Chapter 7.5, Problem 4ACYP

To determine

To Simplify:

  $4\sqrt{8}+3\sqrt{50}$

Answer to Problem 4ACYP

  $\text{23}\sqrt{2}$

Explanation of Solution

Given Information:

Given expression to simplify:

  $4\sqrt{8}+3\sqrt{50}$

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 $a=\sqrt[n]{b}$
• Example : Because ${(-3)}^4=81$ , -3 is a fourth root of 81 and 3 is a principal root. That is $\sqrt[4]{81}=\pm 3$ .
• $\begin{array}{l}\sqrt[n]{b}\\ Here\\ b=radicand\\ n=index\\ \sqrt{}=radicalsign\end{array}$
• The non-negative root is called the Principal Root.
• Suppose n is an integer greater than 1, and a is a real number.

a	n is even	n is odd
a >0	1 unique positive and 1 unique negative real root : $\pm \sqrt[n]{a}$ ; positive root is principal root.	1 unique positive and 0 negative real root : $\sqrt[n]{a}$
a < 0	0 real roots	0 positive and 1 negative real root : $-\sqrt[n]{a}$
a=0	1 real root : $\sqrt[n]{0}=0$	1 real root : $\sqrt[n]{0}=0$

• Pure Imaginary Numbers :
• Square roots of negative real numbers: for any positive real number b, $\sqrt{-b^2}=\sqrt{b^2}\sqrt{-1}$ , or $$ bi where

  $\begin{array}{l}\sqrt{-1}=i\\ i=iota\end{array}$

• When you find an even root of an even power and the result is an odd power , you must use the absolute value of the result to ensure that the answer is nonnegative.
• If the index is odd, there is only one real root. Therefore, there is no principal root when index is odd , and absolute value symbol are not needed.
• Product Property of Radicals: $\begin{array}{l}\text{For any real number a and b and any integer n > 1 , }\\ \sqrt[n]{ab}=\sqrt[n]{a}\cdot \sqrt[n]{b}\text{ , if n is even and a \& b are both nonnegative }\\ \text{or if n is odd}\text{.}\end{array}$
• 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 $4\sqrt{8}+3\sqrt{50}$
• $\begin{array}{l}4\sqrt{8}+3\sqrt{50}\\ =4\sqrt{2^2\cdot 2}+3\sqrt{5^2\cdot 2}\mathrm{...........}[\text{Factor using squares]}\\ \text{=8}\sqrt{2}+15\sqrt{2}\\ =(8+15)\sqrt{2}\mathrm{......................}[\text{add like radicals]}\\ \text{=23}\sqrt{2}\end{array}$
• So, $\text{23}\sqrt{2}$ is the solution.