Chapter 7, Problem 33PT

To determine

To Solve:

  $-2+\sqrt{3m-1}<4$

Answer to Problem 33PT

Solution region is

  $\frac{1}{3}\le m<\frac{37}{3}$

Explanation of Solution

Given Information:

Given inequality:

  $-2+\sqrt{3m-1}<4$

Concept Used :

Solve Radical inequalities:

• If the index of the root is even , identify the values of the variable for which the radicand is nonnegative.
• Solve the inequality algebraically.
• Test values to check.

Calculation:

• Given $-2+\sqrt{3m-1}<4$
• $\begin{array}{l}\text{Since the radicand of a square root must be greater than or equal to zero, }\\ \text{first solve 3m-1}\ge 0\text{ to identify the values of m for which the left hand side}\\ \text{of the inequality is defined}\text{.}\\ 3m-1\ge \mathrm{0....................}[\text{set the radicand }\ge \text{ 0]}\\ 3m\ge 0+\mathrm{1....................}[add\text{ 1 each side]}\\ 3m\ge 1\\ \frac{3m}{3}\ge \frac{1}{3}\mathrm{.....................}[\text{Divide each side by 3]}\\ m\ge \frac{1}{3}\end{array}$
• $\begin{array}{l}Now,solve\\ -2+\sqrt{3m-1}<4\\ \sqrt{3m-1}<4+\mathrm{2..................}[\text{isolate the radical]}\\ \sqrt{3m-1}<6\\ {\left(\sqrt{3m-1}\right)}^2<{\left(6\right)}^2\mathrm{................}[\text{eliminate the radical]}\\ 3m-1<36\\ 3m<36+\mathrm{1.........................}[add\text{ 1 each side]}\\ 3m<37\\ \frac{3m}{3}<\frac{37}{3}\mathrm{............................}[\text{divide 3 each side]}\\ m<\frac{37}{3}\end{array}$

  $\begin{array}{l}Combining\\ m\ge \frac{1}{3}\\ and\\ m<\frac{37}{3}\\ \text{The solution region is }\frac{1}{3}\le m<\frac{37}{3},\text{ the intersection of two regions}\text{.}\end{array}$