A fraction like \(\dfrac{1}{\sqrt{2}}\) is a perfectly valid number, but it isn’t considered a finished algebra answer — by convention, a denominator is supposed to be free of radicals. Getting there always comes down to multiplying the fraction by a cleverly disguised form of \(1\), chosen specifically to turn the denominator into a perfect power. This lesson covers every version of that multiplier you’ll need.
Radicals introduces rationalizing as part of a broader toolkit; this lesson is the focused, example-heavy version, covering single-term denominators, two-term (binomial) denominators, and higher-index roots one at a time.
Why Rationalize a Denominator?
Rationalizing a denominator means rewriting a fraction so that no radical remains on the bottom, without changing the fraction’s value. The tool for doing this without changing the value is always the same: multiply by a fraction equal to \(1\), built from whatever clears the radical.
$$ \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2} $$
\(\dfrac{\sqrt{2}}{\sqrt{2}}\) equals exactly \(1\), so multiplying by it never changes what the fraction is worth — only \(\sqrt{2}\) moved from the denominator to the numerator, where a radical is considered acceptable.
Rationalizing a Single-Term (Monomial) Denominator
When the denominator is a single radical, multiply top and bottom by just enough of that same radical to complete a perfect power underneath the root.
$$ \frac{5}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{5\sqrt{3}}{3} $$
For a square root, one extra copy always finishes the job, since \(\sqrt{a}\cdot\sqrt{a}=a\). For a higher-index root, the multiplier needs to complete the radicand up to a full power of the index — not necessarily an identical second copy, as the next section shows.
Rationalizing a Two-Term (Binomial) Denominator with a Conjugate
When the denominator has two terms and at least one is a radical, a single multiplier can’t clear it — squaring \((a+\sqrt{b})\) directly would still leave a radical behind, since \((a+\sqrt{b})^{2}=a^{2}+2a\sqrt{b}+b\). The fix is multiplying by the conjugate instead: the same two terms with the sign between them flipped.
$$ (a+\sqrt{b})(a-\sqrt{b}) = a^{2} - b $$
This is the difference of squares pattern, and it’s exactly why the conjugate works: squaring a square root always removes its radical, and a sum times its conjugate is defined to produce that exact pattern.
$$ \frac{1}{2+\sqrt{3}} \cdot \frac{2-\sqrt{3}}{2-\sqrt{3}} = \frac{2-\sqrt{3}}{4-3} = \frac{2-\sqrt{3}}{1} = 2-\sqrt{3} $$
Rationalizing Denominators with a Higher-Index Root
For an index-\(n\) root, one extra factor doesn’t necessarily finish clearing the radical — the multiplier needs to bring the radicand’s exponent up to a multiple of \(n\).
$$ \frac{1}{\sqrt[3]{x}} \cdot \frac{\sqrt[3]{x^{2}}}{\sqrt[3]{x^{2}}} = \frac{\sqrt[3]{x^{2}}}{\sqrt[3]{x^{3}}} = \frac{\sqrt[3]{x^{2}}}{x} $$
The original radicand was \(x^{1}\); multiplying by \(\sqrt[3]{x^{2}}\) brought the total exponent under the root up to \(x^{3}\), which is a perfect cube and clears the root entirely. A common mistake is multiplying by another copy of \(\sqrt[3]{x}\) instead, which only reaches \(x^{2}\) — still not a perfect cube.
Worked Example A: A Monomial Square-Root Denominator
Rationalize \(\dfrac{4}{\sqrt{5}}\).
Step 1 — multiply by \(\dfrac{\sqrt{5}}{\sqrt{5}}\):
$$ \frac{4}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{4\sqrt{5}}{5} $$
Answer: \(\dfrac{4\sqrt{5}}{5}\)
Worked Example B: A Monomial Denominator That Needs Simplifying First
Rationalize \(\dfrac{3}{\sqrt{8}}\).
Step 1 — simplify the radical in the denominator first:
$$ \sqrt{8} = 2\sqrt{2} $$
Step 2 — rationalize the simplified denominator:
$$ \frac{3}{2\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{3\sqrt{2}}{4} $$
Answer: \(\dfrac{3\sqrt{2}}{4}\). Simplifying \(\sqrt{8}\) first avoided a much larger multiplier and an extra simplification step at the end.
Worked Example C: A Binomial Denominator with a Conjugate
Rationalize \(\dfrac{6}{5-\sqrt{2}}\).
