9.4 Multiply square roots  (Page 13/3)

Solution

1. ⓐ
   $\begin{array}{cccc}& & & \sqrt{2}·\sqrt{6}\hfill \\ \\ \\ \text{Multiply using the Product Property.}\hfill & & & \sqrt{12}\hfill \\ \\ \\ \text{Simplify the radical.}\hfill & & & \sqrt{4}·\sqrt{3}\hfill \\ \\ \\ \text{Simplify.}\hfill & & & 2\sqrt{3}\hfill \end{array}$
   
   
2. ⓑ
   $\begin{array}{cccc}& & & \left(4\sqrt{3}\right)\left(2\sqrt{12}\right)\hfill \\ \\ \\ \text{Multiply using the Product Property.}\hfill & & & 8\sqrt{36}\hfill \\ \\ \\ \text{Simplify the radical.}\hfill & & & 8·6\hfill \\ \\ \\ \text{Simplify.}\hfill & & & 48\hfill \end{array}$

Notice that in (b) we multiplied the coefficients and multiplied the radicals. Also, we did not simplify $\sqrt{12}$ . We waited to get the product and then simplified.

Solution

$\begin{array}{cccc}& & & \left(6\sqrt{2}\right)\left(3\sqrt{10}\right)\hfill \\ \\ \\ \text{Multiply using the Product Property.}\hfill & & & 18\sqrt{20}\hfill \\ \\ \\ \text{Simplify the radical.}\hfill & & & 18\sqrt{4}·\sqrt{5}\hfill \\ \\ \\ \text{Simplify.}\hfill & & & 18·2·\sqrt{5}\hfill \\ & & & 36\sqrt{5}\hfill \end{array}$

Solution

1. ⓐ
   $\begin{array}{cccc}& & & \left(\sqrt{8x^3}\right)\left(\sqrt{3x}\right)\hfill \\ \\ \\ \text{Multiply using the Product Property.}\hfill & & & \sqrt{24x^4}\hfill \\ \\ \\ \text{Simplify the radical.}\hfill & & & \sqrt{4x^4}·\sqrt{6}\hfill \\ \\ \\ \text{Simplify.}\hfill & & & 2x^2\sqrt{6}\hfill \end{array}$
   
   
2. ⓑ
   $\begin{array}{cccc}& & & \left(\sqrt{20y^2}\right)\left(\sqrt{5y^3}\right)\hfill \\ \\ \\ \text{Multiply using the Product Property.}\hfill & & & \sqrt{100y^5}\hfill \\ \\ \\ \text{Simplify the radical.}\hfill & & & 10y^2\sqrt{y}\hfill \end{array}$

Solution

$\begin{array}{cccc}& & & \left(10\sqrt{6p^3}\right)\left(3\sqrt{18p}\right)\hfill \\ \\ \\ \text{Multiply.}\hfill & & & 30\sqrt{108p^4}\hfill \\ \\ \\ \text{Simplify the radical.}\hfill & & & 30\sqrt{36p^4}·\sqrt{3}\hfill \\ \\ \\ & & & 30·6p^2·\sqrt{3}\hfill \\ \\ \\ & & & 180p^2\sqrt{3}\hfill \end{array}$

Solution

ⓐ
$\begin{array}{cccc}& & & {\left(\sqrt{2}\right)}^2\hfill \\ \\ \\ \text{Rewrite as a product.}\hfill & & & \left(\sqrt{2}\right)\left(\sqrt{2}\right)\hfill \\ \\ \\ \text{Multiply.}\hfill & & & \sqrt{4}\hfill \\ \\ \\ \text{Simplify.}\hfill & & & 2\hfill \end{array}$

ⓑ
$\begin{array}{cccc}& & & {\left(\text{−}\sqrt{11}\right)}^2\hfill \\ \\ \\ \text{Rewrite as a product.}\hfill & & & \left(\text{−}\sqrt{11}\right)\left(\text{−}\sqrt{11}\right)\hfill \\ \\ \\ \text{Multiply.}\hfill & & & \sqrt{121}\hfill \\ \\ \\ \text{Simplify.}\hfill & & & 11\hfill \end{array}$

Solution

ⓐ
$\begin{array}{cccc}& & & \left(2\sqrt{3}\right)\left(8\sqrt{3}\right)\hfill \\ \\ \\ \text{Multiply. Remember,}{\left(\sqrt{3}\right)}^2=3.\hfill & & & 16·3\hfill \\ \\ \\ \text{Simplify.}\hfill & & & 48\hfill \end{array}$

ⓑ
$\begin{array}{cccc}& & & {\left(3\sqrt{6}\right)}^2\hfill \\ \\ \\ \text{Multiply.}\hfill & & & 9·6\hfill \\ \\ \\ \text{Simplify.}\hfill & & & 54\hfill \end{array}$