Existence Conditions for Radicals

only for nonnegative radicands R⁺+{0} when the index of the root is even $$ \sqrt{x} \ \ \ \ \text{Domain: } x \in R^+ + \{0\} $$
for all real numbers R when the index of the root is odd $$ \sqrt[3]{x} \ \ \ \ \text{Domain: } x \in R $$

Accordingly, when the index of a root is even, an explicit domain condition must be imposed on the radical expression.

$$ \text{Domain: } \text{radicand} \ge 0 $$

When the index of the root is odd, the radical is defined for every real value of the variable x, so the domain condition may be omitted.

$$ \text{Domain: } x \in R $$

Nevertheless, it is generally advisable to state the domain explicitly in all cases to avoid ambiguity.

Explanation. A radicand cannot be negative when the root index is even, because no real number raised to an even power can produce a negative result. $$ (-2) \cdot (-2) = +4 $$ Therefore, the radical is not defined in the real numbers $$ \sqrt{-4} = \text{undefined in } R $$ By contrast, a radicand may be negative when the root index is odd, since a negative number raised to an odd power remains negative. $$ (-2) \cdot (-2) \cdot (-2) = -8 $$ Hence $$ \sqrt[3]{-8} = -2 $$

In simpler terms, the existence condition depends on the index of the root.

A Practical Example

Consider the following radical expression

$$ \sqrt{x-1} $$

This is a square root with an even index (2).

An explicit domain condition must be introduced, because the radicand cannot be negative.

The radicand (x-1) must be greater than or equal to zero.

$$ \text{Domain: } x-1 \ge 0 $$

Isolate the independent variable x on the left-hand side

$$ \text{Domain: } x-1 +1 \ge 0 +1 $$

$$ \text{Domain: } x \ge 1 $$

This inequality represents the domain of the radical expression.

Example. If x is negative, for example x = -7 $$ \sqrt{x-1} $$ $$ \sqrt{-7-1} $$ $$ \sqrt{-8} $$ The expression is undefined in the real numbers, because there is no real number whose square is negative. The square of any real number is always nonnegative. $$ (-2) \cdot (-2) = +4 $$

Example 2

Now consider the radical

$$ \sqrt[3]{x-1} $$

Here the root index is odd.

No domain restriction is required, because the radicand may assume any real value, including negative numbers.

Example. If x is negative, for example x = -7 $$ \sqrt[3]{x-1} $$ $$ \sqrt[3]{-7-1} $$ $$ \sqrt[3]{-8} = -2 $$ The expression is well defined, because there exists a real number, -2, whose cube equals the radicand -8. $$ (-2) \cdot (-2) \cdot (-2) = -8 $$ The radical is also defined when the radicand is zero. For example, if x = 1 $$ \sqrt[3]{x-1} = \sqrt[3]{1-1} = \sqrt[3]{0} = 0 $$ since $$ 0^3 = 0 \cdot 0 \cdot 0 = 0 $$

And so on.