Step 1 — multiply by the conjugate of the denominator, over itself:
$$ \frac{6}{5-\sqrt{2}} \cdot \frac{5+\sqrt{2}}{5+\sqrt{2}} $$
Step 2 — expand the denominator using the difference of squares pattern:
$$ (5-\sqrt{2})(5+\sqrt{2}) = 25 - 2 = 23 $$
Step 3 — expand the numerator by distributing:
$$ 6(5+\sqrt{2}) = 30 + 6\sqrt{2} $$
Answer: \(\dfrac{30+6\sqrt{2}}{23}\)
Worked Example D: A Higher-Index Root Denominator
Rationalize \(\dfrac{2}{\sqrt[3]{9}}\).
Step 1 — rewrite \(9\) as \(3^{2}\), so \(2\) more factors of \(3\) are needed to reach \(3^{3}\):
$$ \frac{2}{\sqrt[3]{3^{2}}} \cdot \frac{\sqrt[3]{3}}{\sqrt[3]{3}} $$
Step 2 — multiply out:
$$ \frac{2\sqrt[3]{3}}{\sqrt[3]{3^{3}}} = \frac{2\sqrt[3]{3}}{3} $$
Answer: \(\dfrac{2\sqrt[3]{3}}{3}\)
Common Mistakes to Avoid
- Multiplying only the denominator by the rationalizing factor. The numerator has to be multiplied by the exact same thing, since the multiplier must equal \(1\) — multiplying only one part changes the fraction’s value.
- Squaring a binomial denominator instead of using its conjugate. \((a+\sqrt{b})^{2}\) still contains a radical; only multiplying by the conjugate \((a-\sqrt{b})\) produces a difference of squares.
- Flipping the wrong sign when writing a conjugate. The conjugate of \(5-\sqrt{2}\) is \(5+\sqrt{2}\) — only the sign between the two terms changes, not the sign of either term individually.
- Using an identical copy of the radical for a higher-index root. For \(\sqrt[3]{x}\), multiplying by another \(\sqrt[3]{x}\) only reaches \(x^{2}\) under the root, not a perfect cube; the correct multiplier is \(\sqrt[3]{x^{2}}\).
- Forgetting to distribute across a binomial numerator after multiplying by a conjugate. Every term in the numerator gets multiplied by every term in the conjugate, the same as any other binomial expansion.
- Rationalizing before simplifying the original radical. Skipping the simplification step often leads to a larger, messier multiplier and an unsimplified final answer.
Where This Shows Up Later
- Complex Number Arithmetic. Dividing by a complex number uses the exact same conjugate trick, since \(a+bi\) times its conjugate \(a-bi\) also produces a real, radical-free (and imaginary-free) result.
- The quadratic formula. An answer like \(x = \dfrac{2+\sqrt{3}}{2}\) sometimes needs its denominator rationalized further if it appears inside a larger fraction elsewhere in a problem.
- Limits in calculus. Rationalizing a numerator, using the identical conjugate technique from this lesson, is a standard trick for evaluating a limit that would otherwise divide by zero.
- Trigonometric identities. Exact trig values like \(\dfrac{1}{\sqrt{2}}\) are conventionally rationalized to \(\dfrac{\sqrt{2}}{2}\) before being reported as a final answer.
Practice Problems
Work each problem yourself before opening the answer.
Problem 1. Rationalize \(\dfrac{7}{\sqrt{6}}\).
Show answer
$$ \frac{7}{\sqrt{6}} \cdot \frac{\sqrt{6}}{\sqrt{6}} = \frac{7\sqrt{6}}{6} $$
Answer: \(\dfrac{7\sqrt{6}}{6}\)
Problem 2. Rationalize \(\dfrac{3}{\sqrt{x}}\).
Show answer
$$ \frac{3}{\sqrt{x}} \cdot \frac{\sqrt{x}}{\sqrt{x}} = \frac{3\sqrt{x}}{x} $$
Answer: \(\dfrac{3\sqrt{x}}{x}\)
Problem 3. Rationalize \(\dfrac{5}{\sqrt{12}}\).
Show answer
Step 1 — simplify the denominator first:
$$ \sqrt{12} = 2\sqrt{3} $$
Step 2 — rationalize:
$$ \frac{5}{2\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{5\sqrt{3}}{6} $$
Answer: \(\dfrac{5\sqrt{3}}{6}\)
Problem 4. Rationalize \(\dfrac{1}{3+\sqrt{5}}\).
Show answer
Step 1 — multiply by the conjugate:
$$ \frac{1}{3+\sqrt{5}} \cdot \frac{3-\sqrt{5}}{3-\sqrt{5}} $$
Step 2 — expand the denominator:
$$ (3+\sqrt{5})(3-\sqrt{5}) = 9 - 5 = 4 $$
Answer: \(\dfrac{3-\sqrt{5}}{4}\)
Problem 5. Rationalize \(\dfrac{4}{\sqrt{7}-2}\).
