Solving $3\sqrt{7x-5}-4=8$

$$3\sqrt{7x-5}-4=8$$

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.

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

$$7x-5\ge0$$

The $$\sqrt{7x - 5}$$ is only defined whenever the radicand (i.e.: $$7x-5$$) is non-negative. In other words,

$$7x - 5 \geq 0$$

This is because, when dealing with the field of real numbers, the square root of a negative real number is not defined. As a result, the square root of a negative is actually an imaginary number involving the unit $$i$$.

The equation is clearly equivalent to $$\sqrt{7x-5}=4$$.

There is a restriction on the variable, namely $$7x-5\ge0$$. However, there's no need to check that after squaring the solution satisfies it, because the equation becomes $$7x-5=16$$ and thus, obviously, $$7x-5\ge0$$.

In other cases, the condition needs to be checked. Suppose the equation is $$\sqrt{x+3}=x+1$$ Then the variable should satisfy $$x+3\ge0$$ and $$x+1\ge0$$, so $$x\ge-1$$. After squaring we get $$x+3=x^2+2x+1$$, hence $$x^2+x-2=0$$. This has the roots $$1$$ and $$-2$$; the former is good, the latter isn't.

Your answer is one solution and it is correct. Another answer, taking the value of the radical first, would be $$\frac{-11}{7}$$ but that conflicts with a restriction shown at the end. Without the restriction we could end up showing $$4i=4$$.

$$3\sqrt{7x-5}-4=8$$ $$\sqrt{7x-5}=\frac{8+4}{3}=4$$ Squaring both sides we get $$7x-5=16$$ $$x=\frac{21}{7}=3$$

If our answer is to be real we must have $$7x-5\ge 0\iff7x\ge 5\iff x\ge \frac{5}{7}$$ and this coincides with our solution.