# Number of values in square root in different cases

I have two equations:

1. $x = \sqrt{16}$
2. $x^2 = 16$

In first case I think there will be two value of $x = \pm4$. Because $(-4) \cdot (-4) =( +4) \cdot(+4) = 16$

In the second case I am confused. It can also have two values i.e $\pm 4$.

But somewhere I read that one of the equations returns only one value and it will be $+4$. I can't find which one will return $+4$ and which one both values and the reason behind it. Can any body help me please? Thanks.

• In the second case, $x$ is a "Square Root" of $16$. In the first case, $x$ is the "Principle Square Root" of $16$. The "Principle Square Root" is always non-negative. I suppose that the negative square root would be called the "Secondary Square Root" or perhaps the "Conjugate Square Root"??? Dec 15, 2015 at 23:27

Square root is a mathematical function and as any other function it must have only one value for any given argument.

$f(x)=\sqrt{x}$ has one and only one value for $x=16$ if $\sqrt{x}$ is to be defined at all.

Normally, we talk about branches when we deal with multivalued regions.

If we understand $f(x)=\sqrt{x}$ as the inverse of $f(x)=x^{2}$, we have two branches: $x>0$ and $x<0$. $x>0$ is called principal branch and this one is assumed if nothing else is specified for $f(x)=\sqrt{x}$. Since $x=0$ may belong to both branches we extend the region to $x \ge 0$.

From this perspective $x=\sqrt{16}$ is an assignment or identity, depending on the context, the same as $x=4$. However, $x^2=16$ is an equation that needs to be solved, with $x$ being the unknown. In the case of $x=\sqrt{16}$, $x$ is not an unknown.

1. $x = \sqrt{16} = 4 \neq -4$

2. $x^2 = 16 \iff x= \pm\sqrt{16} = \pm 4$

Your reasoning for the first case actually deals with the second case. In general for $x>0$ we always have $\sqrt{x} > 0$.

The square root symbol when applied to a positive real number $r$ is defined to be the positive value whose square is $r$. That's not something you can say is right or wrong, it's just a definition.

In contrast if we ask for all values $x$ that solve the equation $x^2=16$ then there are two values of $x$ that work, $x=4$ and $x=-4$.

You see, equations can have more than one solution.

In other words in one case we have the common definition of a symbol, which must have a single value, while in the other case we are asking for the solutions of an equation which can be multiple values.