# Square root of x

How many values of $$\sqrt x$$ is possible. Is it 2 or 1?The graph says only one but why not 2 one is positive an the other is negative.

Please explain why is it so?

But Thanks in anticipation.

• Lots of questions like this on the site. Possuble duplicate: math.stackexchange.com/questions/59630/… – Ethan Bolker May 27 '17 at 18:45
• For $x > 0$ (and $x$ a real number) $\sqrt {x}$ is ONE of the two possible square roots. By definition $\sqrt{x}$ is the positive square root. $-\sqrt{x}$ is the negative one. $\sqrt{x}$ is not THE square root of $x$. It is the POSITIVE square root of $x$. – fleablood May 27 '17 at 18:48
• Another dupe Square root confusion? – kingW3 May 27 '17 at 18:49

When $x\ge0$, $\sqrt{x}$ is defined as the non-negative square root of $x$, i.e. the non-negative real number $y$ such that $y^2=x$.

If we take both the positive and negative square roots of $x$ as $\sqrt{x}$, expressions like $\sqrt{4}+\sqrt{9}+\sqrt{16}$ can mean $2+3+4$, $2+3-4$, $2-3+4$, etc. Things become very complicated.

If $x\ge0$, the equation $y^2=x^2$ has two solutions $y=x$ and $y=-x$. Since $x$ is the non-negative one, $\sqrt{x^2}=x$.

If $x<0$, the equation $y^2=x^2$ has two solutions $y=x$ and $y=-x$. Since $-x$ is the non-negative one, $\sqrt{x^2}=-x$.

Therefore, $\sqrt{x^2}=|x|$. That's why we have the V-shaped graph.

The square root function $\sqrt{a}$ is defined as the non-negative solution of the equation $x^2-a=0$.

It is not the same as asking: What are the solutions of $x^2-a=0$? (Which in this case, the negative and positive solution are correct.)

You are viewing the square root function as the inverse function of $f(x)=x^2$. The function $f$ doesn't have an inverse function since it is not a bijection (a function $f$ such that every element $y$ of the codomain has a unique element $x$ in the domain that $f(x)=y$).

The function $f(x)=x^2$ is not a bijection if the domain is the real numbers but the square root function has been created by restricting the domain of $f$ to $[0,+\infty[$ to make it a bijection and finding the inverse of the function with the restricted domain.

$\sqrt{x}$ is not "THE" square root of $x$ because if $x > 0$ (and real) there are two square roots of $x$ and there is no "THE" square root of $x$.

$\sqrt{x}$ is not $\{y \in R| y^2 = x\}$. That is a set that can have two values. $\sqrt{}$ is a function that maps $\sqrt{}:[0,\infty) \rightarrow \mathbb R$ so being a function $\sqrt{x}$ is one specific value.

So what is $\sqrt{x}$? It is the function $\sqrt{x} =$ the non-negative square root of $x$. That is its definition It is defined to never be negative.

In other words, if $x > 0$ then there are TWO square roots $y$ and $w$ so that $y^2 = x$ and $w^2 = x$ and $y = -w$ or in other words one of them will be negative and the other one will be positive. $\sqrt{x}$ is ONE of them; the positive one. $-\sqrt{x}$ is the other negative one.

(And, of course, if $x = 0$ then there aren't two $y^2 = 0$ and $w^2 = 0$ and $y = -w$. There is only one $y = w = 0$ and $0 = -0$ so that $y^2 = w^2 = 0$. So $\sqrt {0} = -\sqrt{0} = 0$.)