Chapter 6.5, Problem 51PPS

To determine

To Simplify:

  $\frac{x+1}{\sqrt{x}-1}$

Answer to Problem 51PPS

  $\frac{(x+1)\sqrt{x}+x+1}{x-1}\text{ or }\frac{(x+1)(\sqrt{x}+1)}{x-1}$

Explanation of Solution

Given Information:

Given expression to simplify:

  $\frac{x+1}{\sqrt{x}-1}$

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: $\begin{array}{l}\text{For any real number a and b and any integer n > 1 , }\\ \sqrt[n]{ab}=\sqrt[n]{a}\cdot \sqrt[n]{b}\text{ , if n is even and a \& b are both nonnegative }\\ \text{or if n is odd}\text{.}\end{array}$
• 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 $\frac{x+1}{\sqrt{x}-1}$
• $\begin{array}{l}\frac{x+1}{\sqrt{x}-1}\\ =\frac{x+1}{\sqrt{x}-1}\cdot \frac{\sqrt{x}+1}{\sqrt{x}+1}\mathrm{.................}[\text{Rationalizing the denominator]}\\ \text{=}\frac{(x+1)(\sqrt{x}+1)}{(\sqrt{x}-1)(\sqrt{x}+1)}\\ =\frac{x\sqrt{x}+x+\sqrt{x}+1}{{(\sqrt{x})}^2-1^2}\mathrm{............}[{\text{Distributive Property \& (a+b)(a-b)=a}}^2-b^2]\\ =\frac{(x+1)\sqrt{x}+x+1}{x-1}\mathrm{.............}[\text{Adding like radicals]}\\ \text{=}\frac{(x+1)\sqrt{x}+x+1}{x-1}\text{ or }\frac{(x+1)(\sqrt{x}+1)}{x-1}\end{array}$
• So, $\frac{(x+1)\sqrt{x}+x+1}{x-1}\text{ or }\frac{(x+1)(\sqrt{x}+1)}{x-1}$ is the solution.