Question

Simplify:

$\left(a\right)2+\sqrt{98}\left(b\right)\frac{6-\sqrt{45}}{3}.$

Short answer

Part a. After simplifying we get,$2+7\sqrt{2}.$

Part b. After simplifying we get,$2-\sqrt{5}.$

Step-by-step solution

Part a Step 1. Rewrite the radicand as a product using the largest perfect square factor 

Accordingly, we can write as

$$\begin{gathered} 2+\sqrt{98} \\ =2+\sqrt{49·2} \end{gathered}$$

Part a Step 2. Rewrite the radical as the product of two radicals

Thus,$$\begin{gathered} 2+\sqrt{49·2} \\ 2+\sqrt{49}·\sqrt{2} \end{gathered}$$

Part a Step 3. Simplify

After simplifying we get,

$2+7\sqrt{2}.$

Part b Step 1. Rewrite the radicand as a product using the largest perfect square factor 

Accordingly, we can write as

$$\begin{gathered} \frac{6-\sqrt{45}}{3} \\ =\frac{6-\sqrt{9·5}}{3} \end{gathered}$$

Part b Step 2. Rewrite the radical as the product of two radicals 

Thus,$$\begin{gathered} \frac{6-\sqrt{9·5}}{3} \\ \frac{6-\sqrt{9}·\sqrt{5}}{3} \end{gathered}$$

Part b Step 3. Simplify 

After simplifying we get,

$$\begin{gathered} \frac{6-\sqrt{9}·\sqrt{5}}{3} \\ =\frac{6-3\sqrt{5}}{3} \\ =\frac{3(2-\sqrt{5})}{3} \\ =2-\sqrt{5} \end{gathered}$$