Chapter 7, Problem 13PT

To determine

To Solve:

  $\sqrt[3]{5m+6}+15=21$

Answer to Problem 13PT

  $m=42$

Explanation of Solution

Given Information:

The given equation is :

  $\sqrt[3]{5m+6}+15=21$

Concept Used :

Solve Radical equations:

• Isolate the radical on one side of the equation.
• Raise each side of the equation to a power equal to the index of the radical to eliminate the radical .
• Solve the resulting polynomial equation.Check your results.
• When solving radical equations, the result may be a number that does not satisfy the original equation . Such a number is called an extraneous solution.
• 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 $a=\sqrt[n]{b}$
• $\begin{array}{l}\sqrt[n]{b}\\ Here\\ b=radicand\\ n=index\\ \sqrt{}=radicalsign\end{array}$
• 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 : $\sqrt[n]{0}=0$

Calculation:

  $\begin{array}{l}Given\\ \sqrt[3]{5m+6}+15=21\\ \Rightarrow \sqrt[3]{5m+6}+15-15=21-\mathrm{15...}\text{[subtract 15 from each side]}\\ \Rightarrow \sqrt[3]{5m+6}=6\\ \Rightarrow {\left(\sqrt[3]{5m+6}\right)}^3=6^3\mathrm{.......}[\text{eliminate the radical by taking each side to the power 3= index, which is odd ]}\\ \Rightarrow 5m+6=216\\ \Rightarrow 5m+6-6=216-\mathrm{6............}[\text{subtract 6 from each side]}\\ \Rightarrow 5m=210\\ \Rightarrow \frac{5m}{5}=\frac{210}{5}\mathrm{...............................}[\text{divide each side by 5]}\\ \Rightarrow m=42\\ So,\text{ m=42 is the solution of the given equation}\text{.}\end{array}$