01Fmultiplyingsurds

= Multiplying & Dividing Surds = toc

Multiplying Algebra
Recall that in algebra, we can multiply __any__ terms together. math \\ \textbf{(a)} \quad 3x \times 5y = 15xy \\ \\ \textbf{(b)} \quad 4a \times 2a = 8a^2 math

Notice that the coefficients get multiplied together normally, and then the pronumerals get multiplied by either pushing them together or producing a squared term.

Multiplying Surds
In the same way, __any__ surds can be multiplied together.


 * 1) multiply the coefficients of the surds together (the numbers in front of the surd)
 * 2) multiply the contents of the root sign together (the numbers inside the surds)
 * 3) simplify if possible

** Example 1 **
math \\ \textbf{(a)} \quad 3\,\sqrt{7} \; \times \; 4\,\sqrt{2} \; = \; 12\,\sqrt{14} \\ \\ \\ \textbf{(b)} \quad 6\,\sqrt{10} \; \times \; \sqrt{10} \;=\; 6\,\sqrt{100} \\ . \qquad \qquad \qquad \qquad \;\; = \; 6 \times 10 \\ . \qquad \qquad \qquad \qquad \;\; = \; 60 math

Dividing Surds
The same rule applies to dividing surds. Treat the coefficients like a fraction and simplify the fraction if possible. Then treat the insides of the surds like a fraction and simplify the fraction if possible.

** Example 2 **
math \textbf{(a)} \quad \dfrac{10\,\sqrt{18}}{2\,\sqrt{3}} = 5\,\sqrt{6} math

math \\ \textbf{(b)} \quad \dfrac{\sqrt{20}}{2} = \dfrac{\sqrt{4} \times \sqrt{5}}{2} \\ \\ . \qquad \qquad = \dfrac{2\,\sqrt{5}}{2} \\ \\ . \qquad \qquad = \sqrt{5} math

math \\ \textbf{(c)} \quad \dfrac{16\,\sqrt{15}}{24\,\sqrt{125}} = \dfrac{2\,\sqrt{3}}{3\,\sqrt{25}} \qquad \textit{cancel inside by 5 and outside by 8} \\ \\ . \qquad \qquad \quad = \dfrac{2\,\sqrt{3}}{3 \times 5} \\ \\ . \qquad \qquad \quad = \dfrac{2\,\sqrt{3}}{15} math

Fractions
To find the square root of a mixed number, first change it into an improper fraction

** Example 3 **
math \sqrt{3\frac{1}{2}} = \sqrt{\dfrac{7}{2}} = \dfrac{\sqrt{7}}{\sqrt{2}} math

Expanding Brackets
Recall, that in algebra, expanding brackets means to multiply the term in front of the bracket by each of the terms inside. math \textbf{(a)} \quad 3 \big( x + 2y \big) = 3x + 6y math

and to expand a pair of brackets we use **FOIL** (First Outside Inside Last) to multiply each of the terms in the first bracket by each of the terms in the second bracket math \\ \textbf{(b)} \quad \big( x + 2 \big)\big( x - 4 \big) = x^2 -4x + 2x - 8 \qquad \textit{then simplify by collecting like terms} \\ \\ . \qquad \qquad \qquad \qquad = x^2 -2x - 8 math

Exactly the same rules apply to expanding brackets with surds

** Example 4 **
math \textbf{(a)} \quad \sqrt{2}\, \big( 3\,\sqrt{5} \; + 7 \big) = 3\,\sqrt{10} \;+\; 7\,\sqrt{2} math

math \textbf{(b)} \quad \big( \sqrt{5} \; - 2 \big)\big( \sqrt{3} \;+ 1 \big) = \sqrt{15} + \sqrt{5} - 2 \;\sqrt{3} \; - 2 math

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