Chapter 7.5, Problem 41PPS

To determine

To Simplify:

  $\frac{6}{\sqrt{3}-\sqrt{2}}$

Answer to Problem 41PPS

  $6\sqrt{3}+6\sqrt{2}$

Explanation of Solution

Given Information:

Given expression to simplify:

  $\frac{6}{\sqrt{3}-\sqrt{2}}$

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 $\frac{6}{\sqrt{3}-\sqrt{2}}$
• $\begin{array}{l}\frac{6}{\sqrt{3}-\sqrt{2}}\\ =\frac{6}{\sqrt{3}-\sqrt{2}}\cdot \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}\mathrm{...............}\text{[Rationalizing the denominator]}\\ \text{=}\frac{6(\sqrt{3}+\sqrt{2})}{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})}\\ =\frac{6\sqrt{3}+6\sqrt{2}}{{(\sqrt{3})}^2-{(\sqrt{2})}^2}\mathrm{...................}[(a+b)(a-b)=a^2-b^2]\\ =\frac{6\sqrt{3}+6\sqrt{2}}{3-2}\\ =6\sqrt{3}+6\sqrt{2}\end{array}$
• So, the solution is $6\sqrt{3}+6\sqrt{2}$