that writing down the equation $$3\sqrt{\color{blue}{7x-5}}-4=8$$ is only meaningful if $\color{blue}{7x-5} \ge 0$. This restricts the possible values of the variable $x$: $$7x-5 \ge 0 \iff 7x \ge 5 \iff x \ge \frac{5}{7}$$
Note that the solution you found, $x=3$, satisfies this condition so it is a valid solution. But you could end up with a "solution" that does not satisfy this condition and then you would discard this answer on the basis of the restriction on $x$.
You can't take the square root of a negative number, so we must have
$$7x-5\ge0$$