We are given the equation:

$\sqrt{2x - 1} + 22 = x$

Let's solve it step by step.

Step 1: Isolate the square root term

First, subtract 22 from both sides to isolate the square root term:

$\sqrt{2x - 1} = x - 22$

Step 2: Square both sides

To eliminate the square root, square both sides of the equation:

$(\sqrt{2x - 1})^2 = (x - 22)^2$

This simplifies to:

$2x - 1 = (x - 22)^2$

Step 3: Expand the right-hand side

Now, expand $(x - 22)^2$:

$2x - 1 = x^2 - 44x + 484$

Step 4: Rearrange the equation

Move all terms to one side to set the equation equal to 0:

$0 = x^2 - 44x + 484 - 2x + 1$

Simplify the terms:

$0 = x^2 - 46x + 485$

Step 5: Solve the quadratic equation

Now, solve the quadratic equation $x^2 - 46x + 485 = 0$. We can use the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Here, $a = 1$, $b = -46$, and $c = 485$. Plugging these into the formula:

$x = \frac{-(-46) \pm \sqrt{(-46)^2 - 4(1)(485)}}{2(1)}$

$x = \frac{46 \pm \sqrt{2116 - 1940}}{2}$

$x = \frac{46 \pm \sqrt{176}}{2}$

$x = \frac{46 \pm 13.2665}{2}$

Step 6: Calculate the two possible solutions

Now calculate the two possible values for $x$:

1. $x = \frac{46 + 13.2665}{2} = \frac{59.2665}{2} \approx 29.6333$
2. $x = \frac{46 - 13.2665}{2} = \frac{32.7335}{2} \approx 16.3668$

Step 7: Check for extraneous solutions

Since we squared both sides earlier, we must check both solutions in the original equation to ensure they are valid.

1. For $x \approx 29.6333$: $\sqrt{2(29.6333) - 1} + 22 = 29.6333$ $\sqrt{59.2666 - 1} + 22 = 29.6333$ $\sqrt{58.2666} + 22 \approx 7.6262 + 22 = 29.6262$ This is a valid solution.

2. For $x \approx 16.3668$: $\sqrt{2(16.3668) - 1} + 22 = 16.3668$ $\sqrt{32.7336 - 1} + 22 = 16.3668$ $\sqrt{31.7336} + 22 \approx 5.638 + 22 = 27.638$ This does not satisfy the original equation, so $x \approx 16.3668$ is an extraneous solution.

Final Answer:

The only valid solution is:

$x \approx 29.6333$

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Would you like more details or clarifications on any step?

Here are 5 related questions to explore:

1. How do you recognize and handle extraneous solutions in square root equations?
2. What is the quadratic formula, and how is it derived?
3. How do you check the solution to a quadratic equation in a real-world problem?
4. Can you explain the process of squaring both sides of an equation more thoroughly?
5. What are some common mistakes when solving square root equations?

Tip: Always check for extraneous solutions when squaring both sides of an equation, as it can introduce solutions that don't work in the original equation.