Why Is the Principal Even Root Nonnegative?

An even root is not always strictly positive: the square root of zero is zero. More precisely, the standard real-valued radical symbol denotes the principal even root, which is nonnegative. This convention explains why $\sqrt{25}=5$, even though both $5$ and $-5$ square to $25$.

The distinction is between evaluating a radical and solving an equation. A radical names one chosen value; an equation may have several solutions.

What an even-root symbol means

For a nonnegative real number $a$ and a positive even integer $n$, the expression $\sqrt[n]{a}$ means the unique nonnegative real number $r$ satisfying

$$r^n=a.$$

For example, $\sqrt{16}=4$ because $4^2=16$, and $\sqrt[4]{16}=2$ because $2^4=16$. The word nonnegative includes zero: $\sqrt{0}=0$. If $a>0$, the principal even root is strictly positive.

There is no real even root of a negative number. No real number raised to an even power can be negative, because the product pairs its negative factors: for example, $(-3)^2=9$, not $-9$.

Why choose the nonnegative root?

The function $x\mapsto x^n$ for even $n$ is not one-to-one on all real numbers. A positive input and its negative have the same even power:

$$(-x)^n=x^n.$$

So if you try to reverse this function on all of $\mathbb{R}$, one output can correspond to two inputs. For instance, both $3$ and $-3$ map to $9$ when squared.

To define a single-valued inverse, restrict the inputs to $x\geq 0$. On that domain, raising to an even power is one-to-one and takes nonnegative values. Its inverse is the principal-root function. This is why the radical $\sqrt[n]{a}$ returns the nonnegative value, rather than an instruction to list every number whose $n$th power is $a$.

This convention makes expressions such as $\sqrt{x}$ well-defined as a function for $x\geq 0$. It also keeps the radical notation consistent: one expression has one value.

Worked examples: a radical and an equation

Consider the expression $\sqrt{36}$. The nonnegative number whose square is $36$ is $6$, so

$$\sqrt{36}=6.$$

Now solve the equation $x^2=36$. Both $6^2=36$ and $(-6)^2=36$, so the real solutions are

$$x=\pm 6.$$

The same distinction applies to higher even powers. The principal fourth root of $81$ is $3$, since $3^4=81$. But the real solutions to $x^4=81$ are $x=3$ and $x=-3$.

At the boundary case, $\sqrt{0}=0$, and $x^2=0$ has just one real solution, $x=0$. Thus “even roots are always positive” is not quite right: they are nonnegative, and they are positive only when the radicand is positive.

How this differs from odd roots

Odd powers preserve the sign of a real number. For example, $3^3=27$ and $(-3)^3=-27$. The function $x\mapsto x^n$ is one-to-one on all real numbers when $n$ is odd, so an odd root does not need the same nonnegative-value convention:

$$\sqrt[3]{-27}=-3.$$

For a positive number, an odd-power equation still has just one real solution. This contrasts with a positive even-power equation, which has two real solutions when the right-hand side is positive.

Common mistakes

• Calling the principal root strictly positive: zero is allowed. The precise word is nonnegative.
• Writing $\sqrt{25}=\pm5$: the radical alone means $5$. Use $\pm5$ when solving $x^2=25$.
• Assuming a negative number has a real even root: it does not. Complex roots are a different setting, where “positive” is not an applicable ordering in the usual real-number sense.
• Applying $\sqrt{x^2}=x$ for every real $x$: the correct identity is $\sqrt{x^2}=|x|$. For example, if $x=-4$, then $\sqrt{x^2}=\sqrt{16}=4$, not $-4$.

When you see an even-root symbol, first check that its input is nonnegative, then take the nonnegative root. When you are solving an even-power equation, check separately for both signs rather than treating the radical as the full solution set.