Chapter 7, Problem 35PT

To determine

To Solve:

  $\sqrt{2x+3}-4\le 5$

Answer to Problem 35PT

Solution region is

  $-\frac{3}{2}\le x\le 39$

Explanation of Solution

Given Information:

Given inequality:

  $\sqrt{2x+3}-4\le 5$

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 $\sqrt{2x+3}-4\le 5$
• $\begin{array}{l}\text{Since the radicand of a square root must be greater than or equal to zero, }\\ \text{first solve 2x+3}\ge 0\text{ to identify the values of x for which the left hand side}\\ \text{of the inequality is defined}\text{.}\\ 2x+3\ge \mathrm{0....................}[\text{set the radicand }\ge \text{ 0]}\\ 2x\ge 0-\mathrm{3....................}[subtract\text{ 3 from each side]}\\ 2x\ge -3\\ \frac{2x}{2}\ge \frac{-3}{2}\mathrm{.....................}[\text{Divide each side by 2]}\\ x\ge -\frac{3}{2}\end{array}$
• $\begin{array}{l}Now,solve\\ \sqrt{2x+3}-4\le 5\\ \sqrt{2x+3}\le 5+\mathrm{4..................}[\text{isolate the radical]}\\ \sqrt{2x+3}\le 9\\ {\left(\sqrt{2x+3}\right)}^2\le {\left(9\right)}^2\mathrm{................}[\text{eliminate the radical]}\\ 2x+3\le 81\\ 2x\le 81-\mathrm{3.........................}[subtract\text{ 3 from each side]}\\ 2x\le 78\\ \frac{2x}{2}\le \frac{78}{2}\mathrm{............................}[\text{divide 2 each side]}\\ x\le 39\end{array}$

  $\begin{array}{l}Combining\\ x\ge -\frac{3}{2}\\ and\\ x\le 39\\ \text{The solution region is }-\frac{3}{2}\le x\le 39,\text{ the intersection of two regions}\text{.}\end{array}$