Show answer
Step 1 — multiply by the conjugate:
$$ \frac{4}{\sqrt{7}-2} \cdot \frac{\sqrt{7}+2}{\sqrt{7}+2} $$
Step 2 — expand the denominator:
$$ (\sqrt{7}-2)(\sqrt{7}+2) = 7 - 4 = 3 $$
Step 3 — expand the numerator:
$$ 4(\sqrt{7}+2) = 4\sqrt{7}+8 $$
Answer: \(\dfrac{4\sqrt{7}+8}{3}\)
Problem 6. Rationalize \(\dfrac{2}{\sqrt[3]{5}}\).
Show answer
Step 1 — multiply by enough extra factors to complete a perfect cube:
$$ \frac{2}{\sqrt[3]{5}} \cdot \frac{\sqrt[3]{5^{2}}}{\sqrt[3]{5^{2}}} $$
Step 2 — simplify:
$$ \frac{2\sqrt[3]{25}}{\sqrt[3]{5^{3}}} = \frac{2\sqrt[3]{25}}{5} $$
Answer: \(\dfrac{2\sqrt[3]{25}}{5}\)
Problem 7. Rationalize \(\dfrac{3}{2\sqrt{5}}\).
Show answer
$$ \frac{3}{2\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{3\sqrt{5}}{10} $$
Answer: \(\dfrac{3\sqrt{5}}{10}\)
Problem 8. Rationalize \(\dfrac{\sqrt{2}}{\sqrt{2}-1}\).
Show answer
Step 1 — multiply by the conjugate:
$$ \frac{\sqrt{2}}{\sqrt{2}-1} \cdot \frac{\sqrt{2}+1}{\sqrt{2}+1} $$
Step 2 — expand the denominator:
$$ (\sqrt{2}-1)(\sqrt{2}+1) = 2 - 1 = 1 $$
Step 3 — expand the numerator:
$$ \sqrt{2}(\sqrt{2}+1) = 2 + \sqrt{2} $$
Answer: \(2+\sqrt{2}\)
Problem 9. Rationalize \(\dfrac{5}{\sqrt[3]{4}}\).
Show answer
Step 1 — rewrite \(4\) as \(2^{2}\); one more factor of \(2\) reaches \(2^{3}\):
$$ \frac{5}{\sqrt[3]{2^{2}}} \cdot \frac{\sqrt[3]{2}}{\sqrt[3]{2}} $$
Step 2 — simplify:
$$ \frac{5\sqrt[3]{2}}{\sqrt[3]{2^{3}}} = \frac{5\sqrt[3]{2}}{2} $$
Answer: \(\dfrac{5\sqrt[3]{2}}{2}\)
Problem 10. Rationalize \(\dfrac{x}{\sqrt{x}+3}\).
Show answer
Step 1 — multiply by the conjugate:
$$ \frac{x}{\sqrt{x}+3} \cdot \frac{\sqrt{x}-3}{\sqrt{x}-3} $$
Step 2 — expand the denominator:
$$ (\sqrt{x}+3)(\sqrt{x}-3) = x - 9 $$
Step 3 — distribute the numerator:
$$ x(\sqrt{x}-3) = x\sqrt{x}-3x $$
Answer: \(\dfrac{x\sqrt{x}-3x}{x-9}\)
Quick Reference
| Denominator looks like | Multiply top and bottom by |
|---|---|
| \(\sqrt{a}\) | \(\sqrt{a}\) |
| \(\sqrt[n]{a^{k}}\) | \(\sqrt[n]{a^{n-k}}\) (whatever completes a perfect \(n\)th power) |
| \(a+\sqrt{b}\) or \(a-\sqrt{b}\) | The conjugate, \(a-\sqrt{b}\) or \(a+\sqrt{b}\) |
| \(\sqrt{a}+\sqrt{b}\) | The conjugate, \(\sqrt{a}-\sqrt{b}\) |
| Before rationalizing at all | Simplify the radical first — it usually shrinks the multiplier needed |
Once a monomial and binomial denominator both feel routine, the exact same conjugate idea reappears in Complex Number Arithmetic for dividing by a complex number. Revisit Simplifying Radicals any time a denominator’s radical needs cleaning up before rationalizing, or Radicals for the full toolkit these two lessons split apart. Build speed with the Rationalizing Denominators Generator, or browse the rest of the Algebra lessons as new ones publish.