Q. 219 - Intermediate Algebra

Question

Simplify: 

$\frac{5}{6}\sqrt{162}+\frac{3}{16}\sqrt{128}$

Step-by-Step Solution

Answer

The value of the given expression is $9\sqrt{2}.$

Step 1. Given Information

The given expression is 

$\frac{5}{6}\sqrt{162}+\frac{3}{16}\sqrt{128}.$

Step 2. Simplify the radicals

Like radicals can be added or subtracted so,

$$\begin{gathered} \frac{5}{6}\sqrt{162}+\frac{3}{16}\sqrt{128} \\ =\left(\frac{5}{6}·\sqrt{81}·\sqrt{2}\right)+\left(\frac{3}{16}·\sqrt{64}·\sqrt{2}\right) \\ =\left(\frac{5}{6}·9·\sqrt{2}\right)+\left(\frac{3}{16}·8·\sqrt{2}\right) \\ =\left(\frac{15}{2}·\sqrt{2}\right)+\left(\frac{3}{2}·\sqrt{2}\right) \\ =9\sqrt{2} \end{gathered}$$