Rationalizing Denominators

In mathematics, it is standard practice to rewrite fractions so that no radical appears in the denominator. The process of eliminating radicals from the denominator is called rationalizing the denominator. This is not just a cosmetic preference — rationalized expressions are easier to compare, easier to add and subtract, and historically easier to compute with. In this section you will learn two techniques: one for monomial denominators (a single radical term) and one for binomial denominators (two terms, at least one with a radical).

Why We Rationalize

Consider the expression $\dfrac{1}{\sqrt{2}}$. It is a perfectly valid number, approximately $0.707$. But in simplified form, mathematicians prefer $\dfrac{\sqrt{2}}{2}$. Why?

1. Standardization. A single canonical form makes it easier to compare answers and check work.
2. Easier addition. Adding $\dfrac{1}{\sqrt{2}} + \dfrac{1}{\sqrt{3}}$ requires finding a common denominator with radicals — messy. After rationalizing, $\dfrac{\sqrt{2}}{2} + \dfrac{\sqrt{3}}{3}$ has rational denominators that combine via ordinary LCD methods.
3. Convention. Standardized tests, textbooks, and answer keys expect rationalized denominators.

Rationalizing Monomial Denominators

A monomial denominator has a single radical term: $\sqrt{a}$, $3\sqrt{b}$, etc.

Strategy: Multiply the numerator and denominator by the radical in the denominator. This uses the fact that $\sqrt{a} \cdot \sqrt{a} = a$.

$\frac{1}{\sqrt{a}} = \frac{1}{\sqrt{a}} \cdot \frac{\sqrt{a}}{\sqrt{a}} = \frac{\sqrt{a}}{a}$

You are multiplying by $\dfrac{\sqrt{a}}{\sqrt{a}} = 1$, so the value does not change.

Example 1: Rationalize $\dfrac{1}{\sqrt{5}}$

Multiply top and bottom by $\sqrt{5}$:

$\frac{1}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{\sqrt{5}}{5}$

Answer: $\dfrac{\sqrt{5}}{5}$

Example 2: Rationalize $\dfrac{6}{\sqrt{3}}$

$\frac{6}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{6\sqrt{3}}{3} = 2\sqrt{3}$

Answer: $2\sqrt{3}$

Example 3: Rationalize $\dfrac{4}{\sqrt{8}}$

Option A — Rationalize first, then simplify:

$\frac{4}{\sqrt{8}} \cdot \frac{\sqrt{8}}{\sqrt{8}} = \frac{4\sqrt{8}}{8} = \frac{\sqrt{8}}{2} = \frac{2\sqrt{2}}{2} = \sqrt{2}$

Option B — Simplify the denominator first:

$\frac{4}{\sqrt{8}} = \frac{4}{2\sqrt{2}} = \frac{2}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{2\sqrt{2}}{2} = \sqrt{2}$

Both methods give the same answer. Simplifying first often makes the arithmetic easier.

Answer: $\sqrt{2}$

Example 4: Rationalize $\dfrac{10}{3\sqrt{2}}$

Multiply by $\dfrac{\sqrt{2}}{\sqrt{2}}$:

$\frac{10}{3\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{10\sqrt{2}}{3 \cdot 2} = \frac{10\sqrt{2}}{6} = \frac{5\sqrt{2}}{3}$

Answer: $\dfrac{5\sqrt{2}}{3}$

Example 5: Rationalize $\dfrac{\sqrt{3}}{\sqrt{7}}$

$\frac{\sqrt{3}}{\sqrt{7}} \cdot \frac{\sqrt{7}}{\sqrt{7}} = \frac{\sqrt{21}}{7}$

Answer: $\dfrac{\sqrt{21}}{7}$

Rationalizing Binomial Denominators — The Conjugate Method

When the denominator has two terms (at least one containing a radical), you cannot simply multiply by $\sqrt{a}$. Instead, multiply by the conjugate of the denominator.

The conjugate of a binomial is the same two terms with the opposite sign between them:

Why this works: Multiplying a binomial by its conjugate produces the difference of squares:

$(a + \sqrt{b})(a - \sqrt{b}) = a^2 - (\sqrt{b})^2 = a^2 - b$

The radical is eliminated from the denominator.

Example 6: Rationalize $\dfrac{1}{3 + \sqrt{2}}$

The conjugate of $3 + \sqrt{2}$ is $3 - \sqrt{2}$. Multiply top and bottom:

$\frac{1}{3 + \sqrt{2}} \cdot \frac{3 - \sqrt{2}}{3 - \sqrt{2}} = \frac{3 - \sqrt{2}}{(3)^2 - (\sqrt{2})^2} = \frac{3 - \sqrt{2}}{9 - 2} = \frac{3 - \sqrt{2}}{7}$

Answer: $\dfrac{3 - \sqrt{2}}{7}$

Example 7: Rationalize $\dfrac{5}{4 - \sqrt{3}}$

Conjugate: $4 + \sqrt{3}$.

$\frac{5}{4 - \sqrt{3}} \cdot \frac{4 + \sqrt{3}}{4 + \sqrt{3}} = \frac{5(4 + \sqrt{3})}{16 - 3} = \frac{20 + 5\sqrt{3}}{13}$

Answer: $\dfrac{20 + 5\sqrt{3}}{13}$

Example 8: Rationalize $\dfrac{6}{\sqrt{5} + \sqrt{3}}$

Conjugate: $\sqrt{5} - \sqrt{3}$.

