The Central Idea
Rationalizing a denominator means rewriting an equivalent expression so that the denominator contains no radical.
The guiding strategy is to multiply by a form of \(1\):
\[ \frac{R}{R}=1, \qquad R\neq0. \]For example,
\[ \frac1{\sqrt2} \]can be multiplied by
\[ \frac{\sqrt2}{\sqrt2}. \]Then
\[ \frac1{\sqrt2} \cdot \frac{\sqrt2}{\sqrt2} = \frac{\sqrt2}{2}. \]Therefore,
\[ \boxed{ \frac1{\sqrt2} = \frac{\sqrt2}{2} }. \]Why Rationalize a Denominator?
A denominator containing a radical is not inherently incorrect.
For example,
\[ \frac1{\sqrt3} \]is a valid exact number.
Rationalization is a conventional simplification that often makes comparison, combination, exact arithmetic, and later algebra easier.
Rationalizing a Single Square-Root Denominator
Consider
\[ \frac5{\sqrt7}. \]Multiply numerator and denominator by \(\sqrt7\):
\[ \frac5{\sqrt7} \cdot \frac{\sqrt7}{\sqrt7}. \]Then
\[ \frac{5\sqrt7}{7}. \]Therefore,
\[ \boxed{ \frac5{\sqrt7} = \frac{5\sqrt7}{7} }. \]Why the Radical Disappears from the Denominator
The denominator becomes
\[ \sqrt7\sqrt7. \]Since
\[ (\sqrt7)^2=7, \]we get
\[ \sqrt7\sqrt7=7. \]The radical has been replaced by a rational number.
Example: Coefficient with a Radical Denominator
Rationalize
\[ \frac{3}{2\sqrt5}. \]Multiply by
\[ \frac{\sqrt5}{\sqrt5}. \]Then
\[ \frac{3\sqrt5}{2\cdot5}. \]Therefore,
\[ \boxed{ \frac{3}{2\sqrt5} = \frac{3\sqrt5}{10} }. \]Example: Simplify Before Rationalizing
Consider
\[ \frac{4}{\sqrt{12}}. \]First simplify the denominator:
\[ \sqrt{12}=2\sqrt3. \]Therefore,
\[ \frac4{\sqrt{12}} = \frac4{2\sqrt3} = \frac2{\sqrt3}. \]Now rationalize:
\[ \frac2{\sqrt3} \cdot \frac{\sqrt3}{\sqrt3} = \frac{2\sqrt3}{3}. \]Thus,
\[ \boxed{ \frac4{\sqrt{12}} = \frac{2\sqrt3}{3} }. \]Radical Fractions Can Sometimes Simplify First
Consider
\[ \frac{\sqrt{18}}{\sqrt2}. \]Use the quotient rule:
\[ \frac{\sqrt{18}}{\sqrt2} = \sqrt9 = 3. \]No rationalization step is needed after simplification.
Therefore,
\[ \boxed{3}. \]What Is a Conjugate?
Two binomials are conjugates when they have the same two terms but opposite signs between them.
Examples include
\[ a+b \]and
\[ a-b. \]For radical expressions,
\[ a+\sqrt b \]and
\[ a-\sqrt b \]form a conjugate pair.
The Difference-of-Squares Identity
Conjugates are useful because
\[ (a+b)(a-b)=a^2-b^2. \]The middle terms cancel.
In particular,
\[ (a+\sqrt b)(a-\sqrt b) = a^2-b. \]The resulting expression contains no square-root term.
Example: Multiply Conjugates
Multiply
\[ (5+\sqrt3)(5-\sqrt3). \]Apply the difference-of-squares identity:
\[ 5^2-(\sqrt3)^2. \] \[ =25-3. \]Therefore,
\[ \boxed{22}. \]Why the Middle Terms Cancel
Expand
\[ (a+b)(a-b). \]Distribute:
\[ a^2-ab+ab-b^2. \]The middle terms satisfy
\[ -ab+ab=0. \]Therefore,
\[ \boxed{a^2-b^2}. \]This cancellation is exactly what makes conjugates useful for rationalization.
Rationalizing a Binomial Denominator
Consider
\[ \frac1{3+\sqrt2}. \]Multiplying by \(\sqrt2/\sqrt2\) would not eliminate the entire radical structure of the denominator.
