The relation of $\sqrt{*}$ being multivalued and the solutions to the equations $x^{2}=4$

Question

After asking a question on the square root yesterday (On the real square root and branches of the complex square root.) I saw a lot of arguments of the form,

If we solve the equation $x^{2}=4$ then the solutions are $\pm 2$.

and

If we want to define the square root as a function then we have to pick a branch. And then we have $\sqrt{4}=2$ and only that.

I suppose that when we solve and equation we can apply a function to both sides and get the answer on sets where the function is well defined. Hence we get $2$ as an answer to the above equation if we pick the principal branch of the root.

Following this logic the above equation have unique solution in a universe where we picked a branch.Hence using a function arising in this way i.e resticting a multivalued function to a branch we "loose" or drop som solutions.

So my question is weather this is complete nonsense or if the multivaluedsness of the root and the multiple solutions to above equation is related.

Answer 1

The word function is important. A function can has one or less values for every input. When we say $\sqrt x$ we define it to be the positive answer to the equation $b^2=x$.

Unlike functions equations can have more than one solutions, so we don't need to choose, the equation $b^2=x$ has 2 solutions, $\pm\sqrt x$, but $\sqrt x$ is defined to be positive.

When we apply the function to an equation we do have 2 values: apply $\sqrt \cdot$ to $b^2=x$ to get $|b|=\sqrt x$, this has one value, this also answer the definition of $\sqrt x$ because it is always positive, but we want to solve for $b$ not for $x$, so let's remember what $|\cdot|$ is:$$|b|=\begin{cases}b&b\ge0\\-b&b<0\end{cases}$$hence the equation $|b|$ has to solutions($\pm |b|$) lets apply this to the result we got($|b|=\sqrt x$) and we get $\sqrt x=\begin{cases}b&b\ge0\\-b&b<0\end{cases}$ we can now find that $b=\pm|b|=\pm\sqrt x$.

So even with the definition of $\sqrt x$ to be positive the answer of the equation is $\pm\sqrt x$, two solutions

Answer 2

There are two square roots of $4$, and we choose one to be "the" square root of $4$, namely $2$, and we say that $\sqrt4 = 2$.

There are still two solutions to $x^2=4$, because it turns out that the two square roots are opposite of each other, i.e. if $a$ is a square root of $b$, then $-a$ is also a square root of $b$.

Therefore, the other square root of $4$ can be expressed as $-\sqrt4$, i.e. $-2$.

Answer 3

Let we be in the region of real numbers and we say about equations with one variable.

A root of the equation it's a number such that if we'll substitute this number instead variable in the equation then we'll get a true statement.

To solve an equation it says to find all set of roots of the equation.

An arithmetic square root of a non-negative number $a$ (we'll write it $\sqrt{a}$) it's a non-negative number $b$ for which $b^2=a$.

Now, if we want to solve an equation $x^2=4$ then we get $\{2,-2\}$ by the definition.

By the way, $\sqrt4=2$ by the definition again.

Maybe do you mean to use $\sqrt{x^2}=\sqrt{4}$?

If so, then we obtain $|x|=2$ and $\{2,-2\}$ again.

Answer 4

The word "root" means solution. The square roots of $n>0$ are the solutions to the equation $x^2=n.$ Since $x^2 = 4$ has two solutions, it is therefore rediculous to call $2$ the square root. It makes perfect sense, however, to call $\sqrt 4$ the positive square root of $4$.

This idea carries over into other problems. The fifth roots of $32$ are the five complex numbers, $z$, satisfying the equation $z^5 = 32$. Since there is only one real-valued solution, $z=2$, we say that the fifth root of $32$ is $2$ and no one will complain about us ignoring the other four solutions. Yet, to be correct, we must say that the positive sixth root of $64$ is $2$.

The point is that, given one solution, it is often much easier to find the other solutions. In trigonometry, for example, the equation $\sin x = \frac 12$ has an infinite number of solutions. Yet, on most scientific calculators, pressing $``\mathbf{inv \phantom{0} sin \phantom{0} 0.5 \phantom{0} ="} $ will return $30$ if the calculator is set for degrees. Knowing this, we can find all of the other solutions.