# MULTIPLICATION OF SURDS

Like common numbers, surds can be multiplied together. The rule is that you multiply whole numbers to whole numbers and multiply surds to surds as illustrated below.

$a\sqrt{b}\times c\sqrt{d}=ac\sqrt{bd}$

This is illustrated below:

πThis post extends our series on surdsπ

Example 1
Simplify $12\sqrt{7}\times3\sqrt{6}$

Solution
$(12\times3)(\sqrt{7\times6})$
$36\sqrt{42}$

Example 2
Evaluate $\sqrt{8}\times\sqrt{15}\times\sqrt{135}\times\sqrt{50}$

Solution
This can be written as
$\sqrt{8\times135\times15\times50}$
$\sqrt{(8\times50)(135\times15)}$
$\sqrt{(400)(2025)}$
$(\sqrt{400})(\sqrt{2025})$
$(20)(45)=900$

Surds in the bracket can be multiplied easily by using the normal algebraic expansion, which is
$(a+\sqrt{b})(c+\sqrt{d})= a(c+\sqrt{d})+\sqrt{b}(c+\sqrt{d})$
$ac+a\sqrt{d}+c\sqrt{b}+\sqrt{bd}$

Example 3
Evaluate $(3\sqrt{2}+1)(\sqrt{2}+3)$

Solution
By expansion
$3\sqrt{2}(\sqrt{2}+3)+1(\sqrt{2}+3)$
$6+9\sqrt{2}+\sqrt{2}+3$
$6+3+10\sqrt{2}$
$9+10\sqrt{2}$

Example 4
Evaluate $(2+\sqrt{5})(5-\sqrt{20})$

Solution
$\sqrt{20}$ can be simplified further, hence
$(2+\sqrt{5})(5-\sqrt{4\times5})$
$(2+\sqrt{5})(5-\sqrt{4}\times\sqrt{5})$
$(2+\sqrt{5})(5-2\sqrt{5})$

By expansion,
$2(5-2\sqrt{5})+\sqrt{5}(5-2\sqrt{5})$
$10-4\sqrt{5}+5\sqrt{5}-2(5)$
$10-4\sqrt{5}+5\sqrt{5}-10$
$10+\sqrt{5}-10$
$\sqrt{5}$

Example 5
Evaluate $(\sqrt{75}-\sqrt{48})^2$

Solution
$(\sqrt{75}-\sqrt{48})(\sqrt{75}-\sqrt{48})$

Now, let further simplify
$(\sqrt{25\times3}-\sqrt{16\times3})(\sqrt{25\times3}-\sqrt{16\times3})$
$(\sqrt{25}\times\sqrt{3}-\sqrt{16}\times\sqrt{3})(\sqrt{25}\times\sqrt{3}-\sqrt{16}\times\sqrt{3})$
$(5\sqrt{3}-4\sqrt{3})(5\sqrt{3}-4\sqrt{3})$

expanding the surds expression
$5\sqrt{3}(5\sqrt{3}-4\sqrt{3})-4\sqrt{3}((5\sqrt{3}-4\sqrt{3})$
$25(\sqrt{9})-20(\sqrt{9})-20(\sqrt{9})+16(\sqrt{9})$
$25(3)-20(3)-20(3)+16(3)$
$75-60-60+48=3$

Example 6
Evaluate $(1+\sqrt{2})^3$

Solution
$(1+\sqrt{2})^3=(1+\sqrt{2})(1+\sqrt{2})^2$
$(1+\sqrt{2})((1+\sqrt{2})(1+\sqrt{2}))$
$(1+\sqrt{2})(1(1+\sqrt{2})+\sqrt{2}(1+\sqrt{2})$
$(1+\sqrt{2})(1+\sqrt{2}+\sqrt{2}+2)$
$(1+\sqrt{2})(3+2\sqrt{2})$

Expanding further
$1(3+2\sqrt{2})+\sqrt{2}((3+2\sqrt{2})$
$3+2\sqrt{2}+3\sqrt{2}+4$
$7+5\sqrt{2}$

Related post
Example 7
Evaluate $(2\sqrt{2}+\sqrt{5})(2\sqrt{2}-\sqrt{5})$

Solution
This is a difference of two squares, hence
$(2\sqrt{2})^2-(\sqrt{5})^2$
$(2)^2(\sqrt{2}^2)-(\sqrt{5})^2$

Remember, Square root cancels square.
$4(2)-5$
$8-5=3$

See you in the comment box π.
Help us grow our readership by sharing this post