# Multiplication of radicals using the distributive property

Learn how to perform multiplication of radicals with some carefully chosen examples.

Example #1

 $$(\sqrt{3} + \sqrt{2}) × (\sqrt{3} - \sqrt{2})$$

Here is how to use the distributive property to do this multiplication.

 $$(\sqrt{3} + \sqrt{2}) × (\sqrt{3} - \sqrt{2})$$
 $$= \sqrt{3}×\sqrt{3} - \sqrt{3}×\sqrt{2} + \sqrt{2}×\sqrt{3} - \sqrt{2}×\sqrt{2}$$
 $$= \sqrt{3}×\sqrt{3} - \sqrt{3}×\sqrt{2} + \sqrt{2}×\sqrt{3}$$
 $$- \sqrt{2}×\sqrt{2}$$
 $$= \sqrt{3 × 3} - \sqrt{3 × 2} + \sqrt{2 × 3} - \sqrt{2 × 2}$$
 $$= \sqrt{9} - \sqrt{6} + \sqrt{6} - \sqrt{4}$$
 $$= \sqrt{9} - \sqrt{4}$$

= 3 - 2

=  1

Did you notice that the multiplication has the format (a + b) x (a - b)?

(a + b) x (a - b) = a2 - b2

Therefore, next time, there is no need to do all this math above.

 $$(\sqrt{3} + \sqrt{2}) × (\sqrt{3} - \sqrt{2}) = (\sqrt{3})^2 - (\sqrt{2})^2$$
 $$Notice \ also \ that (\sqrt{3})^2 = 3$$
 $$In \ general,\sqrt{a} ×\sqrt{a} = (\sqrt{a})^2 = a$$

Example #2

 $$(\sqrt{7} + \sqrt{5}) × (\sqrt{7} - \sqrt{5})$$
 $$= (\sqrt{7})^2 - (\sqrt{5})^2 = 7 - 5 = 2$$

Example #3

 $$(\sqrt{6} + \sqrt{9}) × (\sqrt{6} - \sqrt{9}) = 6 - 9 = -3$$

Example #4

 $$(\sqrt{4} - 2\sqrt{12}) × (\sqrt{4} + \sqrt{12})$$
 $$= \sqrt{4} × \sqrt{4} + \sqrt{4} × \sqrt{12} -2\sqrt{12} ×\sqrt{4} -2 \sqrt{12} × \sqrt{12}$$
 $$= (\sqrt{4} × \sqrt{4} + \sqrt{4} × \sqrt{12} -2\sqrt{12} ×\sqrt{4}$$
 $$-2 \sqrt{12} × \sqrt{12}$$
 $$= 4 + 2 × \sqrt{12} -2\sqrt{12} ×2 -2 × 12$$
 $$= 4 + 2\sqrt{12} -4\sqrt{12} -24$$
 $$= -2\sqrt{12} - 20$$

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