$\frac{6}{\sqrt{5} + \sqrt{3}} \cdot \frac{\sqrt{5} - \sqrt{3}}{\sqrt{5} - \sqrt{3}} = \frac{6(\sqrt{5} - \sqrt{3})}{5 - 3} = \frac{6(\sqrt{5} - \sqrt{3})}{2}$

$= 3(\sqrt{5} - \sqrt{3}) = 3\sqrt{5} - 3\sqrt{3}$

Answer: $3\sqrt{5} - 3\sqrt{3}$

Example 9: Rationalize $\dfrac{2 + \sqrt{3}}{1 - \sqrt{3}}$

Conjugate of denominator: $1 + \sqrt{3}$.

Numerator:

$(2 + \sqrt{3})(1 + \sqrt{3}) = 2(1) + 2\sqrt{3} + \sqrt{3}(1) + \sqrt{3} \cdot \sqrt{3}$

$= 2 + 2\sqrt{3} + \sqrt{3} + 3 = 5 + 3\sqrt{3}$

Denominator:

$(1 - \sqrt{3})(1 + \sqrt{3}) = 1 - 3 = -2$

Result:

$\frac{5 + 3\sqrt{3}}{-2} = \frac{-(5 + 3\sqrt{3})}{2} = \frac{-5 - 3\sqrt{3}}{2}$

Answer: $\dfrac{-5 - 3\sqrt{3}}{2}$

Example 10: Rationalize $\dfrac{\sqrt{7}}{\sqrt{7} - 2}$

Conjugate: $\sqrt{7} + 2$.

Numerator:

$\sqrt{7}(\sqrt{7} + 2) = 7 + 2\sqrt{7}$

Denominator:

$(\sqrt{7})^2 - (2)^2 = 7 - 4 = 3$

Answer: $\dfrac{7 + 2\sqrt{7}}{3}$

When the Numerator Has the Radical

Sometimes a problem asks you to rationalize the numerator instead. This is less common but appears in calculus when working with limits. The technique is identical — multiply by the conjugate of the numerator.

Example 11: Rationalize the numerator of $\dfrac{\sqrt{x + h} - \sqrt{x}}{h}$

Conjugate of the numerator: $\sqrt{x + h} + \sqrt{x}$.

$\frac{(\sqrt{x + h} - \sqrt{x})(\sqrt{x + h} + \sqrt{x})}{h(\sqrt{x + h} + \sqrt{x})} = \frac{(x + h) - x}{h(\sqrt{x + h} + \sqrt{x})} = \frac{h}{h(\sqrt{x + h} + \sqrt{x})} = \frac{1}{\sqrt{x + h} + \sqrt{x}}$

Answer: $\dfrac{1}{\sqrt{x + h} + \sqrt{x}}$

Real-World Application: Electrician — Impedance Ratio

When analyzing AC circuits, an electrician might calculate the ratio of two impedances:

$\text{Ratio} = \frac{R}{\sqrt{R^2 + X_L^2}}$

For $R = 6$ ohms and $X_L = 6$ ohms:

$\text{Ratio} = \frac{6}{\sqrt{36 + 36}} = \frac{6}{\sqrt{72}} = \frac{6}{6\sqrt{2}} = \frac{1}{\sqrt{2}}$

Rationalize:

$\frac{1}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2} \approx 0.707$

This ratio is known as the power factor in electrical engineering. Expressing it as $\dfrac{\sqrt{2}}{2}$ is standard and immediately recognizable to anyone who works with AC circuits. The decimal approximation $0.707$ is the value they would enter into calculations or test equipment.

Common Mistakes to Avoid

1. Multiplying by the conjugate of the numerator instead of the denominator. When rationalizing the denominator, multiply by the conjugate of the denominator. Keep the target straight.
2. Forgetting to multiply the numerator too. When you multiply the denominator by the conjugate, you must multiply the numerator by the same expression — otherwise you change the value of the fraction.
3. Sign errors in the conjugate. The conjugate flips only the sign between the two terms: the conjugate of $3 + \sqrt{5}$ is $3 - \sqrt{5}$, not $-3 - \sqrt{5}$.
4. Not simplifying the result. After rationalizing, always check if the fraction can be reduced. For example, $\dfrac{6\sqrt{3}}{12}$ simplifies to $\dfrac{\sqrt{3}}{2}$.
5. Trying to use the conjugate on a monomial. For a single-term denominator like $\sqrt{5}$, just multiply by $\dfrac{\sqrt{5}}{\sqrt{5}}$. The conjugate technique is for binomials.

Practice Problems

Test your understanding with these problems. Click to reveal each answer.

Key Takeaways

• Rationalizing removes radicals from the denominator to produce a standard simplified form
• For monomial denominators ($\sqrt{a}$), multiply top and bottom by $\sqrt{a}$
• For binomial denominators ($a + \sqrt{b}$), multiply top and bottom by the conjugate ($a - \sqrt{b}$)
• The conjugate method works because $(a + \sqrt{b})(a - \sqrt{b}) = a^2 - b$, which eliminates the radical
• Always simplify the result after rationalizing — reduce fractions and simplify any remaining radicals
• Rationalized forms are standard in math, science, and engineering — tests and answer keys expect them

Return to Algebra for more topics in this section.