1. Roots as the inverse of exponents
Every operation in arithmetic has an inverse. Addition undoes subtraction, multiplication undoes division, and exponents are undone by roots.
If $a^n = b$, then $a$ is an "$n$th root" of $b$, written $a = \sqrt[n]{b}$. The little number above the radical sign is the index; when it's missing, $2$ is implied (the square root). So $\sqrt{9} = 3$, because $3^2 = 9$.
For a non-negative number $b$ and a positive integer $n$, $\sqrt[n]{b}$ is the non-negative number $a$ such that $a^n = b$. The symbol $\sqrt{\phantom{x}}$ is called the radical and the number under it the radicand.
"Non-negative" matters: although $(-3)^2 = 9$ just as much as $3^2 = 9$, by convention $\sqrt{9}$ means the positive square root, $3$. The equation $x^2 = 9$ has two solutions, $\pm 3$, but the symbol $\sqrt{9}$ refers to just the positive one. (The negative one is written $-\sqrt{9}$.)
2. Square roots
The square root, $\sqrt{a}$, is the most important and most common case. It's the answer to "what number squared gives $a$?" — which is the same as asking "what is the side length of a square whose area is $a$?"
Perfect squares are the values of $n^2$ for whole-number $n$: $1, 4, 9, 16, 25, 36, 49, 64, \ldots$. Their square roots are exact whole numbers: $\sqrt{49} = 7$.
For all the in-between numbers, the square root is some real number that's not a whole number — and (as we'll see in §5) usually not even a fraction.
$$ \sqrt{2} \approx 1.41421, \qquad \sqrt{10} \approx 3.16228, \qquad \sqrt{50} \approx 7.07107. $$$\sqrt{-1}$ has no real value, because no real number squared gives a negative. Within the world of real numbers, $\sqrt{-4}$ is undefined. (In a later topic, this gap is filled by the imaginary unit $i$, with $i^2 = -1$ — but that's a story for the complex numbers.)
3. Cube roots and higher
The cube root $\sqrt[3]{a}$ asks "what number cubed gives $a$?" Examples:
$$ \sqrt[3]{27} = 3,\quad \sqrt[3]{125} = 5,\quad \sqrt[3]{8} = 2. $$Unlike square roots, cube roots of negative numbers are perfectly real: $\sqrt[3]{-8} = -2$, because $(-2)^3 = -8$. The cube of a negative number is negative, so the cube root sign just travels through.
The general pattern:
- Even-index roots ($n = 2, 4, 6, \ldots$) require a non-negative radicand. $\sqrt[4]{-16}$ has no real value.
- Odd-index roots ($n = 3, 5, 7, \ldots$) accept any real radicand. $\sqrt[5]{-32} = -2$.
The reason is the same as for the squared-vs-cubed distinction: a negative raised to an even power is positive (the negatives pair up and cancel), so an even root can never produce a negative; a negative raised to an odd power stays negative, so an odd root can.
4. Fractional exponents
Here is the clean, unifying fact:
$$ \sqrt[n]{a} = a^{1/n}. $$A root is a fractional exponent. Why? Because if we want the exponent laws from the previous topic to keep working, this is what they force.
Look: $(a^{1/n})^n = a^{(1/n) \cdot n} = a^1 = a$, by the power-of-a-power rule. So $a^{1/n}$ is precisely the number that, raised to the $n$th power, gives $a$ back — and that's the definition of $\sqrt[n]{a}$. The two expressions describe the same number; the only difference is notation.
This extends to any rational exponent:
$$ a^{m/n} = (a^{1/n})^m = (\sqrt[n]{a})^m \;=\; \sqrt[n]{a^m}. $$So $8^{2/3} = (\sqrt[3]{8})^2 = 2^2 = 4$. Or equivalently, $8^{2/3} = \sqrt[3]{8^2} = \sqrt[3]{64} = 4$. The two routes give the same answer; the first is usually easier because you take the root of a small number first.
Once roots are fractional exponents, you don't need a separate set of rules for radicals. Every move you learned for exponents — product, quotient, power-of-power — applies, including with fractional exponents. The radical sign becomes a notational convenience, not a separate operation.
5. Why $\sqrt{2}$ isn't a fraction
One of the most famous proofs in all of mathematics — old enough that the Pythagoreans are said to have hushed it up — shows that $\sqrt{2}$ cannot be written as a fraction of two whole numbers.
Suppose, for the sake of contradiction, that $\sqrt{2} = \tfrac{p}{q}$, where $\tfrac{p}{q}$ is in lowest terms ($\gcd(p, q) = 1$). Square both sides:
$$ 2 = \frac{p^2}{q^2} \quad\Longrightarrow\quad p^2 = 2q^2. $$So $p^2$ is even — which means $p$ itself is even (the square of an odd number is odd). Write $p = 2k$. Then:
$$ (2k)^2 = 2q^2 \quad\Longrightarrow\quad 4k^2 = 2q^2 \quad\Longrightarrow\quad 2k^2 = q^2. $$So $q^2$ is even, which means $q$ is even too. But then both $p$ and $q$ are divisible by $2$, contradicting our assumption that the fraction was in lowest terms.
