Chapter 7.7, Problem 54PPS

To determine

To Solve:

  $\sqrt{2x+5}-\sqrt{9+x}>0$

Answer to Problem 54PPS

Solution region is

  $x>4$

Explanation of Solution

Given Information:

Given inequality:

  $\sqrt{2x+5}-\sqrt{9+x}>0$

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

$\begin{array}{l}Combining\\ x\ge \frac{-5}{2}\text{ , }x\ge -9\\ and\\ x>4\\ \text{The solution region is }x>4,\text{ the intersection of all the regions}\text{.}\end{array}$