Absolute Value Equations #4

Consider the equation 1 = | sqrt (x + 1) - 2 | + | sqrt (x + 1) - 3 | for values of sqrt(x + 1) in different intervals using the definition of absolute value:

Case 1: (- infinity, 2) The equation becomes 5 - 2 sqrt(x + 1) =1 and sqrt(x + 1) = 2 which is not part.

Case 2: [2, 3) The equation becomes constant 1 = 1 where sqrt(x + 1) takes different values of x + 1 in interval [4, 9). Hence, x = 3, 4, 5, 6, and 7 having automatically 5 solutions.

Case 3: [3, infinity) The equation becomes 2 sqrt(x + 1) - 5 = 1 where sqrt(x + 1) = 3 and x = 8 having one solution.

Hence, there are six positive integral solutions for the said equation.

Let $f(x) = \left|\sqrt{x+1}-2\right| + \left|\sqrt{x+1}-3\right|$ .

For $0 \le x < 3$ :

$\begin{aligned} f(x) & = 2-\sqrt{x+1} + 3-\sqrt{x+1} = 5-2\sqrt{x+1} & \implies 5 \le f(x) < 1 \color{#D61F06}\text{ (No solution)} \end{aligned}$

For $3 \le x < 8$ :

$\begin{aligned} f(x) & = \sqrt{x+1} - 2 + 3-\sqrt{x+1} = 1 & \implies \color{#3D99F6}\text{ (5 positive integer solutions: }x=3, 4, 5, 6, 7) \end{aligned}$

For $x \ge 8$ :

$\begin{aligned} f(x) & = \sqrt{x+1} - 2 + \sqrt{x+1} - 3 = 2\sqrt{x+1} - 5 & \implies f(x) \ge 1 \color{#3D99F6}\text{ (1 positive integer solution: }x=8) \end{aligned}$

Therefore, there are $\boxed 6$ positive integer solutions $3 \le x \le 8$ .

Values between 3 and 8 (both inclusive) satisfy