Chapter 6.5, Problem 43PPS

To determine

To Simplify:

  $\frac{9-2\sqrt{3}}{\sqrt{3}+6}$

Answer to Problem 43PPS

  $\frac{20-7\sqrt{3}}{11}$

Explanation of Solution

Given Information:

Given expression to simplify:

  $\frac{9-2\sqrt{3}}{\sqrt{3}+6}$

Concept Used :

• Definition of nth root:
• Words : For any real numbers a and b , and any positive integer n , if $a^n=b$ then a is the nth root of b . That is $a=\sqrt[n]{b}$
• $\begin{array}{l}\sqrt[n]{b}\\ Here\\ b=radicand\\ n=index\\ \sqrt{}=radicalsign\end{array}$
• Product Property of Radicals: $\begin{array}{l}\text{For any real number a and b and any integer n > 1 , }\\ \sqrt[n]{ab}=\sqrt[n]{a}\cdot \sqrt[n]{b}\text{ , if n is even and a \& b are both nonnegative }\\ \text{or if n is odd}\text{.}\end{array}$
• Quotient Property of Radicals:

$\begin{array}{l}\text{For any real numbers a and b}\ne \text{0 and any integer}\\ \text{n >1,}\\ \sqrt[n]{\frac{a}{b}}=\frac{\sqrt[n]{a}}{\sqrt[n]{b}},if\text{ all roots are defined}\text{.}\end{array}$

• Rationalizing the denominator :

To eliminate radicals from a denominator or fractions from a radicand , use this process.

To rationalize a denominator, multiply the numerator and denominator by a quantity so that the radicand has an exact root.

• Like radical expressions : If the radicals have both the index and radicand identical , then they are like radicals.
• In order to add or subtract , the radicals must be like radicals.

Calculation:

• Given $\frac{9-2\sqrt{3}}{\sqrt{3}+6}$
• $\begin{array}{l}\frac{9-2\sqrt{3}}{\sqrt{3}+6}\\ =\frac{9-2\sqrt{3}}{\sqrt{3}+6}\cdot \frac{\sqrt{3}-6}{\sqrt{3}-6}\mathrm{....................}[\text{Rationalizing the denominator]}\\ \text{=}\frac{(9-2\sqrt{3})(\sqrt{3}-6)}{(\sqrt{3}+6)(\sqrt{3}-6)}\\ =\frac{9\sqrt{3}-54-2\sqrt{3}\cdot \sqrt{3}+12\sqrt{3}}{{(\sqrt{3})}^2-6^2}\mathrm{...}[{\text{Distributive Property \& (a+b)(a-b)=a}}^2-b^2]\\ =\frac{9\sqrt{3}-54-6+12\sqrt{3}}{3-36}\\ =\frac{(9+12)\sqrt{3}-60}{-33}\mathrm{.....................}[\text{Addition of like radicals]}\\ =\frac{21\sqrt{3}-60}{-33}\cdot \frac{-1}{-1}\mathrm{..............}[\text{Multiply and divide by -1]}\\ \text{=}\frac{60-21\sqrt{3}}{33}=\frac{3(20-7\sqrt{3})}{3(11)}\\ =\frac{20-7\sqrt{3}}{11}\end{array}$
• So, the solution is $\frac{20-7\sqrt{3}}{11}$