Is the square root of 4 a constant?

[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!

 

[micromass]

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

 

[HallsofIvy]

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".

 

[bigplanet401]

But (-2)^2 = 4?

The square root of 4 is 2.

 

[russ_watters]

Yes. Those are different statements though.

 

[bigplanet401]

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

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 sort of right?

 

[jack476]

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!

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.

 

[Mark44]

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.

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.

 

[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?

 

[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.

 

[mp3car]

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