# Rationalizing a Denominator

How to rationalize the radicals in a denominator: examples and their solutions.

## Example 1: Simplify √5*xy*/*x*^{2}

Reduce *x*^{2} in the numerator to *x*

and cancel *x* in the denominator.

Split the radical sign.

Divide radicals

The denominator is √*x*,

which is an irrational number.

So this cannot be the answer.

To rationalize the denominator,

multiply (√*x*)/(√*x*).

√5*y*⋅√*x* = √5*xy*

Multiply radicals

√*x*⋅√*x* = *x*

So (given) = √5*xy*/*x*.

The denominator is rationalized.

So this is the answer.

## Example 2: Simplify √0.2

0.2 = 2/10

So √0.2 = √2/10.

Cancel 2

and reduce 10 to 5.

Then (given) = √1/5.

The numerator is 1.

So the radical sign only affects the denominator.

So (given) = 1/√5.

The denominator is √5,

which is an irrational number.

So, to rationalize the denominator,

multiply (√5)/(√5).

√5⋅√5 = 5

So (given) = √5/5.

## Conjugate

The conjugate of [*a* + *b*] is [*a* - *b*].

The conjugate of [*a* - *b*] is [*a* + *b*].

So [*a* + *b*] and [*a* - *b*]

are the conjugates of each other.

Recall that (*a* + *b*)(*a* - *b*) = *a*^{2} - *b*^{2}.

Product of a sum and a difference (*a* + *b*)(*a* - *b*)

The product of a binomial and its conjugate

makes *a*^{2} - *b*^{2}.

By using this property,

you can rationalize a binomial denominator.

## Example 3: Simplify 1/(4 + √3)

The denominator is [4 + √3],

which is an irrational binomial.

The conjugate of [4 + √3] is [4 - √3].

So, to rationalize the denominator,

multiply (4 - √3)/(4 - √3).

(4 + √3)(4 - √3) = 4^{2} - (√3)^{2}

Product of a sum and a difference (*a* + *b*)(*a* - *b*)

4^{2} = 16

(√3)^{2} = 3

Square root - Square of a square root

16 - 3 = 13

So (given) = (4 - √3)/13.

## Example 4: Simplify √2/(5 - √6)

The denominator is [5 - √6],

which is an irrational binomial.

The conjugate of [5 - √6] is [5 + √6].

So, to rationalize the denominator,

multiply (5 + √6)/(5 + √6).

√2⋅(5 + √6) = 5√2 + √2⋅√6

(5 - √6)(5 + √6) = 5^{2} - (√6)^{2}

√2⋅√6 = √2⋅√2⋅√3

5^{2} = 25

(√6)^{2} = 6

√2⋅√2⋅√3 = 2√3

25 - 6 = 19

So (given) = (5√2 + 2√3)/19.

## Example 5: Simplify (*x* - 4)/(√*x* - 2)

The denominator is [√*x* - 2],

which is an irrational binomial.

The conjugate of [√*x* - 2] is [√*x* + 2].

So, to rationalize the denominator,

multiply (√*x* + 2)/(√*x* + 2).

Leave the numerator.

You might use (*x* - 4).

(√*x* - 2)(√*x* + 2) = (√*x*)^{2} - 2^{2}.

(√*x*)^{2} = *x*

2^{2} = 4

Cancel (*x* - 4).

Then (given) = √*x* + 2.

Simplifying rational expressions