# Is the square root of 4 a constant?

1. Apr 9, 2015

### bigplanet401

I don't know what to say. sqrt(4) = +/- 2, but constants are fixed. When written sqrt(4), it seems sqrt(4) is a constant, but not in the latter case. Help!

2. Apr 9, 2015

### micromass

Staff Emeritus
The square root is by definition positive. So $\sqrt{4} = 2$.

3. Apr 9, 2015

### HallsofIvy

Staff Emeritus
You are misunderstanding the notation. There are two numbers, -2 and 2, whose square is 4. But only one of those, 2, is "the square root of four".
And, yes, both "2" and "-2" are "constant".

4. Apr 9, 2015

### bigplanet401

But (-2)^2 = 4?

5. Apr 9, 2015

### Staff: Mentor

Yes. Those are different statements though.

6. Apr 9, 2015

### bigplanet401

Maybe one way of looking at it is to ask if
$$\sqrt{4} = -\sqrt{4}$$
and, if $$\sqrt{4} = \pm 2$$,
$$+(\pm 2) = - (\pm 2)$$

which is a true statement. But this would mean -x = x with x nonzero, which is false. So this means $$\sqrt{4}$$ has only one value. Is this kinda sorta right?

7. Apr 9, 2015

### jack476

No, square root of 4 is 2 and only 2, but 2^2 OR (-2)^2 will be equal to 4.

So if you're confronted with something like √x =2, then x is 4. The square root function cannot return a negative real number.

On the other hand, if you've got x^2 = 4, then both x = 2 and x = -2 are valid solutions to that algebraic equation. If you were to solve that equation by taking the square root of both sides, you would have to include x = -2 along with the x = 2 that would be returned by that operation because, unlike the square root function, the squaring function has values for all real numbers.

8. Apr 9, 2015

### Staff: Mentor

A better way to solve the equation $x^2 = 4$ is to write it as $x^2 - 4 = 0$ or (x - 2)(x + 2) = 0, from which we get x = 2 or x = -2. No funny business with taking the square root of both sides needed.

9. Apr 9, 2015

### Mentallic

If $\sqrt{4}=\pm 2$, then when using the quadratic formula

$$x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

Why would we need the $\pm$ symbol there considering $\sqrt{b^2-4ac}$ should give us the positive and negative value?

10. Apr 10, 2015

### BobG

Maybe what's missing from this thread is the definition of a function.

https://www.mathsisfun.com/definitions/function.html

Accordingly, the arcos of 0.5 is 60 degrees because the range of the arcos function is 0 to 180 degrees. None the less, if you're looking at a circle, there are two angles (-60 and +60) that have a cosine of 0.5.

11. Apr 10, 2015

### mp3car

Man I love this board... :) I'm guessing some of our spouses would have left us by now if we didn't have this forum and phys.org! :)