Q. 207 - Intermediate Algebra

Question

In the following exercises, multiply:

$$\begin{gathered} \left(a\right) {\left(9-\sqrt{6}\right)}^2 \\ \left(b\right) {\left(10+3\sqrt{7}\right)}^2 \end{gathered}$$

Step-by-Step Solution

Answer

Part a. By multiplying we get, $87-18\sqrt{6}.$

Part b. By multiplying we get, $163+60\sqrt{7}.$

Part a Step 1. Multiply the given expression

The given expression is ${\left(9-\sqrt{6}\right)}^2.$

Multiply, using the Product of Binomial Squares Pattern

$$\begin{gathered} {\left(9-\sqrt{6}\right)}^2 \\ =9^2-2·9·\sqrt{6}+{\left(\sqrt{6}\right)}^2 \end{gathered}$$

Part a Step 2. Simplify

By simplifying we get,

$$\begin{gathered} =9^2-2·9·\sqrt{6}+{\left(\sqrt{6}\right)}^2 \\ =81-18\sqrt{6}+6 \\ =87-18\sqrt{6} \left[combine like term\right] \end{gathered}$$

Part b Step 1. Multiply the given expression

The given expression is ${\left(10+3\sqrt{7}\right)}^2.$

Multiply, using the Product of Binomial Squares Pattern

$$\begin{gathered} {\left(10+3\sqrt{7}\right)}^2 \\ ={10}^2+2·10·3\sqrt{7}+{\left(3\sqrt{7}\right)}^2 \end{gathered}$$

Part b Step 2. Simplify

By simplifying we get,

$$\begin{gathered} ={10}^2+2·10·3\sqrt{7}+{\left(3\sqrt{7}\right)}^2 \\ =100+60\sqrt{7}+63 \\ =163+60\sqrt{7} \left[combine like term\right] \end{gathered}$$