Chapter 7, Problem 36PT

To determine

To Solve:

  $\sqrt{b+12}-\sqrt{b}>2$

Answer to Problem 36PT

Solution region is

  $0\le b<4$

Explanation of Solution

Given Information:

Given inequality:

  $\sqrt{b+12}-\sqrt{b}>2$

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 $\sqrt{b+12}-\sqrt{b}>2$
• $\begin{array}{l}\text{Since the radicand of a square root must be greater than or equal to zero, }\\ \text{first solve b+12}\ge 0\text{ and b}\ge \text{0 to identify the values of b for which the left hand side}\\ \text{of the inequality is defined}\text{.}\\ b+12\ge \mathrm{0....................}[\text{set the radicand }\ge \text{ 0]}\\ b\ge 0-\mathrm{12....................}[subtract\text{ 12 from each side]}\\ b\ge -12\text{ and b}\ge \text{0}\end{array}$
• $\begin{array}{l}Now,solve\\ \sqrt{b+12}-\sqrt{b}>2\\ \sqrt{b+12}>2+\sqrt{b}\mathrm{.................}[\text{add }\sqrt{b}\text{ each side]}\\ {\left(\sqrt{b+12}\right)}^2>{\left(2+\sqrt{b}\right)}^2\mathrm{................}[\text{eliminate the radical]}\\ b+12>4+b+4\sqrt{b}\\ b+12-4-b>4+b+4\sqrt{b}-4-b\mathrm{.........................}[subtract\text{ (4+b) from each side]}\\ 8>4\sqrt{b}\\ \frac{8}{4}>\frac{4\sqrt{b}}{4}\mathrm{.............................}[\text{divide each side by 4]}\\ 2>\sqrt{b}\\ {\left(\sqrt{b}\right)}^2<2^2\mathrm{..........................}[\text{eliminate the radical]}\\ b<4\end{array}$

  $\begin{array}{l}Combining\\ b\ge 0\text{ and b}\ge \text{ -12}\\ and\\ b<4\\ \text{The solution region is }0\le b<4,\text{ the intersection of all the regions}\text{.}\\ \end{array}$