Every technique in this lesson traces back to one fact proven in Rational Exponents: a radical is a fractional exponent, \(\sqrt[n]{a} = a^{1/n}\). That single equivalence is why radicals obey the same product and quotient structure as exponents, and it’s what makes every rule below provable instead of just memorizable.
What Is a Radical?
A radical expression has three parts: the radical sign \(\sqrt{\phantom{x}}\), the radicand underneath it (the number or expression being rooted), and the index, a small number in the sign’s notch that says which root to take. \(\sqrt[3]{8}\) has radicand \(8\) and index \(3\), and asks “what number, cubed, gives 8?” — the answer is \(2\), since \(2^{3} = 8\).
When no index is written, it defaults to \(2\), the square root: \(\sqrt{9}\) means \(\sqrt[2]{9} = 3\). A radical always returns the principal (nonnegative) root, matching the same convention used for rational exponents.
A simplifying convention. As in the previous lesson, assume every variable here is positive. That means \(\sqrt{x^{2}} = x\) exactly, with no absolute-value bars to track.
Why the Radical of a Power Returns the Base
This isn’t a separate rule to memorize — it falls directly out of the Power Rule for exponents. Since \(\sqrt[n]{a} = a^{1/n}\):
$$ \left(\sqrt[n]{a}\right)^{n} = \left(a^{1/n}\right)^{n} = a^{(1/n) \cdot n} = a^{1} = a $$
Raising the \(n\)th root of \(a\) back to the \(n\)th power always returns \(a\) — root and power are inverse operations, the same way addition and subtraction are.
Properties of Radicals
| Rule | Statement |
|---|---|
| Product Rule for Radicals | \(\sqrt[n]{ab} = \sqrt[n]{a} \cdot \sqrt[n]{b}\) |
| Quotient Rule for Radicals | \(\sqrt[n]{\dfrac{a}{b}} = \dfrac{\sqrt[n]{a}}{\sqrt[n]{b}}\) |
| Power of a Radical | \(\left(\sqrt[n]{a}\right)^{n} = a\) |
| Radical of a Power | \(\sqrt[n]{a^{n}} = a\) |
| Exponent Form | \(\sqrt[n]{a} = a^{1/n}\) |
| Exponent Form of a Power | \(\sqrt[n]{a^{m}} = a^{m/n}\) |
The Product and Quotient Rules are exactly how every simplification technique in this lesson works — everything below is one of those two rules applied with a specific goal.
Simplifying Radicals
To simplify \(\sqrt[n]{a}\), factor the radicand looking for the largest perfect \(n\)th-power factor, then split it off with the Product Rule:
$$ \sqrt{72} = \sqrt{36 \cdot 2} = \sqrt{36} \cdot \sqrt{2} = 6\sqrt{2} $$
Stopping at a smaller factor still works arithmetically but leaves a second simplification step behind — \(\sqrt{72} = \sqrt{4 \cdot 18} = 2\sqrt{18}\) is true, but \(\sqrt{18}\) still has a perfect-square factor left inside it. Always check the radicand one more time after pulling a factor out.
Example 1: Simplifying Radicals
| Expression | Perfect-Power Factor | Simplified |
|---|---|---|
| \(\sqrt{18}\) | \(9 \cdot 2\) | \(3\sqrt{2}\) |
| \(\sqrt{48}\) | \(16 \cdot 3\) | \(4\sqrt{3}\) |
| \(\sqrt{75}\) | \(25 \cdot 3\) | \(5\sqrt{3}\) |
| \(\sqrt[3]{54}\) | \(27 \cdot 2\) | \(3\sqrt[3]{2}\) |
| \(\sqrt[3]{16}\) | \(8 \cdot 2\) | \(2\sqrt[3]{2}\) |
Combining Like Radicals
Just as only like terms combine in a polynomial, only like radicals — the same index and the same radicand — combine, by adding or subtracting their coefficients:
$$ 3\sqrt{2} + 7\sqrt{2} = 10\sqrt{2} $$
\(3\sqrt{2} + 7\sqrt{3}\) cannot be combined any further; the radicands don’t match. Sometimes two radicals that look unlike become like ones once simplified — \(\sqrt{8} + \sqrt{2} = 2\sqrt{2} + \sqrt{2} = 3\sqrt{2}\) — so always simplify first, then check for like radicals.