Instead, multiply by the conjugate:
\[ \frac{3-\sqrt2}{3-\sqrt2}. \]Example: Rationalize \(1/(3+\sqrt2)\)
Multiply:
\[ \frac1{3+\sqrt2} \cdot \frac{3-\sqrt2}{3-\sqrt2}. \]The numerator becomes
\[ 3-\sqrt2. \]The denominator becomes
\[ (3+\sqrt2)(3-\sqrt2) = 9-2 = 7. \]Therefore,
\[ \boxed{ \frac1{3+\sqrt2} = \frac{3-\sqrt2}{7} }. \]Use the Opposite Sign
If the denominator is
\[ a+\sqrt b, \]multiply by
\[ a-\sqrt b. \]If the denominator is
\[ a-\sqrt b, \]multiply by
\[ a+\sqrt b. \]The conjugate keeps the same terms and reverses only the sign between them.
Example: Denominator with Subtraction
Rationalize
\[ \frac2{5-\sqrt3}. \]Multiply by the conjugate:
\[ \frac{5+\sqrt3}{5+\sqrt3}. \]Then
\[ \frac{2(5+\sqrt3)} {25-3}. \]Therefore,
\[ \frac{10+2\sqrt3}{22}. \]Reduce by \(2\):
\[ \boxed{ \frac{5+\sqrt3}{11} }. \]Conjugates with Two Radical Terms
Conjugates need not contain a rational first term.
For example,
\[ \sqrt5+\sqrt2 \]and
\[ \sqrt5-\sqrt2 \]are conjugates.
Their product is
\[ 5-2=3. \]Example: Rationalize Two Radical Terms
Rationalize
\[ \frac1{\sqrt5+\sqrt2}. \]Multiply by the conjugate:
\[ \frac{\sqrt5-\sqrt2}{\sqrt5-\sqrt2}. \]The denominator becomes
\[ 5-2=3. \]Therefore,
\[ \boxed{ \frac1{\sqrt5+\sqrt2} = \frac{\sqrt5-\sqrt2}{3} }. \]Example: Rationalize a More General Binomial
Rationalize
\[ \frac{4}{2+\sqrt7}. \]Multiply by
\[ \frac{2-\sqrt7}{2-\sqrt7}. \]Then
\[ \frac{4(2-\sqrt7)} {4-7}. \] \[ = \frac{8-4\sqrt7}{-3}. \]Therefore,
\[ \boxed{ \frac{4\sqrt7-8}{3} }. \]Keep Track of Negative Denominators
In the previous example,
\[ 4-7=-3. \]The negative sign may be moved to the numerator:
\[ \frac{8-4\sqrt7}{-3} = \frac{4\sqrt7-8}{3}. \]Clear sign placement helps keep the final form readable.
Rationalizing a Variable Square Root
Suppose \(x>0\). Rationalize
\[ \frac5{\sqrt x}. \]Multiply by
\[ \frac{\sqrt x}{\sqrt x}. \]Then
\[ \frac{5\sqrt x}{x}. \]Therefore,
\[ \boxed{ \frac5{\sqrt x} = \frac{5\sqrt x}{x} }. \]Domain Conditions Remain Important
In
\[ \frac1{\sqrt x}, \]the square root requires
\[ x\geq0, \]while the denominator requires
\[ \sqrt x\neq0. \]Therefore,
\[ \boxed{x>0}. \]Rationalizing the denominator does not change this original domain.
Example: Variable Binomial Denominator
Suppose \(x\geq0\) and
\[ 3-\sqrt x\neq0. \]Rationalize
\[ \frac1{3-\sqrt x}. \]Multiply by the conjugate:
\[ \frac{3+\sqrt x}{3+\sqrt x}. \]Then
\[ \frac{3+\sqrt x}{9-x}. \]Therefore,
\[ \boxed{ \frac1{3-\sqrt x} = \frac{3+\sqrt x}{9-x} }. \]Preserve the Original Excluded Value
In
\[ \frac1{3-\sqrt x}, \]the denominator is zero when
\[ \sqrt x=3, \]so
\[ x=9. \]Thus the original real domain is
\[ x\geq0,\qquad x\neq9. \]The rationalized form must be interpreted with the same restriction.
Higher Roots Require a Different Factor
For a square root,
\[ \sqrt a\cdot\sqrt a=a. \]For a cube root, multiplying by the same radical once may not create a perfect cube.
Instead, choose the factor needed to complete a perfect power.
Example: Rationalize a Cube-Root Denominator
Rationalize
\[ \frac1{\sqrt[3]{2}}. \]The denominator contains one factor of \(2\) under a cube root.
