Square root of 4 equals 2 (or) ±2? [duplicate]

Simple yet much debated question inside my study circle. Please explain with reasons.

• Every non-zero complex number has two square roots, but by convention, "the" square root of a positive real is the positive real square root. Commented May 5, 2018 at 14:24
• It depends on the notation; when one uses the notation $\sqrt 4$ then the root is 2 but when one uses the notation $2^{\frac{1}{2}}$ then it can be any one of/ both of roots depending on the context. Commented May 5, 2018 at 14:25
• The square root of $4$ is a single well-defined number. A square root of $4$ has two choices.
– user65203
Commented May 5, 2018 at 15:15
• math.stackexchange.com/questions/2299199/square-root-of-x?rq=1 Commented May 5, 2018 at 20:22

The word "root" is an old expression roughly equivalent to "solution". The square root of $a$ is the solution(s) to the equation $x^2=a$. There are three cases.

If $a$ is a negative number, there are no real-number solutions.

If $a$ equals $0$, then $x=0$.

If $a$ is a positive number, then $x=\sqrt a$ or $x=-\sqrt a$.

The expression $\sqrt a$ is defined as the positive solution to the equation $x^2=a$. So people often write the solutions to $x^2=a$ as $x=\pm \sqrt a$.

This leads to some confusion some times. Students are often told to solve an equation like $x^2=25$ by "taking the square root of both sides. They then write

• $\sqrt{x^2} = \sqrt{25}$ and then $x=5$ and miss the solution $x=-5$

• or they write $x=\pm 5$ but don't really understand what they are doing

• or they write $x=\pm 5$ because $\sqrt{25}$ "means" $\pm 5$ which gives the correct answer but is an incorrect reason.

I have flagged this question as a duplicate (I don't have enough reputation to cast a close vote) and just in case that link is inaccessible, here's the answer from the original question:

$$\sqrt{\cdot}$$ is the principal square root operator. That means it returns only the principal root -- the positive one. $$\sqrt{64}=8$$. It does NOT equal $$-8$$.

On the other hand, the equation $$64=x^2$$ DOES have $$2$$ solutions: $$x=8$$ or $$x=-8$$. Thus both $$8$$ and $$-8$$ are square roots of $$64$$.

Let's see what happens when we take the principal square root of both sides of this equation: \begin{align}64 &= x^2 \\ \implies \sqrt{64} &= \sqrt{x^2} \\ \implies 8 &= |x| \\ \implies x&=8 \text{ or } x=-8\end{align}

Thus the fact that the principal square root operation throws out the negative root isn't much of a problem as the math still works out correctly.

The answer above will apply for this case, replace all "$$64$$" with "$$4$$", replace "$$8$$" with "$$2$$" and replace "$$-8$$" with "$$-2$$".

$\sqrt 4 = 2$
$\sqrt x$ is always non-negative

• @Blue corrected Commented May 5, 2018 at 14:44

By definition for $x≥0$ $y=\sqrt x$ is that value $y≥0$ such that $y^2=x$.

Then of course also the opposite $z=−y$ satisfy the condition $z^2=(−y)^2=x$ but , by convention/definition, the square root is assumed to be the positive one.

The underneath motivation is that the function $y=x^2$ is not invertible on the whole domain $\mathbb{R}$, then by convention the inverse has been chosen for $x\ge 0$ and that is $y=\sqrt x$.

There is not any other motivation but a matter of preference and convention in order to can define an inverse function for $y=x^2$.