If every square root has positive and negative solutions, then is $-2 = 2\sqrt1$?

Question

Since every square root has 2 possible solutions, one positive and one negative. Then wouldn't that happen every time you have a square root?

Let's say for example: If $x + 1 = 2\sqrt{x+4}$ then $x$ would have 2 possible solutions: $5$, $-3$. Verifying for $5$ is easy, but what about $-3$?

$-3 + 1 = 2\sqrt{-3+4} -2 = 2\sqrt{1} -2 = 2 \cdot \pm\sqrt{1}$, which then makes $-3$ have $2$ solutions: $-2 = 2(1)\implies$ no Solution and $-2 = 2(-1)\implies$ Statement is true.

If the second statement is true then $-3$ would also be a correct solution for the equation $x + 1 = 2\sqrt{x+4}$.

Answer 1

Preface

Your post is a great example of why terminology is important. In mathematics, there exists a large number of different types of "stuff", and different types of stuff have different properties associated with them. For example,

1. you can talk about "the absolute value of a number" in mathematics (i.e., it is sensible to say "the absolute value of $-3$ is $3$, and even the sentence "the absolute value of $3$ is $-3$" is a valid sentence, in the sense that it has a truth value, but of course the truth value is "false", since the sentence is false) 2.or, you can talk about "the validity of an equation" (you could say, for example, that $3+5=8$ is a valid equation)
2. but you cannot talk about "the validity of a number" (i.e., "is $3$ a valid number?" is a nonsensical question), or about "the absolute value of an equation".

Note that this is not that unusual, and mathematics is not special in this regard. Other fields of study have the same limitations. Even regular life has them. You can speak of a size of an object or the pitch of a sound, but you can't sensibly talk about the size of a sound of the pitch of an object.

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Main point:

1. A square root is does not have "solutions". A square root is a function. Functions do not have solutions. Functions have values. Equations have solutions.
2. The equation $x^2 = 2$ has two solutions.
3. The function $f(x)=\sqrt{x}$ has one value at $2$, and that value is $\sqrt{2}$, which is a positive number approximately equal to $1.41$.

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Now on to your concrete example.

Taking your equation, it can first be shown that if the equation $x+1=2\sqrt{x+4}$, is true then $x$ is equal to either $5$ or $-3$. Note, and this is very important, this does not mean that $5$ and $3$ are both solutions to the original equation!. It just means that they are the only possible solutions, i.e. that we have proven that all other numbers are certainly not the solutions.

Once we have the two candidates, the easiest way to check if either is an actual solution is to check whether either of them is an actual solution.

• For $5$, as you said, the check shows that it is a solution, since substituting $x=5$ into the equation yields $5+1=2\sqrt{5+4}$ which simplifies to $6=6$ which is a true statement.
• For $-3$, substituting the value yields $-3+1=2\sqrt{-3+4}$ which simplifies to $-2=2\cdot\sqrt{1}$. Taking into account that $\sqrt{1}=1$, this means that substituting $x=-3$ yields the equation $-2=2$ which is a false statement, therefore, $-3$ is not a solution of the original equation.