Chapter 6.5, Problem 44PPS

To determine

To Simplify:

  $\frac{2\sqrt{2}+2\sqrt{5}}{\sqrt{5}+\sqrt{2}}$

Answer to Problem 44PPS

2

Explanation of Solution

Given Information:

Given expression to simplify:

  $\frac{2\sqrt{2}+2\sqrt{5}}{\sqrt{5}+\sqrt{2}}$

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 $\frac{2\sqrt{2}+2\sqrt{5}}{\sqrt{5}+\sqrt{2}}$
• $\begin{array}{l}\frac{2\sqrt{2}+2\sqrt{5}}{\sqrt{5}+\sqrt{2}}\\ =\frac{2\sqrt{2}+2\sqrt{5}}{\sqrt{5}+\sqrt{2}}\cdot \frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}-\sqrt{2}}\mathrm{.................................}\text{[Rationalizing the denominator]}\\ \text{=}\frac{(2\sqrt{2}+2\sqrt{5})(\sqrt{5}-\sqrt{2})}{(\sqrt{5}+\sqrt{2})(\sqrt{5}-\sqrt{2})}\\ =\frac{2\sqrt{2}\cdot \sqrt{5}-2\sqrt{2}\cdot \sqrt{2}+2\sqrt{5}\cdot \sqrt{5}-2\sqrt{5}\cdot \sqrt{2}}{{(\sqrt{5})}^2-{(\sqrt{2})}^2}\mathrm{..}{\text{[Distributive property \& (a+b)(a-b)=a}}^2-b^2]\\ =\frac{2\sqrt{10}-4+10-2\sqrt{10}}{5-2}\\ =\frac{(2-2)\sqrt{10}+6}{3}\mathrm{..........................................}[\text{Subtracting like radicals]}\\ \text{=}\frac{6}{3}=2\end{array}$
• So, the solution is 2