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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.

Example 1: 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.

Example 2: 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} Both terms are like radicals (same radicand 3). Combine the coefficients: 23+63=832\sqrt{3} + 6\sqrt{3} = 8\sqrt{3}

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

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Updated 2026-06-26

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