The contradiction means our supposition was wrong: no such fraction exists. $\sqrt{2}$ is irrational.
A real number that cannot be written as a fraction of two integers. The square root of any whole number that isn't a perfect square is irrational. So are $\pi$ and $e$. Their decimal expansions never terminate and never repeat.
The Pythagoreans believed every length could be expressed as a ratio of whole numbers — that was practically a religious commitment. The discovery that $\sqrt{2}$ (the diagonal of a unit square!) could not be so expressed was reportedly so disturbing that the discoverer, Hippasus, was supposedly drowned at sea to keep the secret. The story is probably apocryphal, but the cultural impact was real: it forced an expansion of what "number" meant.
6. Simplifying and operating on radicals
The two algebraic facts that handle almost every radical manipulation:
$$ \sqrt{a \cdot b} = \sqrt{a} \cdot \sqrt{b}, \qquad \sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}. $$(For non-negative $a$ and $b$, with $b \neq 0$ in the quotient.) These are just the power-of-a-product and power-of-a-quotient rules, applied at exponent $\tfrac{1}{2}$.
Simplifying
To simplify a square root, factor out the largest perfect square hiding inside the radicand:
$$ \sqrt{50} = \sqrt{25 \cdot 2} = \sqrt{25} \cdot \sqrt{2} = 5\sqrt{2}. $$The form $5\sqrt{2}$ is the canonical one — a whole-number coefficient times a square root with no perfect-square factors left inside.
Multiplying
$$ \sqrt{2} \cdot \sqrt{8} = \sqrt{16} = 4. $$Combine the radicands under one root, then simplify.
Adding and subtracting
You can only add or subtract radicals that have the same radicand — the radical part plays the role of a "variable":
$$ 3\sqrt{2} + 5\sqrt{2} = 8\sqrt{2}, \qquad 7\sqrt{3} - 2\sqrt{3} = 5\sqrt{3}. $$You can't combine $\sqrt{2} + \sqrt{3}$ into a single radical — they're just two distinct irrational numbers, the same way $x + y$ is two distinct unlike terms.
7. Common pitfalls
The "freshman's dream" again. Roots distribute over products and quotients, never over sums. $\sqrt{9 + 16} = \sqrt{25} = 5$, but $\sqrt{9} + \sqrt{16} = 3 + 4 = 7$. They are different numbers.
$\sqrt{x^2} = |x|$, not $x$. If $x = -3$, then $\sqrt{(-3)^2} = \sqrt{9} = 3$, not $-3$. The radical symbol is committed to the non-negative answer, even when the input has been squared from a negative.
$\sqrt{-4}$ has no real value. If a calculation produces a negative under a square root, either you've made an error, or you've stumbled into territory that needs complex numbers. Either way, stop and check the setup.
$\sqrt{a}$ is the inverse of squaring, which means it's $a^{1/2}$, not $a^2$. The two are opposites: squaring $\sqrt{9}$ takes you back to $9$, but squaring $9$ gives $81$. Always check which direction the operation is taking you.
8. Worked examples
Example 1 · Simplify $\sqrt{72}$
Find the biggest perfect square that divides $72$: $36$.
$$ \sqrt{72} = \sqrt{36 \cdot 2} = 6\sqrt{2}. $$Answer: $\boxed{6\sqrt{2}}$.
Example 2 · Evaluate $\sqrt[3]{-125}$
Cube roots accept negative radicands. Find the number whose cube is $-125$: $(-5)^3 = -125$.
$$ \sqrt[3]{-125} = -5. $$Example 3 · Evaluate $16^{3/4}$
Rewrite the fractional exponent as a root of a power (or a power of a root):
$$ 16^{3/4} = (16^{1/4})^3 = (\sqrt[4]{16})^3 = 2^3 = 8. $$Answer: $\boxed{8}$.
Example 4 · Simplify $3\sqrt{5} + 2\sqrt{20}$
First simplify $\sqrt{20}$: $\sqrt{20} = \sqrt{4 \cdot 5} = 2\sqrt{5}$.
Substitute back:
$$ 3\sqrt{5} + 2 \cdot 2\sqrt{5} = 3\sqrt{5} + 4\sqrt{5} = 7\sqrt{5}. $$Answer: $\boxed{7\sqrt{5}}$.
Example 5 · Simplify $\sqrt{18} \cdot \sqrt{2}$
Combine under one radical: $\sqrt{18 \cdot 2} = \sqrt{36} = 6$.
Answer: $\boxed{6}$. (Nice when the product becomes a perfect square.)