Example 2: Combining and Multiplying Radicals
| Expression | Rule Applied | Result |
|---|---|---|
| \(3\sqrt{2} + 7\sqrt{2}\) | Combine like radicals | \(10\sqrt{2}\) |
| \(5\sqrt{3} - 2\sqrt{3}\) | Combine like radicals | \(3\sqrt{3}\) |
| \(\sqrt{3} \cdot \sqrt{12}\) | Product Rule | \(6\) |
| \((2\sqrt{5})(3\sqrt{5})\) | Multiply coefficients, then the Product Rule | \(30\) |
| \(\sqrt{2} \cdot \sqrt{8}\) | Product Rule | \(4\) |
Rationalizing Denominators
A denominator with a single radical term clears by multiplying top and bottom by that same radical:
$$ \frac{1}{\sqrt{3}} = \frac{1}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{3} $$
A denominator with a radical binomial, like \(2 + \sqrt{5}\), needs its conjugate instead — multiplying by the same radical alone won’t clear a sum. The conjugate of \(a + \sqrt{b}\) is \(a - \sqrt{b}\), and their product is always a clean difference of squares, \(a^{2} - b\), with no radical left at all.
Example 3: Rationalizing Denominators
| Expression | Multiply By | Result |
|---|---|---|
| \(\dfrac{1}{\sqrt{3}}\) | \(\dfrac{\sqrt{3}}{\sqrt{3}}\) | \(\dfrac{\sqrt{3}}{3}\) |
| \(\dfrac{5}{\sqrt{2}}\) | \(\dfrac{\sqrt{2}}{\sqrt{2}}\) | \(\dfrac{5\sqrt{2}}{2}\) |
| \(\dfrac{2}{\sqrt{5}-1}\) | \(\dfrac{\sqrt{5}+1}{\sqrt{5}+1}\) | \(\dfrac{\sqrt{5}+1}{2}\) |
| \(\dfrac{4}{3+\sqrt{2}}\) | \(\dfrac{3-\sqrt{2}}{3-\sqrt{2}}\) | \(\dfrac{12-4\sqrt{2}}{7}\) |
Worked Example A: Simplifying and Combining
Simplify \(\sqrt{200} - 3\sqrt{8} + \sqrt{18}\).
Step 1 — simplify each radical separately:
$$ \sqrt{200} = \sqrt{100 \cdot 2} = 10\sqrt{2}, \qquad 3\sqrt{8} = 3\left(2\sqrt{2}\right) = 6\sqrt{2}, \qquad \sqrt{18} = 3\sqrt{2} $$
Step 2 — combine, since every term is now a like radical:
$$ 10\sqrt{2} - 6\sqrt{2} + 3\sqrt{2} = 7\sqrt{2} $$
Nothing can be combined until every radical is simplified first — three unlike-looking radicals turned out to be the same radical in disguise.
Worked Example B: Rationalizing with a Conjugate
Simplify \(\dfrac{6}{2+\sqrt{3}}\).
Step 1 — multiply by the conjugate of the denominator:
$$ \frac{6}{2+\sqrt{3}} \cdot \frac{2-\sqrt{3}}{2-\sqrt{3}} $$
Step 2 — the denominator becomes a difference of squares:
$$ (2+\sqrt{3})(2-\sqrt{3}) = 2^{2} - \left(\sqrt{3}\right)^{2} = 4 - 3 = 1 $$
Step 3 — distribute the numerator:
$$ 6(2-\sqrt{3}) = 12 - 6\sqrt{3} $$
Since the denominator simplified all the way to \(1\), the final answer is just \(12 - 6\sqrt{3}\).
Worked Example C: Radicals with Variables
Simplify \(\sqrt{50x^{3}}\).
Step 1 — factor out the largest perfect-square factor:
$$ 50x^{3} = 25x^{2} \cdot 2x $$
Step 2 — apply the Product Rule:
$$ \sqrt{25x^{2} \cdot 2x} = \sqrt{25x^{2}} \cdot \sqrt{2x} = 5x\sqrt{2x} $$
Common Mistakes to Avoid
- Stopping at a partial factorization. \(\sqrt{72} = 2\sqrt{18}\) is technically true but incomplete — \(\sqrt{18}\) still has a perfect-square factor. Always use the largest perfect-power factor, or re-check what’s left.
- Adding radicals with different radicands or indices. \(\sqrt{2} + \sqrt{3}\) stays exactly as it is; it never becomes \(\sqrt{5}\).
- Multiplying radicals but forgetting the coefficients. \((2\sqrt{3})(5\sqrt{2}) = 10\sqrt{6}\), not \(\sqrt{6}\) — the coefficients multiply too.
- Assuming a radical distributes over a sum. \(\sqrt{a+b} \neq \sqrt{a} + \sqrt{b}\). Test it: \(\sqrt{9+16} = \sqrt{25} = 5\), but \(\sqrt{9} + \sqrt{16} = 3 + 4 = 7\).
- Leaving a radical in the denominator. A radical answer isn’t fully simplified until the denominator is rationalized.
