Do you write plus/minus if a variable squares equals the square root of a number?

Question

For example, if I have $x^2 = \sqrt{49}$. I know that $7$ is the number, but as my final answer, do I write that $x = +\sqrt{7}$ and $-\sqrt{7}$ or just $x = \sqrt{7}$?

Answer 1

For any $a>0$, if you have an equation $x^{2}=a$, then note that both $(\sqrt{a})^{2}=a$ and $(-\sqrt{a})^{2}=a$. Thus, both $a$ and $-a$ are solutions, and hence, they must both be considered. In your case, you have $x^{2}=\sqrt{49}=7$. Therefore, both $x=\sqrt{7}$ and $x=-\sqrt{7}$ satisfy this equation.

Answer 2

Note that $x^2=7 \implies (x-\sqrt{7})(x+\sqrt{7})=0$.

Therefore, there are two roots, $x=\pm \sqrt{7}$.

Answer 3

It depends on which field you expect/require the answer to be in:

If the answer is allowed to be a complex number, then the equation $x^2 = \sqrt{49}$ can be rewritten as $x^2 = 7$ OR $x^2 = -7$, which collectively yield four values: $x = +\sqrt{7}$, $x = -\sqrt{7}$, $x = +i\sqrt{7}$, and $x = -i\sqrt{7}$.

If the answer is required to be a real number, then the equation $x^2 = \sqrt{49}$ can be rewritten as $x^2 = 7$ OR $x^2 = -7$, which, since $x^2 = -7$ has no real solution, yields only two values: $x = +\sqrt{7}$ and $x = -\sqrt{7}$.

If the answer is required to be a rational number, or a number in an extension of the rationals which does not contain $\sqrt{7}$, then of course there is no solution.