Q. 175 - Intermediate Algebra

Question

In the following exercises, simplify.

a.$\sqrt{48}+\sqrt{27}$

b.$\sqrt[3]{54}+\sqrt[3]{128}$

c.$6\sqrt[4]{5}-\frac{3}{2}\sqrt[4]{320}$

Step-by-Step Solution

Answer

On simplification, we get $-\sqrt{2},5\sqrt[3]{3},-\sqrt[4]{5}$ respectively

Part (a) Step 1: Given information

Given the expression $\sqrt{72}-\sqrt{98}$

Part (a) Step 2: Simplifying the radical and combine like terms

Simplifying radical and combining we get

$$\begin{gathered} \sqrt{72}-\sqrt{98}=\sqrt{36·2}-\sqrt{49·2} \\ \sqrt{72}-\sqrt{98}=\sqrt{36}·\sqrt{2}-\sqrt{49}·\sqrt{2} \\ \sqrt{72}-\sqrt{98}=6\sqrt{2}-7\sqrt{2} \\ \sqrt{72}-\sqrt{98}=-\sqrt{2} \end{gathered}$$

Part (b) Step 1: Given information

Given expression $\sqrt[3]{54}+\sqrt[3]{128}$

Part (b) Step 2: Simplifying the radical and combine like terms

Simplifying radical and combining we get

$$\begin{gathered} \sqrt[3]{24}+\sqrt[3]{81}=\sqrt[3]{8·3}+\sqrt[3]{27·3} \\ \sqrt[3]{24}+\sqrt[3]{81}=\sqrt[3]{8}·\sqrt[3]{3}+\sqrt[3]{27}·\sqrt[3]{3} \\ \sqrt[3]{24}+\sqrt[3]{81}=2\sqrt[3]{3}+3\sqrt[3]{3} \\ \sqrt[3]{24}+\sqrt[3]{81}=2\sqrt[3]{3}+3\sqrt[3]{3} \\ \sqrt[3]{24}+\sqrt[3]{81}=5\sqrt[3]{3} \end{gathered}$$

Part (c) Step 1: Given information

Given expression $6\sqrt[4]{5}-\frac{3}{2}\sqrt[4]{320}$

Part (c) Step 2: Simplifying radical and combining

Simplifying, we get

$$\begin{gathered} \frac{1}{2}\sqrt[4]{80}-\frac{2}{3}\sqrt[4]{405}=\frac{1}{2}\sqrt[4]{16·5}-\frac{2}{3}\sqrt[4]{81·5} \\ \frac{1}{2}\sqrt[4]{80}-\frac{2}{3}\sqrt[4]{405}=\frac{1}{2}\sqrt[4]{16}·\sqrt[4]{5}-\frac{2}{3}\sqrt[4]{81}·\sqrt[4]{5} \\ \frac{1}{2}\sqrt[4]{80}-\frac{2}{3}\sqrt[4]{405}=\frac{1}{2}·2·\sqrt[4]{5}-\frac{2}{3}·3·\sqrt[4]{5} \\ \frac{1}{2}\sqrt[4]{80}-\frac{2}{3}\sqrt[4]{405}=\frac{2}{2}·\sqrt[4]{5}-\frac{6}{3}·\sqrt[4]{5} \\ \frac{1}{2}\sqrt[4]{80}-\frac{2}{3}\sqrt[4]{405}=\sqrt[4]{5}-2\sqrt[4]{5} \\ \frac{1}{2}\sqrt[4]{80}-\frac{2}{3}\sqrt[4]{405}=-\sqrt[4]{5} \end{gathered}$$