Chapter 7.7, Problem 56PPS

To determine

To Solve:

  $\sqrt{3x+9}-2<7$

Answer to Problem 56PPS

Solution region is

  $-3\le x<24$

Explanation of Solution

Given Information:

Given inequality:

  $\sqrt{3x+9}-2<7$

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 $\sqrt{3x+9}-2<7$
• $\begin{array}{l}\text{Since the radicand of a square root must be greater than or equal to zero, }\\ \text{first solve 3x+9}\ge 0\text{ to identify the values of }x\text{ for which the left hand side}\\ \text{of the inequality is defined}\text{.}\\ 3x+9\ge \mathrm{0....................}[\text{set the radicand }\ge \text{ 0]}\\ 3x\ge 0-\mathrm{9....................}[subtract\text{ 9 from each side]}\\ 3x\ge -9\\ \frac{3x}{3}\ge \frac{-9}{3}\mathrm{.....................}[\text{Divide each side by 3]}\\ x\ge -3\end{array}$
• $\begin{array}{l}Now,solve\\ \sqrt{3x+9}-2<7\\ \sqrt{3x+9}<7+\mathrm{2.................................}[\text{add 2 each side]}\\ \sqrt{3x+9}<9\\ {\left(\sqrt{3x+9}\right)}^2<{\left(9\right)}^2\mathrm{................}[\text{eliminate the radical]}\\ 3x+9<81\\ 3x<81-\mathrm{9.........................}[subtract\text{ 9 from each side]}\\ 3x<72\\ \frac{3x}{3}<\frac{72}{3}\mathrm{........................................}[\text{divide 3 each side]}\\ x<24\end{array}$

$\begin{array}{l}Combining\\ x\ge -3\\ and\\ x<24\\ \text{The solution region is }-3\le x<24,\text{ the intersection of two regions}\text{.}\end{array}$