- Getting the conjugate’s sign backward. The conjugate of \(a - \sqrt{b}\) is \(a + \sqrt{b}\) — only the middle sign flips, not both terms.
Practice Problems
Work each problem yourself before opening the answer.
Problem 1. Simplify \(\sqrt{48}\).
Show answer
$$ \sqrt{48} = \sqrt{16 \cdot 3} = \sqrt{16} \cdot \sqrt{3} = 4\sqrt{3} $$
Answer: \(4\sqrt{3}\)
Problem 2. Simplify \(\sqrt{98}\).
Show answer
$$ \sqrt{98} = \sqrt{49 \cdot 2} = \sqrt{49} \cdot \sqrt{2} = 7\sqrt{2} $$
Answer: \(7\sqrt{2}\)
Problem 3. Simplify \(4\sqrt{2} + 9\sqrt{2}\).
Show answer
$$ 4\sqrt{2} + 9\sqrt{2} = (4+9)\sqrt{2} = 13\sqrt{2} $$
Answer: \(13\sqrt{2}\)
Problem 4. Simplify \(\sqrt{5} \cdot \sqrt{20}\).
Show answer
$$ \sqrt{5} \cdot \sqrt{20} = \sqrt{5 \cdot 20} = \sqrt{100} = 10 $$
Answer: \(10\)
Problem 5. Simplify \((4\sqrt{3})(2\sqrt{6})\).
Show answer
Step 1 — multiply the coefficients and the radicands separately:
$$ (4 \cdot 2)\left(\sqrt{3} \cdot \sqrt{6}\right) = 8\sqrt{18} $$
Step 2 — simplify the resulting radical:
$$ 8\sqrt{18} = 8\left(3\sqrt{2}\right) = 24\sqrt{2} $$
Answer: \(24\sqrt{2}\)
Problem 6. Rationalize \(\dfrac{3}{\sqrt{5}}\).
Show answer
$$ \frac{3}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{3\sqrt{5}}{5} $$
Answer: \(\dfrac{3\sqrt{5}}{5}\)
Problem 7. Rationalize \(\dfrac{5}{4-\sqrt{3}}\).
Show answer
Step 1 — multiply by the conjugate:
$$ \frac{5}{4-\sqrt{3}} \cdot \frac{4+\sqrt{3}}{4+\sqrt{3}} $$
Step 2 — the denominator becomes a difference of squares:
$$ 4^{2} - \left(\sqrt{3}\right)^{2} = 16 - 3 = 13 $$
Step 3 — distribute the numerator:
$$ 5(4+\sqrt{3}) = 20 + 5\sqrt{3} $$
Answer: \(\dfrac{20+5\sqrt{3}}{13}\)
Problem 8. Simplify \(\sqrt{72x^{5}}\).
Show answer
$$ 72x^{5} = 36x^{4} \cdot 2x $$
$$ \sqrt{36x^{4} \cdot 2x} = \sqrt{36x^{4}} \cdot \sqrt{2x} = 6x^{2}\sqrt{2x} $$
Answer: \(6x^{2}\sqrt{2x}\)
Problem 9. Evaluate \((2+\sqrt{5})(2-\sqrt{5})\).
Show answer
$$ (2+\sqrt{5})(2-\sqrt{5}) = 2^{2} - \left(\sqrt{5}\right)^{2} = 4 - 5 = -1 $$
Answer: \(-1\). A binomial times its own conjugate always eliminates the radical, even when the result is negative.
Problem 10. Simplify \(\sqrt[3]{24}\).
Show answer
$$ \sqrt[3]{24} = \sqrt[3]{8 \cdot 3} = \sqrt[3]{8} \cdot \sqrt[3]{3} = 2\sqrt[3]{3} $$
Answer: \(2\sqrt[3]{3}\)
Quick Reference
| Situation | Move |
|---|---|
| Radicand has a perfect \(n\)th-power factor | Pull it out with the Product Rule |
| Same index and radicand | Combine by adding/subtracting coefficients |
| Multiplying radicals | Multiply coefficients together, radicands together, then simplify |
| One radical term in the denominator | Multiply by that same radical over itself |
| A radical binomial in the denominator | Multiply by its conjugate |
| Different radicands, or a sum inside the radical | No rule applies; simplify separately instead |
Once these techniques feel automatic, they carry forward directly into solving radical equations and, later, into Complex Numbers, where the same conjugate trick clears an imaginary denominator instead of a radical one. Build speed with the Radical Generator, browse the rest of the Algebra lessons as they publish, revisit Rational Exponents or Integer Exponents if any rule above felt unfamiliar, or explore the site’s calculators to check your own arithmetic against a worked result.