Chapter 6.7, Problem 50PPS

To determine

To Solve:

  $1+\sqrt{5x-2}>4$

Answer to Problem 50PPS

Solution region is $x>\frac{11}{5}$

Explanation of Solution

Given Information:

Given inequality:

  $1+\sqrt{5x-2}>4$

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 $1+\sqrt{5x-2}>4$

$\begin{array}{l}\text{Since the radicand of a square root must be greater than or equal to zero, }\\ \text{first solve }5x-2\ge 0\text{ to identify the values of }x\text{ for which the left hand side}\\ \text{of the inequality is defined}\text{.}\\ 5x-2\ge \mathrm{0....................}[\text{set the radicand }\ge \text{ 0]}\\ \text{5}x\ge 0+\mathrm{2....................}[\text{add 2 each side]}\\ \frac{5x}{5}\ge \frac{2}{5}\mathrm{........................}[\text{divide each side by 5]}\\ x\ge \frac{2}{5}\\ \end{array}$

$\begin{array}{l}Now,solve\\ 1+\sqrt{5x-2}>4\\ \sqrt{5x-2}>4-\mathrm{1.................}[\text{isolate the radical]}\\ \sqrt{5x-2}>3\\ {\left(\sqrt{5x-2}\right)}^2>3^2\mathrm{................}[\text{eliminate the radical]}\\ \text{5}x-2>9\\ 5x>9+\mathrm{2.........................}[\text{add 2 each side]}\\ 5x>11\\ x>\frac{11}{5}\mathrm{.............................}[\text{divide each side by 5]}\end{array}$

$\begin{array}{l}Combining\\ x\ge \frac{2}{5}\\ and\\ x>\frac{11}{5}\\ \text{The solution region is }x>\frac{11}{5},\text{ the intersection of the two regions}\text{.}\\ \text{We can check by taking a value from the region :}\\ \text{x=}\frac{18}{5}\\ \text{1+}\sqrt{5x-2}=1+\sqrt{5\cdot \frac{18}{5}-2}=1+4=5>4\\ So,\text{ the value from the region satisfies the inequality}\text{.}\\ \text{Hence, the solution region is }x>\frac{11}{5}\end{array}$