Divide Square Roots and
Rationalize the Denominator
Division with radicals is very similar to multiplication. If we think
about division as reducing fractions, we can reduce the coefficients
outside the radicals and reduce the radicands inside the radicals.
Example 1:
15√108
20√2
3√54
4
3√2∙3∙3∙3
4
3∙3√6
4
Reduce:
15
20
=
3
4
and
√108 √54
=
1
√2
Factor the radicand
Simplify the square root
9 √6
4
It is considered bad practice to have a radical in the denominator of the
final answer. If there is a radical in the denominator, we will rationalize
it, or clear out any radicals in the denominator.
Rationalizing the denominator is done by multiplying both the
numerator and denominator by the radical in the denominator. This will
create a pair in the denominator and, thus, simplify to a non-radical.
Modified from Beginning and Intermediate Algebra, by Tyler Wallace, CC-BY 2010. Licensed under a
Creative Commons Attribution 3.0 Unported License (http://creativecommons.org/licenses/by/3.0)
Example 2:
√6
√5
The fraction cannot be reduced nor simplified
√6
√5
∙
√5
√5
Multiply the numerator and denominator by √5
√30
√25
Simplify √25
√30
5
Example 3:
6√14
12√22
√7
2√11
√7
2√11
√77
2√121
√77
2 ∙11
Reduce:
∙
√11
√11
6
12
=
1
2
and
√14
√7
=
√22 √11
Multiply the numerator and denominator by √11
Simplify √121
Multiply
√77
22
Modified from Beginning and Intermediate Algebra, by Tyler Wallace, CC-BY 2010. Licensed under a
Creative Commons Attribution 3.0 Unported License (http://creativecommons.org/licenses/by/3.0)
Example 4:
18√6𝑥 3 𝑦 4
8√10𝑥𝑦 6
9√3𝑥 2
Reduce:
4√5𝑦 2
9𝑥√3
4𝑦 √5
9𝑥√15
4𝑦 √25
9𝑥√15
4𝑦 ∙ 5
8
=
9
4
and
√6𝑥 3 𝑦 4
√10𝑥𝑦 6
=
√3𝑥 2
√5𝑦 2
Simplify √𝑥 2 and √𝑦 2
4𝑦 √5
9𝑥√3
18
∙
√5
√5
Multiply the numerator and denominator by √5
Simplify √25
Multiply
9𝑥√15
20𝑦
If there is a binomial in the denominator, we can’t just multiply the
numerator and the denominator by the radical to clear out the radical in
the denominator. Instead we have to multiply the numerator and the
denominator by the conjugate of the denominator. Recall from the
discussion of multiplication, that whenever we multiply conjugates the
two middle terms have a sum of zero, resulting in a product containing
no square root. Thus, we need only multiply the first and last terms.
Modified from Beginning and Intermediate Algebra, by Tyler Wallace, CC-BY 2010. Licensed under a
Creative Commons Attribution 3.0 Unported License (http://creativecommons.org/licenses/by/3.0)
2
Example 5:
2
√3 − 5
√3 − 5
∙
√3 + 5
√3 + 5
2 (√3 + 5)
(√3 − 5)(√3 + 5)
2 ∙ √3 + 2 ∙ 5
√3 ∙ √3 − 5 ∙5
Multiply the numerator and denominator by the conjugate
Distribute 2 in the numerator
Multiply first and last terms in the denominator
2√3 + 10
√9 − 25
2√3 + 10
3 − 25
2√3 + 10
−22
2(√3 + 5)
−2(11)
√3 + 5
−11
Simplify √9
Subtract 3–25 in the denominator
Factor the numerator and denominator
Reduce each term by a factor of 2
Modified from Beginning and Intermediate Algebra, by Tyler Wallace, CC-BY 2010. Licensed under a
Creative Commons Attribution 3.0 Unported License (http://creativecommons.org/licenses/by/3.0)
√15
√5 + √3
Example 6:
√15
√5 + √3
∙
√5 − √3
√5 − √3
Multiply the numerator and denominator by the conjugate
√15(√5 − √3)
Distribute the √15 in the numerator
(√5 + √3)(√5 − √3)
√15 ∙ √5 − √15 ∙ √3
Multiply first and last terms in the denominator
√5 ∙ √ 5 − √ 3 ∙ √ 3
√3 ∙ 5 ∙ 5 − √3 ∙ 3 ∙ 5
√25 − √9
5√3 − 3√5
5−3
5√3 − 3√5
2
Factor the radicands in the numerator
Simplify all of the square roots
Subtract 5-3 in the denominator
Modified from Beginning and Intermediate Algebra, by Tyler Wallace, CC-BY 2010. Licensed under a
Creative Commons Attribution 3.0 Unported License (http://creativecommons.org/licenses/by/3.0)