Textbook
For a real number $a\ge0$, the square root $\sqrt a$ is the unique nonnegative number whose square equals $a$. Thus $\sqrt9=3$, although the equation $x^2=9$ has two solutions, 3 and −3. The root symbol specifies a selected value rather than the full solution set.
For an integer $k\ge2$ and $a>0$, the positive root $\sqrt[k]{a}$ equals $a^{1/k}$. A negative real number has no real root of even order. An odd-order real root also exists for negative inputs: $\sqrt[3]{-8}=-2$. State the number domain before calculating.
For every real $x$, $\sqrt{x^2}=|x|$ because the selected root is nonnegative. Canceling the root and square to obtain $x$ without assuming $x\ge0$ is incorrect. At $x=-3$, the expression equals 3. A zero radicand has root zero for every integer order $k\ge2$. The order’s parity affects negative inputs but does not change this zero case.
Definitions and further reading: Basic Analysis I.
In plain language
A square root recovers a side from an area: area 9 gives the nonnegative side length 3. The equation asking which numbers square to 9 also permits −3, but the root symbol selects one nonnegative value. An odd-order root may be negative, so the building analogy does not describe every root. Rejecting negative lengths need not evict perfectly lawful cube roots from mathematics.
Example
The example tests candidates for input 9 and distinguishes the equation’s solutions from the square-root value. Squaring verifies a candidate, but selecting the root also requires nonnegativity.
Roots: Recovering a Power’s Input
Positive candidate
The square of 3 is 9, so this candidate works.
Step 1 of 3: Positive candidate
Read all steps
Positive candidate
The square of 3 is 9, so this candidate works.
- State
- 3current3current9current
Negative candidate
The square of −3 is also 9: the equation has two solutions.
- State
- −3current−3current9current
Selecting the root
The root symbol selects the nonnegative value 3.
- State
- −3excluded3included
Candidate $-3$ passes the squaring test but fails nonnegativity. The second check excludes it from the value of $\sqrt9$ without denying that it solves the equation. The two results differ: the equation has solutions $x\in\{-3,3\}$, while the expression has value $\sqrt9=3$. The sign $\pm$ describes the solutions; writing $\sqrt9=\pm3$ would violate the selected value’s uniqueness.
Related concepts: Exponentiation.