To create \(2^3\), we need two additional factors of \(2\):
\[ \sqrt[3]{4}. \]Multiply:
\[ \frac1{\sqrt[3]{2}} \cdot \frac{\sqrt[3]{4}}{\sqrt[3]{4}}. \]The denominator becomes
\[ \sqrt[3]{8}=2. \]Therefore,
\[ \boxed{ \frac1{\sqrt[3]{2}} = \frac{\sqrt[3]{4}}{2} }. \]Example: Another Cube-Root Denominator
Rationalize
\[ \frac3{\sqrt[3]{5^2}}. \]The denominator contains \(5^2\) inside a cube root.
One additional factor of \(5\) completes the cube:
\[ \sqrt[3]{5}. \]Multiply:
\[ \frac{3\sqrt[3]{5}} {\sqrt[3]{5^3}}. \]Therefore,
\[ \boxed{ \frac{3\sqrt[3]{5}}{5} }. \]General Strategy for Higher-Root Denominators
Suppose a denominator contains
\[ \sqrt[n]{a^m} \]
with \(0
Multiply by enough additional factors to make the exponent inside the
radical a multiple of \(n\).
A natural choice is
Then
Rationalizing Denominators with Coefficients
Consider
\[ \frac7{3\sqrt2}. \]Only the radical part needs to be rationalized.
Multiply by
\[ \frac{\sqrt2}{\sqrt2}. \]Then
\[ \frac{7\sqrt2}{6}. \]Therefore,
\[ \boxed{ \frac7{3\sqrt2} = \frac{7\sqrt2}{6} }. \]Rationalizing After Simplifying a Fraction
Consider
\[ \frac{6\sqrt2}{4\sqrt3}. \]First reduce the coefficients:
\[ \frac{3\sqrt2}{2\sqrt3}. \]Rationalize:
\[ \frac{3\sqrt2}{2\sqrt3} \cdot \frac{\sqrt3}{\sqrt3} = \frac{3\sqrt6}{6}. \]Reduce:
\[ \boxed{ \frac{\sqrt6}{2} }. \]Conjugates Also Simplify Some Numerators
Conjugates are not used only in denominators.
The product
\[ (\sqrt7+\sqrt3)(\sqrt7-\sqrt3) \]simplifies immediately to
\[ 7-3=4. \]Recognizing conjugate structure is useful throughout algebra.
Conjugates and Difference of Squares
The identity
\[ (u+v)(u-v)=u^2-v^2 \]is the central algebraic mechanism.
If
\[ v=\sqrt b, \]then
\[ v^2=b, \]so the radical disappears.
This is why conjugates are specifically suited to square-root binomials.
Example: Rationalizing with a Radical Numerator
Rationalize
\[ \frac{\sqrt3}{2+\sqrt5}. \]Multiply by the conjugate:
\[ \frac{2-\sqrt5}{2-\sqrt5}. \]The numerator becomes
\[ 2\sqrt3-\sqrt{15}. \]The denominator becomes
\[ 4-5=-1. \]Therefore,
\[ \boxed{ \sqrt{15}-2\sqrt3 }. \]Sometimes the Denominator Becomes One
Consider
\[ \frac1{\sqrt2+1}. \]Multiply by
\[ \frac{\sqrt2-1}{\sqrt2-1}. \]The denominator becomes
\[ 2-1=1. \]Therefore,
\[ \boxed{ \frac1{\sqrt2+1} = \sqrt2-1 }. \]Rationalization Preserves Value
Rationalizing a denominator does not approximate or alter the value.
It multiplies by \(1\), written in a useful form.
For example,
\[ \frac1{\sqrt2} \]and
\[ \frac{\sqrt2}{2} \]are exactly equal.
Checking a Rationalization Numerically
Since
\[ \sqrt2\approx1.414, \]we have
\[ \frac1{\sqrt2} \approx0.707. \]Also,
\[ \frac{\sqrt2}{2} \approx0.707. \]The decimal check confirms the exact algebraic equivalence.
Common Error: Multiplying Only the Denominator
To rationalize
\[ \frac1{\sqrt3}, \]it is incorrect to change only the denominator to \(3\).
Multiplication must affect numerator and denominator equally:
\[ \frac1{\sqrt3} \cdot \frac{\sqrt3}{\sqrt3}. \]Otherwise the value of the fraction changes.
Common Error: Using the Same Binomial Instead of the Conjugate
To rationalize
\[ \frac1{2+\sqrt3}, \]multiplying by
\[ \frac{2+\sqrt3}{2+\sqrt3} \]produces
\[ (2+\sqrt3)^2, \]which still contains a radical term.
