Square Root of 16: Solving for -2 with Step-by-Step Explanation

[brycenrg]

Homework Statement

2-squareroot(16) = -2

Homework Equations

The Attempt at a Solution

Why is there not two answers? I thought the squareroot of something always has two answers.

 

[ShayanJ]

The square root is supposed to be a function, and a function is not allowed to give more than one output for a given input. So the convention is that $\sqrt{b^2}=+b$. That's the reason every time you solve a quadratic equation, you need to use $\pm$, because you need both positive and negative roots but the square root only gives the positive one.

 

[PeroK]

Homework Statement

2-squareroot(16) = -2

Homework Equations

The Attempt at a Solution

Why is there not two answers? I thought the squareroot of something always has two answers.

The equation

$x^2 = 4$

has two solutions.

$x = \pm \sqrt{4} = \pm 2$

The square root of $4$, denoted by $\sqrt{4}$ is the positive solution and is equal to $2$.

It is not the case that $\sqrt{4}=\pm2$.

 

[ehild]

The square root is supposed to be a function, and a function is not allowed to give more than one output for a given input. So the convention is that $\sqrt{b^2}=+b$. .

$\sqrt{b^2}=+b$
Correctly: $\sqrt{b^2}=|b|$

 

[Kaura]

Yeah there is the caveat when dealing with square roots in terms of functions that the absolute value or positive root is accepted as the output of the function
However you are correct that square roots typically have two solutions
I suppose it depends on the question or situation

 

[Ray Vickson]

Yeah there is the caveat when dealing with square roots in terms of functions that the absolute value or positive root is accepted as the output of the function
However you are correct that square roots typically have two solutions
I suppose it depends on the question or situation

You need to distinguish between two related concepts: (1) the concept of a square root as a solution or solutions of an equation; and (2) the concept of the square root as a mathematical function. When you ask for a square root of 4, there are two possible values, +2 and -2. If you ask for the square root of 4 there is only one value, +2. In any programming language I know of, or in any spreadsheet or on any scientific calculator, when you enter "sqrt" or equivalent, you always get a the single value $\sqrt{b^2} = |b|$.

 

[HallsofIvy]

Yeah there is the caveat when dealing with square roots in terms of functions that the absolute value or positive root is accepted as the output of the function
However you are correct that square roots typically have two solutions
I suppose it depends on the question or situation

a "square root" doesn't have a "solution" because a square root isn't a problem! A quadratic equation typically has two solutions.

 

[arochon]

The first thing we need to do is define the word "solution". Most people would say that "solution" means "answer", but that is a synonym rather than a definition. My definition is "anything that makes the mathematical statement true." So when we look at the problem $$x^2=4$$ the values that make this true are both 2 and -2. When we look at $$\sqrt{4} = x$$ then only 2 is a solution. I think the confusion comes in that the steps used to solve these problems both include the concept of square roots - but the steps used to find a solution are not the same as the solution itself.

 

[FactChecker]

It's the convention of the '$\sqrt {}$' symbol that it only means the positive value. In practice, you should always put a '$\pm$' in front unless you have ruled out one of the values. The question as stated has ruled out the negative square root value.

 

[kuruman]

A related issue comes up when one has to calculate, for example, the electric potential inside a spherical charge distribution of radius $R$. Part of what one has to evaluate is an integral that results in a term$$A= \left[ \sqrt{x^2} \right]_{r-R}^{r+R} $$where $r$ is an arbitrary point in space. This needs to be done formally. Clearly, it evaluates to $$A=\sqrt{(r+R)^2} -\sqrt{(r-R)^2}$$but where do you go from here? The first term is a positive quantity, $\sqrt{(r+R)^2}=r+R$. The second term requires a bit of thought. It needs to be subtracted from the first term which means that $\sqrt{(r-R)^2}=|r-R|$. This results in two possibilities $$
A= \begin{cases}
2 R & \text{if } r \geq R \\
2r & \text{if } r < R.
\end{cases}$$

 

[Mark44]

It's the convention of the '$\sqrt {}$' symbol that it only means the positive value.

Yes.

In practice, you should always put a '$\pm$' in front unless you have ruled out one of the values.

No, one should not always do this.
If you're solving and equation such as $x^2 = 4$, then there will be two solutions, $\pm\sqrt 4 = \pm 2$. On the other hand, $\sqrt 9 = 3$, and $\pm$ should not be used.

 

[FactChecker]

On the other hand, $\sqrt 9 = 3$, and $\pm$ should not be used.

If there is a reason to rule one solution out, you should rule it out. Otherwise, all solutions should be considered. In this case, the absence of a '$\pm$' on the left side gives the reason to not consider the negative solution on the right side.

 

[Mark44]

On the other hand, √9=3, and ± should not be used.

If there is a reason to rule one solution out, you should rule it out. Otherwise, all solutions should be considered. In this case, the absence of a '±±' on the left side gives the reason to not consider the negative solution on the right side.

As you said earlier, by convention the √ symbol denotes the positive value only. I hope you are not saying that there are circumstances in which, say, √9=±39=±3 would be correct.

 

[FactChecker]

As you said earlier, by convention the $\sqrt{}$ symbol denotes the positive value only. I hope you are not saying that there are circumstances in which, say, $\sqrt 9 = \pm 3$ would be correct.

For the third time, no.

 

[Mark44]

OK, but please understand that what you wrote was confusing.
I wrote

On the other hand, $\sqrt 9 = 3$, and $\pm$ should not be used.

, which you quoted, and in your reply you said

If there is a reason to rule one solution out, you should rule it out. Otherwise, all solutions should be considered. In this case, the absence of a '$\pm$' on the left side gives the reason to not consider the negative solution on the right side.

A simpler way to look at this is that $\sqrt 9$ simplifies to a single number.

 

[LCKurtz]

All these answers to a 4 year old post by someone who hasn't been active for 2 years.

 

[Mark44]

Question was asked and answered, and the OP hasn't been seen for 3 years, so I'm closing this thread.