Ytukyg

The square root sign explicitly denotes the nonnegative root of a nonnegative real number. If you want to indicate the negative root, you use $-sqrt{r}$. If you want to indicate all square roots of the number, that is written as $pmsqrt{r}$.

In summary, the square root symbol does not denote two values. By convention, it is the nonnegative value. One could say philosophically that this is arbitrary, but there are some obvious advantages to letting it be the nonnegative root. For one, the positive root is used more often.

$sqrt{}:[0,infty)tomathbb R$ prescribed by: $$xmapstotext{ non-negative }ytext{ satisfying }y^2=x$$ is a function because this non-negative $y$ with $y^2=x$ is unique.

If $f:left(-infty,frac34right]to[0,infty)$ is a function prescribed by: $$xmapsto 3-4x$$ then also the composition $$sqrt{}circ f:left(-infty,frac34right]tomathbb R$$ is a function.

Its prescription is denoted as: $$xmapstosqrt{3-4x}$$

There is no confusion in finding the values of $$y=sqrt {4-3x}$$ for $xin (-infty,4/3]$.

You do not have $pm $ in front of the radical sign which means you simply pick the positive square root.

Note that for example $$sqrt {16}=4$$ and $$-sqrt {16} =-4$$

You pretty much answer the question yourself, albeit unknowingly.

I thought any equation that has a square root sign is not a function, because of the ± sign in front of it?

Yes, you're absolutely right. If it has the $pm$ sign, we can't call it a function, but what you wrote was

$y=sqrt{4-3x}$

which does NOT have the $pm$ sign in front.

And the consensus is that when we don't have a sign in front of the radical sign, we always interpret it to mean the non-negative root. Thus, it is a function.

Don't confuse the symbol $sqrt{a}$ with the solutions of the equation $x^2=a$.

They are distinct things: $sqrt{a}$ is, by commonly agreed convention, the unique nonnegative real number $b$ such that $b^2=a$ (provided such $b$ exists, which it does when $age0$).

With this convention, the solutions of $x^2=a$, for $age0$, are $sqrt{a}$ and $-sqrt{a}$. This is often written as $pmsqrt{a}$, but I find the symbol misleading.

The confusion between the two concepts has, unfortunately, been perpetuated for a long time. Be assured that all professional mathematicians have no doubt to agree with the convention stated above.

My high school textbook tried to clear up this confusion by introducing a new symbol (a square root with a kind of hook at the end of the vinculum) for denoting “both square roots” at the same time. This is of course even more misleading: assuming the radical symbol stands for the modified one, how many values should the expression
$$
sqrt{4}+sqrt{9}+sqrt{16}
$$
denote? Eight, possibly. Does this have practical uses? None.

Using $sqrt{a}$ to denote the unique nonnegative real number whose square is $a$ is handy and has several nice properties, the main one being that, for $x,yge0$,
$$
sqrt{xyvphantom{l}}=sqrt{xvphantom{ly}},sqrt{yvphantom{l}}
$$

When the square root is used alone, it indicates only the principle square root, a non-negative value.

It’s a common misconception to think $x = sqrt n$ is the same as $x^2 = n$. However, the first has only one solution, while the second has two: $x = pmsqrt{n}$.

For instance, $x = sqrt 4$ means only the positive square root of $4$, so you have $x = color{blue}{+}2$.

However, the solutions of $x^2 = 4$ are $x = color{blue}{pm 2}$, so the squared equation has two solutions.

Likewise, $y = sqrt{4-3x}$ is a function because you take only the positive root.

On the other hand, if you had $y^2 = 4-3x$, you would be correct. This would be the equivalent to having $y = pmsqrt{4-3x}$. Note that the original function does not have the $pm$ sign, which makes all the difference.