The useful multiplier is the conjugate:
\[ \boxed{2-\sqrt3}. \]Common Error: Changing Both Signs in a Conjugate
The conjugate of
\[ 4+\sqrt5 \]is
\[ \boxed{4-\sqrt5}, \]not
\[ -4-\sqrt5. \]Keep the terms the same and reverse only the sign between them.
Common Error: Forgetting to Simplify the Result
Suppose rationalization produces
\[ \frac{6\sqrt3}{12}. \]The process is not complete until the numerical fraction is reduced:
\[ \boxed{\frac{\sqrt3}{2}}. \]Common Error: Using a Square-Root Method for a Cube Root
Consider
\[ \frac1{\sqrt[3]{2}}. \]Multiplying by another
\[ \sqrt[3]{2} \]gives
\[ \sqrt[3]{4} \]in the denominator, which is still irrational.
We need
\[ \sqrt[3]{4} \]as the multiplier so that the denominator becomes
\[ \sqrt[3]{8}=2. \]Common Error: Forgetting Original Domain Restrictions
Consider
\[ \frac1{\sqrt{x}-2}. \]The real square root requires
\[ x\geq0. \]The denominator also requires
\[ \sqrt{x}\neq2, \]so
\[ x\neq4. \]Rationalization preserves these restrictions.
A Reliable Procedure for a Single Square-Root Denominator
- Simplify the radical first if possible.
- Identify the radical factor in the denominator.
- Multiply numerator and denominator by that square root.
- Use \((\sqrt a)^2=a\).
- Reduce numerical factors.
- Preserve the original domain restrictions.
A Reliable Procedure for a Binomial Square-Root Denominator
- Identify the denominator binomial.
- Write its conjugate by reversing the sign between the terms.
- Multiply numerator and denominator by the conjugate.
- Use the difference-of-squares identity in the denominator.
- Distribute and simplify the numerator.
- Reduce any common numerical factors.
- Preserve original restrictions.
A Reliable Procedure for Higher-Root Denominators
- Factor the radicand if useful.
- Identify how many copies of each factor are already present.
- Determine how many additional copies are needed to form a perfect \(n\)th power.
- Multiply numerator and denominator by the corresponding \(n\)th root.
- Evaluate the perfect \(n\)th root in the denominator.
- Simplify the numerator and any numerical factors.
Example: Full Single-Radical Workflow
Rationalize
\[ \frac{5}{\sqrt{20}}. \]Simplify:
\[ \sqrt{20}=2\sqrt5. \]Then
\[ \frac5{2\sqrt5}. \]Multiply by
\[ \frac{\sqrt5}{\sqrt5}. \]We obtain
\[ \frac{5\sqrt5}{10}. \]Reduce:
\[ \boxed{ \frac{\sqrt5}{2} }. \]Example: Full Conjugate Workflow
Rationalize
\[ \frac3{4+\sqrt7}. \]Use the conjugate
\[ 4-\sqrt7. \]Then
\[ \frac3{4+\sqrt7} \cdot \frac{4-\sqrt7}{4-\sqrt7}. \]The denominator is
\[ 16-7=9. \]The numerator is
\[ 12-3\sqrt7. \]Therefore,
\[ \frac{12-3\sqrt7}{9}. \]Reduce by \(3\):
\[ \boxed{ \frac{4-\sqrt7}{3} }. \]Example: Full Higher-Root Workflow
Rationalize
\[ \frac2{\sqrt[3]{12}}. \]Factor
\[ 12=2^2\cdot3. \]To create perfect cubes, we need one more factor of \(2\) and two more factors of \(3\):
\[ 2\cdot3^2=18. \]Multiply by
\[ \frac{\sqrt[3]{18}}{\sqrt[3]{18}}. \]The denominator becomes
\[ \sqrt[3]{216}=6. \]Therefore,
\[ \frac{2\sqrt[3]{18}}{6} = \boxed{ \frac{\sqrt[3]{18}}{3} }. \]Practice 1: Single Radical
Rationalize:
\[ \frac1{\sqrt5}. \]Practice 2: Coefficient and Radical
Rationalize:
\[ \frac4{3\sqrt2}. \]Practice 3: Conjugate
Rationalize:
\[ \frac1{2+\sqrt3}. \]Practice 4: Conjugate with Subtraction
Rationalize:
\[ \frac2{3-\sqrt5}. \]Practice 5: Two Radical Terms
Rationalize:
\[ \frac1{\sqrt7+\sqrt2}. \]Practice 6: Simplify Then Rationalize
Simplify completely:
\[ \frac6{\sqrt{27}}. \]Practice 7: Cube Root
Rationalize:
\[ \frac1{\sqrt[3]{3}}. \]Practice 8: Variable Radical
Suppose \(x>0\). Rationalize:
\[ \frac2{\sqrt x}. \]Practice 9: Variable Conjugate
Suppose \(x\geq0\) and \(x\neq16\). Rationalize:
\[ \frac1{4-\sqrt x}. \]Practice 10: Identify the Error
A student rationalizes
\[ \frac1{3+\sqrt2} \]by multiplying by
\[ \frac{3+\sqrt2}{3+\sqrt2}. \]Explain why this does not accomplish the goal and give the correct multiplier.
