Learn Before
Example

Simplifying 5(7+25)\sqrt{5}(7 + 2\sqrt{5}) and 6(2+18)\sqrt{6}(\sqrt{2} + \sqrt{18})

Distribute a square root across a binomial where the resulting products may simplify to integers or yield like radicals that combine.

5(7+25)\sqrt{5}(7 + 2\sqrt{5}):

Distribute 5\sqrt{5} to each term:

57+525=75+255\sqrt{5} \cdot 7 + \sqrt{5} \cdot 2\sqrt{5} = 7\sqrt{5} + 2 \cdot \sqrt{5} \cdot \sqrt{5}

Since 55=5\sqrt{5} \cdot \sqrt{5} = 5, the second term becomes 25=102 \cdot 5 = 10:

75+10=10+757\sqrt{5} + 10 = 10 + 7\sqrt{5}

The result is 10+7510 + 7\sqrt{5}. When a square root is multiplied by itself, the Product Property yields the radicand as an integer.

6(2+18)\sqrt{6}(\sqrt{2} + \sqrt{18}):

Distribute 6\sqrt{6}:

62+618=12+108\sqrt{6} \cdot \sqrt{2} + \sqrt{6} \cdot \sqrt{18} = \sqrt{12} + \sqrt{108}

Simplify each radical by extracting perfect square factors: 12=43=23\sqrt{12} = \sqrt{4} \cdot \sqrt{3} = 2\sqrt{3} and 108=363=63\sqrt{108} = \sqrt{36} \cdot \sqrt{3} = 6\sqrt{3}:

23+632\sqrt{3} + 6\sqrt{3}

Both terms are like radicals (same radicand 33). Combine the coefficients: 2+6=82 + 6 = 8:

6(2+18)=83\sqrt{6}(\sqrt{2} + \sqrt{18}) = 8\sqrt{3}

In part ⓐ, distributing a square root times itself produces an integer. In part ⓑ, distributing produces two radicals that — after simplification — share the same radicand and combine into a single term.

0

1

Updated 2026-04-21

Contributors are:

Who are from:

Tags

OpenStax

Elementary Algebra @ OpenStax

Ch.9 Roots and Radicals - Elementary Algebra @ OpenStax

Algebra

Math

Prealgebra

Related
Learn After