We are tasked with finding the real solutions of the equation:

$\sqrt{4x + 1} - \sqrt{x - 2} = 3$

Step 1: Isolate one square root

To begin, isolate one of the square root terms. Let's isolate $\sqrt{4x + 1}$:

$\sqrt{4x + 1} = \sqrt{x - 2} + 3$

Step 2: Square both sides

Now, square both sides to eliminate the square root on the left-hand side:

$(\sqrt{4x + 1})^2 = (\sqrt{x - 2} + 3)^2$

This simplifies to:

$4x + 1 = (\sqrt{x - 2})^2 + 2 \cdot \sqrt{x - 2} \cdot 3 + 3^2$

Expanding the right-hand side:

$4x + 1 = (x - 2) + 6\sqrt{x - 2} + 9$

Simplify further:

$4x + 1 = x + 7 + 6\sqrt{x - 2}$

Step 3: Isolate the square root term

Next, move all terms not involving the square root to the left-hand side:

$4x + 1 - x - 7 = 6\sqrt{x - 2}$

Simplify:

$3x - 6 = 6\sqrt{x - 2}$

Step 4: Divide by 6

Divide both sides of the equation by 6:

$\frac{3x - 6}{6} = \sqrt{x - 2}$

Simplify:

$\frac{x - 2}{2} = \sqrt{x - 2}$

Step 5: Square both sides again

Square both sides to eliminate the remaining square root:

$\left(\frac{x - 2}{2}\right)^2 = (x - 2)$

Expanding the left-hand side:

$\frac{(x - 2)^2}{4} = x - 2$

Step 6: Multiply by 4

Multiply both sides by 4 to eliminate the denominator:

$(x - 2)^2 = 4(x - 2)$

Step 7: Solve the quadratic equation

Rearrange this into a quadratic equation:

$(x - 2)^2 - 4(x - 2) = 0$

Factor this expression:

$(x - 2)[(x - 2) - 4] = 0$

Simplify the factors:

$(x - 2)(x - 6) = 0$

Thus, the solutions are:

$x - 2 = 0 \quad \text{or} \quad x - 6 = 0$

$x = 2 \quad \text{or} \quad x = 6$

Step 8: Check for extraneous solutions

We need to check both solutions in the original equation because squaring both sides can introduce extraneous solutions.

1. For $x = 2$: $\sqrt{4(2) + 1} - \sqrt{2 - 2} = \sqrt{9} - 0 = 3$ This satisfies the equation.

2. For $x = 6$: $\sqrt{4(6) + 1} - \sqrt{6 - 2} = \sqrt{25} - \sqrt{4} = 5 - 2 = 3$ This also satisfies the equation.

Final Answer:

The real solutions are $x = 2$ and $x = 6$.

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Would you like more details or further clarification?

Here are 5 related questions:

1. How do we know when squaring both sides introduces extraneous solutions?
2. What is the geometric interpretation of this problem?
3. Can the same process be used to solve equations with higher powers?
4. How do we solve square root equations with more than two terms?
5. What happens if no real solutions exist in such square root equations?

Tip: Always check your solutions when squaring both sides in an equation, as extraneous solutions may arise.