Chapter 7.7, Problem 60PPS

To determine

To Solve:

  $\sqrt{2}-\sqrt{b+6}\le -\sqrt{b}$

Answer to Problem 60PPS

Solution region is

  $0\le b\le 2$

Explanation of Solution

Given Information:

Given inequality:

  $\sqrt{2}-\sqrt{b+6}\le -\sqrt{b}$

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 $\sqrt{2}-\sqrt{b+6}\le -\sqrt{b}$
• $\begin{array}{l}\text{Since the radicand of a square root must be greater than or equal to zero, }\\ \text{first solve b+6}\ge 0\text{ and b}\ge \text{0 to identify the values of }x\text{ for which the left hand side}\\ \text{of the inequality is defined}\text{.}\\ b+6\ge \mathrm{0....................}[\text{set the radicand }\ge \text{ 0]}\\ b\ge -\mathrm{6....................}[subtract\text{ 6 from each side]}\\ and\\ b\ge 0\end{array}$
• $\begin{array}{l}Now,solve\\ \sqrt{2}-\sqrt{b+6}\le -\sqrt{b}\\ \sqrt{b}-\sqrt{b+6}\le -\sqrt{2}\mathrm{.................................}[add\sqrt{b}\text{ and subtract }\sqrt{2}\text{ each side] }\\ {\left(\sqrt{b}-\sqrt{b+6}\right)}^2\le {\left(-\sqrt{2}\right)}^2\mathrm{................}[\text{eliminate the radical]}\\ b+b+6-2\sqrt{b^2+6b}\le 2\\ 2b-2\sqrt{b^2+6b}\le -\mathrm{4.........................}[subtract\text{ 6 from each side]}\\ b-\sqrt{b^2+6b}\le -\mathrm{2.............................}[\text{divide each side by 2]}\\ -\sqrt{b^2+6b}\le -2-b\mathrm{...........................}[\text{subtract 'b' from each side]}\\ {\left(-\sqrt{b^2+6b}\right)}^2\le {\left(-2-b\right)}^2\mathrm{.................}[\text{eliminate radical ]}\\ b^2+6b\le b^2+4+4b\\ 6b\le 4+4b\mathrm{.......................................}[{\text{subtract b}}^{2}from\text{ each side]}\\ \text{2b}\le \text{4}\mathrm{.............................................}\text{[subtract 4b from each side]}\\ \text{b}\le \text{2}\mathrm{...............................................}\text{[divide each side by 2]}\end{array}$

$\begin{array}{l}Combining\\ b\ge -6\&b\ge 0\\ and\\ b\le 2\\ \text{The solution region is }0\le b\le 2,\text{ the intersection of two regions}\text{.}\end{array}$