Is the square root of 4 a constant?

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Discussion Overview

The discussion centers around the nature of the square root of 4, specifically whether it can be considered a constant. Participants explore the implications of the square root notation, the definitions involved, and the distinctions between different mathematical contexts.

Discussion Character

  • Debate/contested
  • Conceptual clarification
  • Mathematical reasoning

Main Points Raised

  • Some participants argue that sqrt(4) = +/- 2, suggesting that while it seems like a constant, it is not fixed in all contexts.
  • Others assert that by definition, the square root is positive, leading to the conclusion that sqrt(4) = 2.
  • One participant clarifies that while both 2 and -2 are solutions to the equation x^2 = 4, the square root function specifically returns only 2.
  • Another participant discusses the implications of the square root function and the squaring function, emphasizing that the square root cannot yield negative results.
  • A later reply questions the necessity of the +/- symbol in the quadratic formula, suggesting that sqrt(b^2-4ac) should inherently provide both values.
  • One participant introduces the concept of functions and their definitions, using the arcos function as an example to illustrate that multiple angles can correspond to the same value in different contexts.

Areas of Agreement / Disagreement

Participants express differing views on whether sqrt(4) should be considered a constant, with some asserting it is definitively 2 while others maintain that it encompasses both 2 and -2 in different contexts. The discussion remains unresolved with multiple competing perspectives.

Contextual Notes

There are limitations in the discussion regarding the definitions of functions and the implications of mathematical notation, which are not fully explored or agreed upon by participants.

bigplanet401
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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!
 
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The square root is by definition positive. So ##\sqrt{4} = 2##.
 
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".
 
But (-2)^2 = 4?

The square root of 4 is 2.
 
Last edited by a moderator:
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?
 
bigplanet401 said:
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.
 
jack476 said:
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.
 
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
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
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! :)
 
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