Practice Answers
Practice 1
\[ \frac1{\sqrt5} \cdot \frac{\sqrt5}{\sqrt5} = \boxed{ \frac{\sqrt5}{5} }. \]Practice 2
\[ \frac4{3\sqrt2} \cdot \frac{\sqrt2}{\sqrt2} = \frac{4\sqrt2}{6}. \]Reduce:
\[ \boxed{ \frac{2\sqrt2}{3} }. \]Practice 3
Multiply by the conjugate:
\[ \frac{2-\sqrt3}{2-\sqrt3}. \]The denominator becomes
\[ 4-3=1. \]Therefore,
\[ \boxed{2-\sqrt3}. \]Practice 4
Multiply by
\[ \frac{3+\sqrt5}{3+\sqrt5}. \]Then
\[ \frac{2(3+\sqrt5)}{9-5} = \frac{6+2\sqrt5}{4}. \]Reduce:
\[ \boxed{ \frac{3+\sqrt5}{2} }. \]Practice 5
Use the conjugate
\[ \sqrt7-\sqrt2. \]The denominator becomes
\[ 7-2=5. \]Therefore,
\[ \boxed{ \frac{\sqrt7-\sqrt2}{5} }. \]Practice 6
Simplify:
\[ \sqrt{27}=3\sqrt3. \]Then
\[ \frac6{3\sqrt3} = \frac2{\sqrt3}. \]Rationalize:
\[ \frac2{\sqrt3} \cdot \frac{\sqrt3}{\sqrt3} = \boxed{ \frac{2\sqrt3}{3} }. \]Practice 7
Multiply by
\[ \frac{\sqrt[3]{9}}{\sqrt[3]{9}}. \]Then
\[ \frac{\sqrt[3]{9}}{\sqrt[3]{27}} = \boxed{ \frac{\sqrt[3]{9}}{3} }. \]Practice 8
\[ \frac2{\sqrt x} \cdot \frac{\sqrt x}{\sqrt x} = \boxed{ \frac{2\sqrt x}{x} }. \]Practice 9
Multiply by the conjugate:
\[ \frac{4+\sqrt x}{4+\sqrt x}. \]Therefore,
\[ \boxed{ \frac{4+\sqrt x}{16-x} }. \]The original restrictions remain
\[ x\geq0,\qquad x\neq16. \]Practice 10
Multiplying by the same binomial produces
\[ (3+\sqrt2)^2, \]which still contains a radical term.
The correct multiplier is the conjugate:
\[ \boxed{ \frac{3-\sqrt2}{3-\sqrt2} }. \]What to Remember
Rationalization rewrites an exact expression by multiplying by a useful form of \(1\).
For a single square-root denominator,
\[ \boxed{ \frac1{\sqrt a} \cdot \frac{\sqrt a}{\sqrt a} = \frac{\sqrt a}{a} } \]when the original expression is defined.
For a binomial square-root denominator, use the conjugate:
\[ \boxed{ (u+v)(u-v)=u^2-v^2 }. \]The essential habits are:
- simplify radicals before rationalizing when useful;
- multiply numerator and denominator by the same nonzero expression;
- use the matching square root for a single square-root denominator;
- use the conjugate for a square-root binomial denominator;
- use difference of squares to eliminate radical middle terms;
- for higher roots, multiply by enough factors to complete a perfect power;
- reduce the final expression completely;
- preserve every original domain restriction.
The central habit is: choose a multiplier that turns the denominator into a perfect power while multiplying the entire fraction by exactly \(1\).
Where We Go Next
Radical expressions can now be simplified and rewritten into standard exact forms. The final step in this section is to solve equations in which a variable appears inside a radical.
The next lesson, Radical Equations and Extraneous Solutions, develops methods for isolating radicals, raising both sides to appropriate powers, solving the resulting equations, and rejecting solutions that do not satisfy the original radical equation.